Complete Basic Trigonometry Proofs - Trigonometry

Card 0 of 36

Question

Using trigonometric identities determine whether the following is valid:

$\tan2A = \frac{2\sin \left(-A\right ) \cos \left(-A\right )}{\cos^{2}\left(-A\right ) + \sin^{2}\left(-A\right)}$

Answer

We can choose either side to work with to attempt to obtain the equivalency. Here we will work with the right side as it is the more complex. First, we want to eliminate the negative angles using the appropriate relations. Sine is odd and therefore, the negative sign comes out front. Cosine is even which is interpreted by dropping the negative out of the equation:

$\tan 2A = \frac{-2\sin A\cos A}{\cos^{2}A + \sin^{2}A}$

The squaring of the sine in the denominator makes the sine term positive, i.e.

$\sin^{2}\left(-A \right ) = \left(\sin-A \right )\times\left(\sin -A \right ) = -\sin A \times -\sin A = \sin^{2}A$

The numerator is the double angle formula for sine:

$\tan 2A = \frac{-\sin 2A}{\cos^{2}A + \sin^{2}A}$

The denominator is recognized to be the pythagorean theorem as it applies to trigonometry:

$\tan 2A = \frac{-2\sin 2A}{1}$

The final reduced equation is:

$\tan 2A = -2\sin 2A$

Thus proving that the equivalence is false.

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Question

You can derive the formula $1+\tan^2\theta=\sec^2\theta$ by dividing the formula $\sin^2\theta+\cos^2\theta=1$ by which of the following functions?

Answer

The correct answer is $\cos^2\theta$ . Rather than memorizing all three Pythagorean Relationships, you can memorize only $\sin^2\theta+\cos^2\theta=1$ , then simply divide all terms by $\cos^2\theta$ to get the formula that relates $\tan^2\theta$ and $\sec^2\theta$ . Alternatively, you can divide all terms of $\sin^2\theta+\cos^2\theta=1$ by $\sin^2\theta$ to get the formula that relates $\cot^2\theta$ and $\csc^2\theta$ . The former is demonstrated below.

$\sin^2\theta+\cos^2\theta=1$

$\frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}$

$\tan^2\theta+1=\sec^2\theta$

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Question

You can derive the formula $1+\cot^2\theta=\csc^2\theta$ by dividing the formula $\sin^2\theta+\cos^2\theta=1$ by which of the following functions?

Answer

The correct answer is $\sin^2\theta$ . Rather than memorizing all three Pythagorean Relationships, you can memorize only $\sin^2\theta+\cos^2\theta=1$ , then simply divide all terms by $\sin^2\theta$ to get the formula that relates $\cot^2\theta$ and $\csc^2\theta$ . Alternatively, you can divide all terms of $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$ to get the formula that relates $\tan^2\theta$ and $\sec^2\theta$ . The former is demonstrated below.

$\sin^2\theta+\cos^2\theta=1$

$\frac{\sin^2\theta}{\sin^2\theta}+\frac{\cos^2\theta}{\sin^2\theta}=\frac{1}{\sin^2\theta}$

$1+\cot^2\theta=\csc^2\theta$

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Question

One popular way to simplify trigonometric expressions is to put the entire expression in terms of only and functions.

Answer

Getting every term in an expression in terms of sine and cosine functions is a popular way to verify trigonometric identities or complete trigonometry proofs. These two trig functions are more commonly used over their counterparts secant, cosecant, tangent, and cotangent. Moreover, getting all terms of an expression in strictly sine and cosine may help you to spot and then substitute $\sin^2\theta+\cos^2\theta=1$ , or it may help you spot other functions that can be reduced or simplified. Other general techniques to aid in verifying trigonometric identities are:

Know the eight basic relationship and recognize alternative forms of each. These are: $\csc\theta=\frac{1}{\sin\theta}$ , $\sec\theta=\frac{1}{\cos\theta}$ , $\cot\theta=\frac{1}{\tan\theta}$ , $\tan\theta=\frac{\sin\theta}{\cos\theta}$ , $\cot\theta=\frac{\cos\theta}{\sin\theta}$ , $\sin^2\theta+\cos^2\theta=1$ , $1+\tan^2\theta=\sec^2\theta$ , and $1+\cot^2\theta=\csc^2\theta$ .

