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Trigonometry : Pythagorean Identities

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Example Questions

Example Question #1 : Trigonometric Identities

Simplify $\frac{1}{1-\cos^2\theta}$ .

Possible Answers:

$\frac{1}{1-\cos^2\theta}$

$\sec^2\theta$

$\sin\theta\cos\theta$

$\csc^2\theta$

Correct answer:

$\csc^2\theta$

Explanation:

To simplify, recognize that $1-\cos^2\theta$ is a reworking on $\sin^2+\cos^2=1$ , meaning that $1-\cos^2\theta=\sin^2\theta$ .

Plug that into our given equation:

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

Remember that $\csc\theta=\frac{1}{\sin\theta}$ , so $\frac{1}{\sin^2\theta}=\csc^2\theta$ .

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Example Question #2 : Trigonometric Identities

Simplify $\frac{1-\cos^2\theta}{\sin\theta\cos\theta}$ .

Possible Answers:

$\cot\theta$

$\cos\theta$

$\sec^2\theta$

$\tan\theta$

$\csc\theta-\cos\theta$

Correct answer:

$\tan\theta$

Explanation:

Recognize that $1-\cos^2\theta$ is a reworking on $\sin^2+\cos^2=1$ , meaning that $1-\cos^2\theta=\sin^2\theta$ .

Plug that in to our given equation:

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

Notice that one of the $\sin\theta$ 's cancel out.

$\frac{\sin\theta}{\cos\theta}=\tan\theta$ .

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Example Question #3 : Pythagorean Identities

$\frac{1}{(cos (33^{\circ}))^{2}} - (tan (33^{\circ}))^{2} = ?$

Possible Answers:

$\frac{5\sqrt{2}}{2}$

$3\sqrt{3}$

-1

Correct answer:

Explanation:

$\frac{1}{(cos (33^{\circ}))^{2}} - (tan (33^{\circ}))^{2} = (sec (33^{\circ}))^{2}- (tan (33^{\circ}))^{2}$

Recall the Pythagorean Identity:

$1 + (tan x)^{2} = (sec x)^{2}$

We can rearrange the terms:

$1 = (sec x)^{2} - (tan x)^{2}$

This is exactly what our original equation looks like, so the answer is 1.

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Example Question #3 : Pythagorean Identities

Simplify the equation using identities:

$\cot^{2}x\cos x \sin x+ \cos x \sin x$

Possible Answers:

$1-\csc x$

$\cos^{2}x$

$\sin x$

$\cot x$

Correct answer:

$\cot x$

Explanation:

There are a couple valid strategies for solving this problem. The simplest is to first factor out $\cos x\sin x$ from both sides. This leaves us with:

$\cot^{2}x\cos x\sin x + \cos x \sin x = (\cos x \sin x)(cot^{2}x + 1)$

Next, substitute with the known identity $\cot^{2}x + 1=\csc^{2}x$ to get:

$\cos x\sin x (\csc^{2}x)$

From here, we can eliminate the quadratic by converting:

$\csc^{2}x= \frac{1}{\sin^{2}x}$

giving us

$\frac{\cos x\sin x}{\sin^{2}x}=\frac{\cos x}{\sin x}= \cot x$

Thus,

$\cot^{2}x\cos x\sin x+\cos x\sin x = \cot x$

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Example Question #3 : Trigonometric Identities

Simplify the expression: $(1+\cos a)(1-\cos a)$

Possible Answers:

$1-\cos^{2}a$

$\cos a$

The equation cannot be further simplified.

$\sin a$

$\sin^{2}a$

Correct answer:

$\sin^{2}a$

Explanation:

The expression $(1-\cos a)(1+\cos a)$ represents a difference of squares. In this case, the product is $1-\cos^{2}a$ (remember that 1 is also a perfect square).

One Pythagoran identity for trigonometric functions is:

$1-\cos^{2}a = \sin^{2}a$

Thus, we can say that the most simplified version of the expression is $\sin^{2}a$ .

