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Trigonometry : Identities of Doubled Angles

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Example Questions

Example Question #1 : Identities Of Doubled Angles

Simplify the function below:

$f\left ( x\right ) = \frac{\cos^2x - \sin^2 x}{1 - \cos^2 x}$

Possible Answers:

$1 - 2\sec^2 x$

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

$\tan 2x$

$\csc^2 x -2$

Correct answer:

$\csc^2 x -2$

Explanation:

We need to use the following formulas:

a) $\cos 2x = 2\cos^2x - 1 = 1 - 2\sin^2x = \cos^2x - \sin^2x$

b) $\sin^2x + \cos^2 x = 1$

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

We can simplify $f\left ( x\right )$ as follows:

$f\left ( x\right ) = \frac{\cos^2x - \sin^2 x}{1 - \cos^2 x} = \frac{1 - 2\sin^2x}{\sin^2x} = \frac{1}{\sin^2x} - 2 = \csc^2x - 2$

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Example Question #1 : Identities Of Doubled Angles

Given $\sin \left ( x\right ) = a$ , what is $\cos \left ( 2x\right )$ in terms of ?

Possible Answers:

1 - 2a^2

$\frac{1}{2}\left ( a + 1\right )$

a^2 - 1

$\frac{a^2}{2}$

Correct answer:

1 - 2a^2

Explanation:

To solve this problem, we need to use the formula:

$\cos\left ( 2x\right ) = 1 - 2\sin^2\left (x\right )$

Substituting $\sin \left ( x\right ) = a$ , we get

$\cos\left ( 2x\right ) = 1 - 2a^2$

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Example Question #1 : Identities Of Doubled Angles

Using trigonometric identities, determine whether the following is valid:

$\frac{2\cos A}{1 - \cos 2A} = \csc A \cot A$

Possible Answers:

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

True

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

False

Uncertain

Correct answer:

True

Explanation:

In order to prove this trigonometric equation we can work with either the left or right side of the equation and attempt to make them equal. We will choose to work with the left side of the equation. First we separate the fractional term:

$\frac{2}{1-\cos 2A} \cos A = \csc A \cot A$

We separated the fractional term because we notice we have a double angle. Recalling our trigonometric identities, the fractional term is the inverse of the power reducing formula for sine.

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

Now separating out the sine terms:

$\frac{1}{\sin A}\frac{\cos A}{\sin A} = \csc A \cot A$

Now recalling the basic identities:

$\csc A \cot A = \csc A \cot A$

Using the trigonometric identities we have proven that the equation is true.

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Example Question #1 : Identities Of Doubled Angles

Using trigonometric identities determine whether the following is true:

$\csc A \sec A = \frac{2}{\sin 2A}$

Possible Answers:

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

Uncertain

False

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

True

Correct answer:

True

Explanation:

We choose which side to work with in the given equation. Selecting the right hand side since it contains a double angle we attempt to use the double angle formula to determine the equivalence:

$\csc A \sec A = \frac{2}{2\sin A \cos A}$

Next we reduce and split the fraction as follows:

$\csc A \sec A = \frac{1}{\sin A}\frac{1}{\cos A}$

Recalling the basic identities:

$\csc A \sec A = \csc A \sec A$

This proves the equivalence.

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Example Question #5 : Identities Of Doubled Angles

Using a double angle formula, find the value of

$\sin \left(2*\frac{\pi}{3}\right)$ .

Possible Answers:

$\frac{2\sqrt3}{3}$

$\frac{1}{2}$

$\frac{\sqrt{3}}{2}$

$-\frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{3}$

Correct answer:

$\frac{\sqrt{3}}{2}$

Explanation:

The formula for a doubled angle with sine is $\sin(2a)=2\sin (a)\cos(a)$

Plug in our given value and solve.

$\sin(2*\frac{\pi}{3})=2\sin (\frac{\pi}{3})\cos(\frac{\pi}{3})$

$\sin(2*\frac{\pi}{3})=2*\frac{\sqrt{3}}{2}*\frac{1}{2}$

Combine our terms.

$\sin(2*\frac{\pi}{3})=\frac{2\sqrt{3}}{4}$

$\sin(2*\frac{\pi}{3})=\frac{\sqrt{3}}{2}$

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Example Question #6 : Identities Of Doubled Angles

Find the value of $cos(2\theta)$ if $sin(\theta)= \frac{2}{3}$ , and if the value of cos(x) is less than zero.