Understand how to add and subtract fractions, reduce fractions, and transform fractions into equivalent fractions

Know how to factor, and know special product techniques (i.e. difference of squares)

Only work on one side of the equation at a time

Choose the side of the equation that looks more complicated and attempt to transform it into the other side of the equation

Alternatively, you may transform each side of the equation into the same form. If doing this, you may need to do scratch work, then go back and nicely organize the two transformed sides into one clean looking verification.

Avoid substitutions that introduce radicals.

Multiply the numerator and denominator of a fraction by the conjugate of either.

Simplify a square root of a fraction by using conjugates to transform it into the quotient of a perfect squares.

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Question

Fill in the four blanks, naming the appropriate trigonometric functions, and fill in the two question marks to complete the proofs showing that $\tan\theta=\frac{\sin\theta}{\cos\theta}$ and $\cot\theta=\frac{\cos\theta}{\sin\theta}$ .

For any angle $\theta$ :

$(\theta)=\frac{y}{r}$ , $\theta=\frac{x}{r}$ , $\theta=\frac{y}{x}$ , and $\theta=\frac{x}{y}$ where is any point on the terminal side of $\theta$ and is units from the origin.

Then, $\tan\theta=?=\frac{\frac{y}{r}}{\frac{x}{r}}=\frac{\sin\theta}{\cos\theta}$ and $\cot\theta=?=\frac{\frac{x}{r}}{\frac{y}{r}}=\frac{\cos\theta}{\sin\theta}$

Answer

This question is asking us to fill in the first four blanks by adding the appropriate trigonometric function that satisfies each. By defining an angle $\theta$ on the coordinate plane, and labelling the distances x, y, and r, we can ask ourselves which trig functions relate each of the following sides together. Many students use the acronym SOHCAHTOA to remember these. Using the diagram below, we can see that $\sin\theta=\frac{opp}{hyp}=\frac{y}{r}$ , $\cos\theta=\frac{adj}{hyp}=\frac{x}{r}$ , and $\tan\theta=\frac{opp}{adj}=\frac{y}{x}$ . Knowing each of these three answers, the fourth answer (based on the available answer choices) must be $\cot\theta=\frac{x}{y}$ .

Next, we need to fill in the two question marks. To do this, simply substitute the info from the first part of the proof to get that $\tan\theta=\frac{y}{x}=\frac{\frac{y}{r}}{\frac{x}{r}}$ and $\cot\theta=\frac{x}{y}=\frac{\frac{x}{r}}{\frac{y}{r}}$ .

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Question

Which of the following might be an effective first step in verifying the following identity? Assume that you are going to attempt to convert the left hand side into the right hand side.

$\sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}=\frac{1}{\sec x+\tan x}$

Answer

While it is important to understand that there is more than one correct way to verify an identity, certain procedures best apply to certain expressions and will allow you to verify an identity more simply than other methods.

The correct answer to how to best begin this verification is "Simplify a square root of a fraction by using conjugates to transform it into the quotient of perfect squares." This technique will be effective because while the left side of the equation has a square root and the right hand side does not, eliminating that square root will be crucial to verifying the identity. The verification of this identity is demonstrated below:

$\sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}=\sqrt{\frac{(\sec x-\tan x)\cdot (\sec x+\tan x)}{(\sec x+\tan x)\cdot (\sec x+\tan x)}}=\sqrt{\frac{\sec^2 x-\tan^2 x}{(\sec x+\tan x)^2}}$ $=\sqrt{\frac{1}{(\sec x+\tan x)^2}}=\frac{1}{\sec x+\tan x}$

Let's also explore why the incorrect answers are incorrect.