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Example Question #3 : Pythagorean Identities

If theta is in the second quadrant, and $sin(\theta)= \frac{3}{5}$ , what is $cos(\theta)$ ?

Possible Answers:

$-\frac{3}{5}$

$\frac{5}{3}$

$\frac{4}{5}$

$-\frac{4}{5}$

$\frac{3}{5}$

Correct answer:

$-\frac{4}{5}$

Explanation:

Write the Pythagorean Identity.

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

Substitute the value of $sin(\theta)$ and solve for $cos(\theta)$ .

$(\frac{3}{5})^2+cos^2(\theta)=1$

$cos^2(\theta)=1-(\frac{3}{5})^2$

$cos^2(\theta)=1-(\frac{3}{5})^2= \frac{25}{25}-\frac{9}{25}=\frac{16}{25}$

$cos(\theta)=\pm \sqrt{\frac{16}{25}}=\pm \frac{4}{5}$

Since the cosine is in the second quadrant, the correct answer is:

$cos(\theta)=-\frac{4}{5}$

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Example Question #1 : Pythagorean Identities

For which values of is the following equation true?

sin(x)cos(x)tan(x) =1-cos^2(x)

Possible Answers:

$0^{\circ}\leq x \leq 180^{\circ}$

$0^{\circ}\leq x \leq 90^{\circ}$

$\mathbb{R}=(-\infty, \infty )$

$0^{\circ}\leq x <360^{\circ}$

$x = 90^{\circ}, 180^{\circ}, 270^{\circ}$

Correct answer:

$\mathbb{R}=(-\infty, \infty )$

Explanation:

According to the Pythagorean identity

sin^2x+cos^2x=1 ,

the right hand side of this equation can be rewritten as sin^2x . This yields the equation

sinxcosxtanx =sin^2x .

Dividing both sides by sinx yields:

cosxtanx=sinx .

Dividing both sides by cosx yields:

$tanx=\frac{sinx}{cosx}$ .

This is precisely the definition of the tangent function; since the domain of tanx consists of all real numbers, the values of which satisfy the original equation also consist of all real numbers. Hence, the correct answer is

$\mathbb{R}=(-\infty, \infty )$ .

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Example Question #1 : Pythagorean Identities

$\sin^2x + \cos^2x + \tan^2x =$

Possible Answers:

$\sec^2x$

$\tan^2x-\cot^2x$

$\csc^2x$

$1-\cos^2x$

Correct answer:

$\sec^2x$

Explanation:

By the Pythagorean identity, the first two terms simplify to 1:

$\sin^2x+\cos^2x+\tan^2x=1+\tan^2x$ .

Dividing the Pythagorean identity by $\cos^2x$ allows us to simplify the right-hand side.

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

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

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Example Question #1 : Trigonometry

What is sin^2x+cos^2x equal to?

Possible Answers:

tan(x)

Correct answer:

Explanation:

Step 1: Recall the trigonometric identity that has sine and cosine in it...

sin^2x+cos^2x=1

The sum is equal to 1.

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Example Question #2 : Trigonometric Identities

Given $\sin(\theta)=\frac{1}{4}$ , what is $\cos(\theta)$ ?

Possible Answers:

$\cos(\theta)=\pm15/4$

$\cos(\theta)=\pm\sqrt{\frac{15}{4}}$

$\cos(\theta)=15/4$

$\cos(\theta)=15/16$

Correct answer:

$\cos(\theta)=\pm\sqrt{\frac{15}{4}}$

Explanation:

Using the Pythagorean Identity

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

one can solve for $\cos^2(\theta)$ by plugging in (1/4)^2 for $\sin^2$ .

Solving for $\cos^2(\theta)$ , you get it equal to 15/16 .

Taking the square root of both sides will get the correct answer of

$\cos(\theta)=\pm\sqrt{\frac{15}{4}}$ .

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Trigonometry : Pythagorean Identities

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