Apply Basic and Definitional Identities - Trigonometry

Card 0 of 32

Which of the following identities is incorrect?

Tap to see back →

The true identity is $\small \cos(-x)=\cos(x)$ because cosine is an even function.

Which of the following trigonometric identities is INCORRECT?

Tap to see back →

Cosine and sine are not reciprocal functions.

$sin (x) = $\frac{1}{csc (x)}$$ and $cos (x) = $\frac{1}{sec (x)}$$

Using the trigonometric identities prove whether the following is valid:

Tap to see back →

We begin with the left hand side of the equation and utilize basic trigonometric identities, beginning with converting the inverse functions to their corresponding base functions:

Next we rewrite the fractional division in order to simplify the equation:

In fractional division we multiply by the reciprocal as follows:

If we reduce the fraction using basic identities we see that the equivalence is proven:

Which of the following is the best answer for ?

Tap to see back →

Write the Pythagorean identity.

Substract from both sides.

The other answers are incorrect.

State $\tan x$ in terms of sine and cosine.

Tap to see back →

The definition of tangent is sine divided by cosine.

$\tan x=\frac{\sin x}{\cos x}$$

Simplify.

$\sin(-x)\cos(-x)\tan(-x)$

Tap to see back →

Using these basic identities:

$\sin(-x)=-\sin x$

$\cos(-x)=\cos(x)$

$\tan(-x)=-\tan x$

we find the original expression to be

$(-\sin x)(\cos x)(-\tan x)$

which simplifies to

$\sin x\cos x\tan x$ .

Further simplifying:

$\sin x\cos x$\frac{\sin x}{\cos x}$$

The cosines cancel, giving us

$\sin ^2 x$

Express $\frac{1}{\sec x \csc x \tan x}$ in terms of only sines and cosines.

Tap to see back →

The correct answer is . Begin by substituting $$\frac{1}{\sec x}$=\cos x$ , $$\frac{1}{\csc x}$=\sin x$ , and $\frac{1}{\tan x}$=\frac{\cos x}{\sin x}$ . This gives us:

$$\frac{1}{\sec x\csc x\tan x}$=\frac{\cos x\sin x\cos x}{\sin $x}$=\cos^2$ x$ .

Express $\sec x\tan x+\frac{\cot x}{\csc x}$$ in terms of only sines and cosines.

Tap to see back →

To solve this problem, use the identities $\sec x =\frac{1}{\cos x}$$ , $\tan x=\frac{\sin x}{\cos x}$$ , $\cot x =\frac{\cos x}{\sin x}$$ , and $$\frac{1}{\csc x}$=\sin x$ . Then we get

$\sec x\tan x+\frac{\cot x}{\csc x}$$

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

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

Which of the following identities is incorrect?

Tap to see back →

The true identity is $\small \cos(-x)=\cos(x)$ because cosine is an even function.

Which of the following trigonometric identities is INCORRECT?

Tap to see back →

Cosine and sine are not reciprocal functions.

$sin (x) = $\frac{1}{csc (x)}$$ and $cos (x) = $\frac{1}{sec (x)}$$

Using the trigonometric identities prove whether the following is valid:

Tap to see back →

We begin with the left hand side of the equation and utilize basic trigonometric identities, beginning with converting the inverse functions to their corresponding base functions:

Next we rewrite the fractional division in order to simplify the equation:

In fractional division we multiply by the reciprocal as follows:

If we reduce the fraction using basic identities we see that the equivalence is proven:

Which of the following is the best answer for ?

Tap to see back →

Write the Pythagorean identity.

Substract from both sides.

The other answers are incorrect.

State $\tan x$ in terms of sine and cosine.

Tap to see back →

The definition of tangent is sine divided by cosine.

$\tan x=\frac{\sin x}{\cos x}$$

Simplify.

$\sin(-x)\cos(-x)\tan(-x)$

Tap to see back →

Using these basic identities:

$\sin(-x)=-\sin x$

$\cos(-x)=\cos(x)$

$\tan(-x)=-\tan x$

we find the original expression to be

$(-\sin x)(\cos x)(-\tan x)$

which simplifies to

$\sin x\cos x\tan x$ .

Further simplifying:

$\sin x\cos x$\frac{\sin x}{\cos x}$$

The cosines cancel, giving us

$\sin ^2 x$

Express $\frac{1}{\sec x \csc x \tan x}$ in terms of only sines and cosines.

Tap to see back →

The correct answer is . Begin by substituting $$\frac{1}{\sec x}$=\cos x$ , $$\frac{1}{\csc x}$=\sin x$ , and $\frac{1}{\tan x}$=\frac{\cos x}{\sin x}$ . This gives us:

$$\frac{1}{\sec x\csc x\tan x}$=\frac{\cos x\sin x\cos x}{\sin $x}$=\cos^2$ x$ .

Express $\sec x\tan x+\frac{\cot x}{\csc x}$$ in terms of only sines and cosines.

Tap to see back →

To solve this problem, use the identities $\sec x =\frac{1}{\cos x}$$ , $\tan x=\frac{\sin x}{\cos x}$$ , $\cot x =\frac{\cos x}{\sin x}$$ , and $$\frac{1}{\csc x}$=\sin x$ . Then we get

$\sec x\tan x+\frac{\cot x}{\csc x}$$

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

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

Which of the following identities is incorrect?

Tap to see back →

The true identity is $\small \cos(-x)=\cos(x)$ because cosine is an even function.

Which of the following trigonometric identities is INCORRECT?

Tap to see back →

Cosine and sine are not reciprocal functions.

$sin (x) = $\frac{1}{csc (x)}$$ and $cos (x) = $\frac{1}{sec (x)}$$

Using the trigonometric identities prove whether the following is valid:

Tap to see back →

We begin with the left hand side of the equation and utilize basic trigonometric identities, beginning with converting the inverse functions to their corresponding base functions:

Next we rewrite the fractional division in order to simplify the equation:

In fractional division we multiply by the reciprocal as follows:

If we reduce the fraction using basic identities we see that the equivalence is proven:

Which of the following is the best answer for ?

Tap to see back →

Write the Pythagorean identity.

Substract from both sides.

The other answers are incorrect.