Firstly, while getting everything in terms of sines and cosines is a popular strategy in verifying trig identities, it will actually introduce more fractions, while failing to get rid of the square root function. Next, this problem does not lend itself to substituting a Pythagorean Identity. Pythagorean Identities are most helpful when you can use them to replace something like $\sin^2 x+\cos^2 x$ with . They can also be helpful if you want to replace something like $\sin^2 x$ with $1-\cos^2 x$ (while this seems more complicated to replace one term with two terms, if all of the rest of the terms in the expression were all cosines, this could be helpful.) Next, consider factoring. While often a helpful strategy, this method lends itself best to when you have polynomial terms such as $(\cos^2x-4)$ , which could be factored into $(\cos x+2)(\cos x-2)$ . This problem does not have this type of expression, so this isn't helpful to us. Finally, reducing fractions can be a great strategy when there are fractions to reduce. Do not fall into the trap of thinking that you are able to cancel the $\sec x$ or $\tan x$ expressions on the left hand side of the original equations. If you wanted to split this up, the only way to properly do it would be to get $\sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}=\sqrt{\frac{\sec x}{\sec x+\tan x}-\frac{\tan x}{\sec x+\tan x}}$ . Notice how large this expression is, and notice that it is getting more complicated rather than simpler and closer to $\frac{1}{\sec x+\tan x}$ . Often times, when trying to verify trig identities, you may start with one approach and find it not fruitful, and need to go back to the start and try again. Always remember that there may be another way to approach the problem; be flexible and be willing to experiment and try more than one approach!

Compare your answer with the correct one above

Question

Which of the following might be an effective first step in verifying the following identity? Assume that you are going to attempt to convert the left hand side into the right hand side.

$\frac{\sin\theta}{\csc \theta}+\frac{\cos \theta}{\sec\theta}=1$

Answer

While it is important to understand that there is more than one correct way to verify an identity, certain procedures best apply to certain expressions and will allow you to verify an identity more simply than other methods.

The correct answer to how to best begin this verification is "Get everything in terms of sines and cosines." This technique will be effective because left side presently does not have any like terms and it is an easy way to create like terms, which may then combine, reduce, cancel each other out, or simplify. The verification of this identity is demonstrated below:

$\frac{\sin\theta}{\csc\theta}+\frac{\cos\theta}{\sec\theta}=\sin\theta\cdot \frac{1}{\csc\theta}+\cos\theta\cdot \frac{1}{\sec\theta}=\sin\theta\cdot \sin\theta+\cos\theta\cdot \cos\theta$ $=\sin^2\theta+\cos^2\theta=1$

Let's also explore why the incorrect answers are incorrect.

In order to substitute a Pythagorean Identity, we usually want to see either $\sin^2\theta+\cos^2\theta$ , $\sin^2\theta$ , or $\cos^2\theta$ . None of these are present in our expression, so this is a non-starter. Next, let's consider the answer choice: "Multiply the numerator and denominator of a fraction by the conjugate of each other." Conjugates are applicable when the numerator or denominator are comprised of two or more terms. In our expression, each numerator and denominator consists of one term only, so this strategy isn't pertinent to this verification. Likewise, there are not any terms present that we can factor, and there are not any terms present that we can reduce.

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Question

Which of the following might be an effective first step in verifying the following identity? Assume that you are going to attempt to convert the left hand side into the right hand side.

$\frac{\sin x}{1+\cos x}=\frac{1-\cos x}{\sin x}$

Answer

While it is important to understand that there is more than one correct way to verify an identity, certain procedures best apply to certain expressions and will allow you to verify an identity more simply than other methods.

The correct answer to how to best begin this verification is "Multiply the numerator and denominator of a fraction by the conjugate of each other." This technique will be effective because initially, the denominators are pretty different, but furthermore, it will help turn $1+\cos x$ into $1-\cos^2 x$ , which is a Pythagorean identity that can be replaced by $\sin^2 x$ . You can see this demonstrated in full in the full verification below:

$\frac{\sin x}{1+\cos x}=\frac{1-\cos x}{1-\cos x}\cdot \frac{\sin x}{1+\cos x}=\frac{(1-\cos x)\sin x}{1-\cos^2 x}=\frac{(1-\cos x)\sin x}{\sin^2x}$ $=\frac{1-\cos x}{\sin x}$

Let's also explore why the incorrect answers are incorrect.

