Apply Basic and Definitional Identities

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Trigonometry › Apply Basic and Definitional Identities

Questions 1 - 8

1

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

$\cos^2 x$

$\cos x \sin^2 x$

$\sin x \cos^2 x$

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

$\sin^2x$

Explanation

The correct answer is $\cos^2 x$ . 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:

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

2

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

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

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

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

$\sin x +1$

$\cos x +1$

Explanation

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

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

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

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

3

Which of the following trigonometric identities is INCORRECT?

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

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

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

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

$cot (x) = \frac{cos (x)}{sin (x)}$

Explanation

Cosine and sine are not reciprocal functions.

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

4

Using the trigonometric identities prove whether the following is valid:

$\frac{\csc^{2}A}{\sec^{2}A} = \cot^{2}A$

True

False

Uncertain

Only in the range of: $0< A < \frac{\pi}{2}$

Only in the range of: $0 < A < 2\pi$

Explanation

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:

$\frac{\frac{1}{\sin^{2}A}}{\frac{1}{\cos^{2}A}} = \cot^{2}A$

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

$\frac{1}{\sin^{2}A}\div\frac{1}{\cos^{2}A} = \cot^{2}A$

In fractional division we multiply by the reciprocal as follows:

$\frac{1}{\sin^{2}A}\times \cos^{2}A = \cot^{2}A$

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

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

$\cot^{2}A = \cot^{2}A$

5

Which of the following is the best answer for $cos^2(\theta)$ ?

$1-sin^2(\theta)$

$-sin^2(\theta)$

$tan^2(\theta)-1$

$-sec^2(\theta)$

$-cot^2(\theta)$

Explanation

Write the Pythagorean identity.

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

Substract $sin^2(\theta)$ from both sides.

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

The other answers are incorrect.

6

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

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

$\frac{1}{\sin x \cos x}$

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

$\sin x \cos x$

$1+\sin x$

Explanation

The definition of tangent is sine divided by cosine.

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

7

Simplify.

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

$\sin^2 x$

$-\sin^2 x$

$\tan^2 x$

$-\tan^2 x$

$-\sin x\cos x$

Explanation

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

$\sin ^2 x$

8

Which of the following identities is incorrect?

$\mathrm{cos}(-x)=-\mathrm{cos}(x)$

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

$\small \sec(-x)=\sec(x)$

$\small \csc(-x)=-\csc(x)$

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

Explanation

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

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