Firstly, while getting everything in terms of sines and cosines is a popular strategy in verifying trig identities, it won't help us here because everything is already in terms of sines and cosines. Secondly, in order to substitute a Pythagorean Identity, we usually want to see either $\sin^2\theta+\cos^2\theta$ , $\sin^2\theta$ , or $\cos^2\theta$ . None of these are present in our expression from the get-go, so this is a non-starter (though you may note that we used one of these in the middle of the proof. Finally, there are not any terms present that we can factor, and there are not any terms present that we can reduce.

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Question

Which of the following might be an effective first step in verifying the following identity? Assume that you are going to attempt to convert the left hand side into the right hand side.

$\tan^4\theta+\tan^2\theta=\sec^4\theta-\sec^2\theta$

Answer

While it is important to understand that there is more than one correct way to verify an identity, certain procedures best apply to certain expressions and will allow you to verify an identity more simply than other methods.

The correct answer to how to best begin this verification is factor the expression. No matter which side you choose to begin with, you have the ability to factor since $\tan^2\theta$ goes into both $\tan^4\theta$ and $\tan^2\theta$ and since $\sec^2\theta$ goes into both $\sec^4\theta$ and $-\sec^2\theta$ . That said, be sure to perform work on only one side of the equation; once you've selected one, do not be tempted to modify the other! The verification of this identity is demonstrated below:

$\tan^4\theta+\tan^2\theta=\tan^2\theta(\tan^2\theta+1)=\tan^2\theta\sec^2\theta=(\sec^2\theta-1)\sec^2\theta$ $=\sec^4\theta-\sec^2\theta$

Let's also explore why the incorrect answers are incorrect.

Firstly, while getting everything in terms of sines and cosines is a popular strategy in verifying trig identities, it will actually introduce more fractions. We can also ask ourselves "Why is getting everything in terms of sines and cosines so helpful?" It's helpful because it consolidates expressions that may have 3 to 6 different trig functions down to having only 1 or 2 trig functions (sine, cosine, both). However, in our problem, to begin, we only have two different trig functions (tangent and secant). There are ways to solve this problem such that you could begin with converting to sine and cosine as your first step, but, it would be a messier and less concise way to verify this trig identity.

The next incorrect answer offers to "combine like terms." However, there are no like terms present to combine.

The next option offers to "Multiply the numerator and denominator of a fraction by the conjugate of each other." This is not relevant here, because we do not have any fractions present. Likewise, the option to "reduce the fraction" is not helpful here.

Notice that there is no answer choice offering to "Substitute a Pythagorean Identity," and this is intentional, because you can absolutely verify this identity by substituting the identity $\tan^2\theta=\sec^2\theta-1$ . A brief proof is below, and is a good example of how there can be more than one correct way to verify a trig identity!

$\tan^4\theta+\tan^2\theta=\tan^2\theta\cdot \tan^2\theta+\tan^2\theta=(\sec^2\theta-1)\cdot (\sec^2\theta-1)+(\sec^2\theta-1)$ $=\sec^4\theta-2\sec^2\theta+1+\sec^2\theta-1=\sec^4\theta-\sec^2\theta$

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Question

Using trigonometric identities determine whether the following is valid:

$\tan2A = \frac{2\sin \left(-A\right ) \cos \left(-A\right )}{\cos^{2}\left(-A\right ) + \sin^{2}\left(-A\right)}$

Answer

We can choose either side to work with to attempt to obtain the equivalency. Here we will work with the right side as it is the more complex. First, we want to eliminate the negative angles using the appropriate relations. Sine is odd and therefore, the negative sign comes out front. Cosine is even which is interpreted by dropping the negative out of the equation:

$\tan 2A = \frac{-2\sin A\cos A}{\cos^{2}A + \sin^{2}A}$

The squaring of the sine in the denominator makes the sine term positive, i.e.

$\sin^{2}\left(-A \right ) = \left(\sin-A \right )\times\left(\sin -A \right ) = -\sin A \times -\sin A = \sin^{2}A$

The numerator is the double angle formula for sine:

$\tan 2A = \frac{-\sin 2A}{\cos^{2}A + \sin^{2}A}$

The denominator is recognized to be the pythagorean theorem as it applies to trigonometry:

$\tan 2A = \frac{-2\sin 2A}{1}$

The final reduced equation is:

$\tan 2A = -2\sin 2A$

Thus proving that the equivalence is false.

Compare your answer with the correct one above

Question

You can derive the formula $1+\tan^2\theta=\sec^2\theta$ by dividing the formula $\sin^2\theta+\cos^2\theta=1$ by which of the following functions?

Answer

The correct answer is $\cos^2\theta$ . Rather than memorizing all three Pythagorean Relationships, you can memorize only $\sin^2\theta+\cos^2\theta=1$ , then simply divide all terms by $\cos^2\theta$ to get the formula that relates $\tan^2\theta$ and $\sec^2\theta$ . Alternatively, you can divide all terms of $\sin^2\theta+\cos^2\theta=1$ by $\sin^2\theta$ to get the formula that relates $\cot^2\theta$ and $\csc^2\theta$ . The former is demonstrated below.

$\sin^2\theta+\cos^2\theta=1$

$\frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}$

$\tan^2\theta+1=\sec^2\theta$

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Question

You can derive the formula $1+\cot^2\theta=\csc^2\theta$ by dividing the formula $\sin^2\theta+\cos^2\theta=1$ by which of the following functions?

Answer

The correct answer is $\sin^2\theta$ . Rather than memorizing all three Pythagorean Relationships, you can memorize only $\sin^2\theta+\cos^2\theta=1$ , then simply divide all terms by $\sin^2\theta$ to get the formula that relates $\cot^2\theta$ and $\csc^2\theta$ . Alternatively, you can divide all terms of $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$ to get the formula that relates $\tan^2\theta$ and $\sec^2\theta$ . The former is demonstrated below.

$\sin^2\theta+\cos^2\theta=1$

$\frac{\sin^2\theta}{\sin^2\theta}+\frac{\cos^2\theta}{\sin^2\theta}=\frac{1}{\sin^2\theta}$

$1+\cot^2\theta=\csc^2\theta$

Compare your answer with the correct one above

Question

One popular way to simplify trigonometric expressions is to put the entire expression in terms of only and functions.

Answer

Getting every term in an expression in terms of sine and cosine functions is a popular way to verify trigonometric identities or complete trigonometry proofs. These two trig functions are more commonly used over their counterparts secant, cosecant, tangent, and cotangent. Moreover, getting all terms of an expression in strictly sine and cosine may help you to spot and then substitute $\sin^2\theta+\cos^2\theta=1$ , or it may help you spot other functions that can be reduced or simplified. Other general techniques to aid in verifying trigonometric identities are:

Know the eight basic relationship and recognize alternative forms of each. These are: $\csc\theta=\frac{1}{\sin\theta}$ , $\sec\theta=\frac{1}{\cos\theta}$ , $\cot\theta=\frac{1}{\tan\theta}$ , $\tan\theta=\frac{\sin\theta}{\cos\theta}$ , $\cot\theta=\frac{\cos\theta}{\sin\theta}$ , $\sin^2\theta+\cos^2\theta=1$ , $1+\tan^2\theta=\sec^2\theta$ , and $1+\cot^2\theta=\csc^2\theta$ .

Understand how to add and subtract fractions, reduce fractions, and transform fractions into equivalent fractions

Know how to factor, and know special product techniques (i.e. difference of squares)

Only work on one side of the equation at a time

Choose the side of the equation that looks more complicated and attempt to transform it into the other side of the equation

Alternatively, you may transform each side of the equation into the same form. If doing this, you may need to do scratch work, then go back and nicely organize the two transformed sides into one clean looking verification.

Avoid substitutions that introduce radicals.

Multiply the numerator and denominator of a fraction by the conjugate of either.

Simplify a square root of a fraction by using conjugates to transform it into the quotient of a perfect squares.

Compare your answer with the correct one above

Question

Fill in the four blanks, naming the appropriate trigonometric functions, and fill in the two question marks to complete the proofs showing that $\tan\theta=\frac{\sin\theta}{\cos\theta}$ and $\cot\theta=\frac{\cos\theta}{\sin\theta}$ .

For any angle $\theta$ :

$(\theta)=\frac{y}{r}$ , $\theta=\frac{x}{r}$ , $\theta=\frac{y}{x}$ , and $\theta=\frac{x}{y}$ where is any point on the terminal side of $\theta$ and is units from the origin.

Then, $\tan\theta=?=\frac{\frac{y}{r}}{\frac{x}{r}}=\frac{\sin\theta}{\cos\theta}$ and $\cot\theta=?=\frac{\frac{x}{r}}{\frac{y}{r}}=\frac{\cos\theta}{\sin\theta}$

Answer

This question is asking us to fill in the first four blanks by adding the appropriate trigonometric function that satisfies each. By defining an angle $\theta$ on the coordinate plane, and labelling the distances x, y, and r, we can ask ourselves which trig functions relate each of the following sides together. Many students use the acronym SOHCAHTOA to remember these. Using the diagram below, we can see that $\sin\theta=\frac{opp}{hyp}=\frac{y}{r}$ , $\cos\theta=\frac{adj}{hyp}=\frac{x}{r}$ , and $\tan\theta=\frac{opp}{adj}=\frac{y}{x}$ . Knowing each of these three answers, the fourth answer (based on the available answer choices) must be $\cot\theta=\frac{x}{y}$ .

Next, we need to fill in the two question marks. To do this, simply substitute the info from the first part of the proof to get that $\tan\theta=\frac{y}{x}=\frac{\frac{y}{r}}{\frac{x}{r}}$ and $\cot\theta=\frac{x}{y}=\frac{\frac{x}{r}}{\frac{y}{r}}$ .

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Question

Which of the following might be an effective first step in verifying the following identity? Assume that you are going to attempt to convert the left hand side into the right hand side.

$\sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}=\frac{1}{\sec x+\tan x}$

Answer

While it is important to understand that there is more than one correct way to verify an identity, certain procedures best apply to certain expressions and will allow you to verify an identity more simply than other methods.

The correct answer to how to best begin this verification is "Simplify a square root of a fraction by using conjugates to transform it into the quotient of perfect squares." This technique will be effective because while the left side of the equation has a square root and the right hand side does not, eliminating that square root will be crucial to verifying the identity. The verification of this identity is demonstrated below:

$\sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}=\sqrt{\frac{(\sec x-\tan x)\cdot (\sec x+\tan x)}{(\sec x+\tan x)\cdot (\sec x+\tan x)}}=\sqrt{\frac{\sec^2 x-\tan^2 x}{(\sec x+\tan x)^2}}$ $=\sqrt{\frac{1}{(\sec x+\tan x)^2}}=\frac{1}{\sec x+\tan x}$

Let's also explore why the incorrect answers are incorrect.

Firstly, while getting everything in terms of sines and cosines is a popular strategy in verifying trig identities, it will actually introduce more fractions, while failing to get rid of the square root function. Next, this problem does not lend itself to substituting a Pythagorean Identity. Pythagorean Identities are most helpful when you can use them to replace something like $\sin^2 x+\cos^2 x$ with . They can also be helpful if you want to replace something like $\sin^2 x$ with $1-\cos^2 x$ (while this seems more complicated to replace one term with two terms, if all of the rest of the terms in the expression were all cosines, this could be helpful.) Next, consider factoring. While often a helpful strategy, this method lends itself best to when you have polynomial terms such as $(\cos^2x-4)$ , which could be factored into $(\cos x+2)(\cos x-2)$ . This problem does not have this type of expression, so this isn't helpful to us. Finally, reducing fractions can be a great strategy when there are fractions to reduce. Do not fall into the trap of thinking that you are able to cancel the $\sec x$ or $\tan x$ expressions on the left hand side of the original equations. If you wanted to split this up, the only way to properly do it would be to get $\sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}=\sqrt{\frac{\sec x}{\sec x+\tan x}-\frac{\tan x}{\sec x+\tan x}}$ . Notice how large this expression is, and notice that it is getting more complicated rather than simpler and closer to $\frac{1}{\sec x+\tan x}$ . Often times, when trying to verify trig identities, you may start with one approach and find it not fruitful, and need to go back to the start and try again. Always remember that there may be another way to approach the problem; be flexible and be willing to experiment and try more than one approach!

Compare your answer with the correct one above

Question

Which of the following might be an effective first step in verifying the following identity? Assume that you are going to attempt to convert the left hand side into the right hand side.

$\frac{\sin\theta}{\csc \theta}+\frac{\cos \theta}{\sec\theta}=1$

Answer

While it is important to understand that there is more than one correct way to verify an identity, certain procedures best apply to certain expressions and will allow you to verify an identity more simply than other methods.

The correct answer to how to best begin this verification is "Get everything in terms of sines and cosines." This technique will be effective because left side presently does not have any like terms and it is an easy way to create like terms, which may then combine, reduce, cancel each other out, or simplify. The verification of this identity is demonstrated below:

$\frac{\sin\theta}{\csc\theta}+\frac{\cos\theta}{\sec\theta}=\sin\theta\cdot \frac{1}{\csc\theta}+\cos\theta\cdot \frac{1}{\sec\theta}=\sin\theta\cdot \sin\theta+\cos\theta\cdot \cos\theta$ $=\sin^2\theta+\cos^2\theta=1$

Let's also explore why the incorrect answers are incorrect.

In order to substitute a Pythagorean Identity, we usually want to see either $\sin^2\theta+\cos^2\theta$ , $\sin^2\theta$ , or $\cos^2\theta$ . None of these are present in our expression, so this is a non-starter. Next, let's consider the answer choice: "Multiply the numerator and denominator of a fraction by the conjugate of each other." Conjugates are applicable when the numerator or denominator are comprised of two or more terms. In our expression, each numerator and denominator consists of one term only, so this strategy isn't pertinent to this verification. Likewise, there are not any terms present that we can factor, and there are not any terms present that we can reduce.

Compare your answer with the correct one above

Question

Which of the following might be an effective first step in verifying the following identity? Assume that you are going to attempt to convert the left hand side into the right hand side.

$\frac{\sin x}{1+\cos x}=\frac{1-\cos x}{\sin x}$

Answer

While it is important to understand that there is more than one correct way to verify an identity, certain procedures best apply to certain expressions and will allow you to verify an identity more simply than other methods.

The correct answer to how to best begin this verification is "Multiply the numerator and denominator of a fraction by the conjugate of each other." This technique will be effective because initially, the denominators are pretty different, but furthermore, it will help turn $1+\cos x$ into $1-\cos^2 x$ , which is a Pythagorean identity that can be replaced by $\sin^2 x$ . You can see this demonstrated in full in the full verification below:

$\frac{\sin x}{1+\cos x}=\frac{1-\cos x}{1-\cos x}\cdot \frac{\sin x}{1+\cos x}=\frac{(1-\cos x)\sin x}{1-\cos^2 x}=\frac{(1-\cos x)\sin x}{\sin^2x}$ $=\frac{1-\cos x}{\sin x}$

Let's also explore why the incorrect answers are incorrect.

Firstly, while getting everything in terms of sines and cosines is a popular strategy in verifying trig identities, it won't help us here because everything is already in terms of sines and cosines. Secondly, in order to substitute a Pythagorean Identity, we usually want to see either $\sin^2\theta+\cos^2\theta$ , $\sin^2\theta$ , or $\cos^2\theta$ . None of these are present in our expression from the get-go, so this is a non-starter (though you may note that we used one of these in the middle of the proof. Finally, there are not any terms present that we can factor, and there are not any terms present that we can reduce.

Compare your answer with the correct one above

Question

Which of the following might be an effective first step in verifying the following identity? Assume that you are going to attempt to convert the left hand side into the right hand side.

$\tan^4\theta+\tan^2\theta=\sec^4\theta-\sec^2\theta$

Answer

While it is important to understand that there is more than one correct way to verify an identity, certain procedures best apply to certain expressions and will allow you to verify an identity more simply than other methods.

The correct answer to how to best begin this verification is factor the expression. No matter which side you choose to begin with, you have the ability to factor since $\tan^2\theta$ goes into both $\tan^4\theta$ and $\tan^2\theta$ and since $\sec^2\theta$ goes into both $\sec^4\theta$ and $-\sec^2\theta$ . That said, be sure to perform work on only one side of the equation; once you've selected one, do not be tempted to modify the other! The verification of this identity is demonstrated below:

$\tan^4\theta+\tan^2\theta=\tan^2\theta(\tan^2\theta+1)=\tan^2\theta\sec^2\theta=(\sec^2\theta-1)\sec^2\theta$ $=\sec^4\theta-\sec^2\theta$

Let's also explore why the incorrect answers are incorrect.

Firstly, while getting everything in terms of sines and cosines is a popular strategy in verifying trig identities, it will actually introduce more fractions. We can also ask ourselves "Why is getting everything in terms of sines and cosines so helpful?" It's helpful because it consolidates expressions that may have 3 to 6 different trig functions down to having only 1 or 2 trig functions (sine, cosine, both). However, in our problem, to begin, we only have two different trig functions (tangent and secant). There are ways to solve this problem such that you could begin with converting to sine and cosine as your first step, but, it would be a messier and less concise way to verify this trig identity.

The next incorrect answer offers to "combine like terms." However, there are no like terms present to combine.

The next option offers to "Multiply the numerator and denominator of a fraction by the conjugate of each other." This is not relevant here, because we do not have any fractions present. Likewise, the option to "reduce the fraction" is not helpful here.

Notice that there is no answer choice offering to "Substitute a Pythagorean Identity," and this is intentional, because you can absolutely verify this identity by substituting the identity $\tan^2\theta=\sec^2\theta-1$ . A brief proof is below, and is a good example of how there can be more than one correct way to verify a trig identity!

$\tan^4\theta+\tan^2\theta=\tan^2\theta\cdot \tan^2\theta+\tan^2\theta=(\sec^2\theta-1)\cdot (\sec^2\theta-1)+(\sec^2\theta-1)$ $=\sec^4\theta-2\sec^2\theta+1+\sec^2\theta-1=\sec^4\theta-\sec^2\theta$

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Question

Using trigonometric identities determine whether the following is valid:

$\tan2A = \frac{2\sin \left(-A\right ) \cos \left(-A\right )}{\cos^{2}\left(-A\right ) + \sin^{2}\left(-A\right)}$

Answer

We can choose either side to work with to attempt to obtain the equivalency. Here we will work with the right side as it is the more complex. First, we want to eliminate the negative angles using the appropriate relations. Sine is odd and therefore, the negative sign comes out front. Cosine is even which is interpreted by dropping the negative out of the equation:

$\tan 2A = \frac{-2\sin A\cos A}{\cos^{2}A + \sin^{2}A}$

The squaring of the sine in the denominator makes the sine term positive, i.e.

$\sin^{2}\left(-A \right ) = \left(\sin-A \right )\times\left(\sin -A \right ) = -\sin A \times -\sin A = \sin^{2}A$

The numerator is the double angle formula for sine:

$\tan 2A = \frac{-\sin 2A}{\cos^{2}A + \sin^{2}A}$

The denominator is recognized to be the pythagorean theorem as it applies to trigonometry:

$\tan 2A = \frac{-2\sin 2A}{1}$

The final reduced equation is:

$\tan 2A = -2\sin 2A$

Thus proving that the equivalence is false.

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Question

You can derive the formula $1+\tan^2\theta=\sec^2\theta$ by dividing the formula $\sin^2\theta+\cos^2\theta=1$ by which of the following functions?

Answer

The correct answer is $\cos^2\theta$ . Rather than memorizing all three Pythagorean Relationships, you can memorize only $\sin^2\theta+\cos^2\theta=1$ , then simply divide all terms by $\cos^2\theta$ to get the formula that relates $\tan^2\theta$ and $\sec^2\theta$ . Alternatively, you can divide all terms of $\sin^2\theta+\cos^2\theta=1$ by $\sin^2\theta$ to get the formula that relates $\cot^2\theta$ and $\csc^2\theta$ . The former is demonstrated below.

$\sin^2\theta+\cos^2\theta=1$

$\frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}$

$\tan^2\theta+1=\sec^2\theta$

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