Possible Answers:

$-\frac{4}{9}$

$\frac{1}{9}$

$-\frac{\sqrt5}{3}$

$\frac{\sqrt{5}}{9}$

$\frac{5}{9}$

Correct answer:

$\frac{1}{9}$

Explanation:

Write the Pythagorean Identity.

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

Substitute the value of $sin(\theta)= \frac{2}{3}$ and solve for $(cos\theta)$ .

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

$cos^2(\theta)=1-\frac{4}{9}= \frac{5}{9}$

$cos(\theta)=\pm \sqrt \frac{5}{9}$

Since the $(cos\theta)$ must be less than zero, choose the negative sign.

$cos(\theta)=-\frac{\sqrt{5}}{3}$

Write the double-angle identity of $cos(2\theta)$ .

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

Substitute the known values.

$(-\frac{\sqrt{5}}{3})^2-(\frac{2}{3})^2= \frac{5}{9}-\frac{4}{9}=\frac{1}{9}$

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Example Question #7 : Identities Of Doubled Angles

Suppose is an angle in the third quadrant, such that:

$tan\ X = 0.83$

What is the value of $cos\ 2X$ ?

Possible Answers:

0.32

0.14

0.18

-0.21

-0.08

Correct answer:

0.18

Explanation:

We can exploit the following trigonometric identity:

sec^2 X = tan^2X + 1 = (0.83)^2 + 1 = 1.6889

Then we can do:

$cos^2X = \frac{1}{sec^2X} = \frac{1}{1.6889} = 0.5921$

With this value we can conveniently find our solution to be:

$cos \ 2X = 2cos^2 X -1 = 2 * 0.5921 - 1 = 0.1842 \approx 0.18$

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Example Question #8 : Identities Of Doubled Angles

What is the period of $f(x) = 2sin(\pi x)cos(\pi x)$ ?

Possible Answers:

$\pi$

$2\pi$

0.5

Correct answer:

Explanation:

The key here is to double-angle identity for sin(2ax) to simplify the function.

2sin(ax)cos(ax) = sin(2ax)

In this case, $a = \pi$ , which means...

$2sin(\pi x)cos(\pi x) = sin(2\pi x)$

From there, we can use the fact that the period of sin(bx) or cos(bx) is $\frac{2\pi}{b}$ . Consequently,

$\frac{2\pi}{2\pi} = 1$

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Example Question #1 : Identities Of Doubled Angles

Expand the following expression using double-angle identities.

$\sin 4x$

Possible Answers:

$\sin^4 x$

$4\sin x \cos ^3 x-4\sin^3\cos x$

$2\sin 2x \cos x$

$2 \cos^2 x \sin x-2\sin^2 x \cos x$

Correct answer:

$4\sin x \cos ^3 x-4\sin^3\cos x$

Explanation:

Since $\sin 2x = 2\sin x\cos x$

and $\sin (4x) = \sin (2(2x))$ ,

then $\sin 4x = 2 \sin 2x \cos 2x$ .

Here we have to use the double-angle identities for both sine and cosine,

$\sin 2x = 2\sin x\cos x$ and $\cos 2x = \cos^2 x -\sin^2 x$ .

Using these identities:

$2 \sin 2x \cos 2x = 2(2\sin x \cos x)(\cos^2 x-\sin^2 x)$

Using the distributive property:

$4\sin x \cos^3 x - 4\sin^3 x \cos x$

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Example Question #1 : Identities Of Doubled Angles

Simplify the function below:

$f\left ( x\right ) = \frac{1}{\cos^2x} - \frac{1-\cos2x}{1 + \cos2x}$

Possible Answers:

$\tanx$ $\tan x$

$\cos x$

$\tan^2x$

Correct answer:

Explanation:

We need to use the following formulas:

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

b) $\sin^2x = \frac{1}{2}\left ( 1-\cos 2x\right )$

c) $\cos^2x = \frac{1}{2}\left ( 1+\cos 2x\right )$

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

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

We can simplify $f\left ( x\right )$ as follows:

$f\left ( x\right ) = \frac{1}{\cos^2x} - \frac{1-\cos2x}{1 + \cos2x} = \sec^2x - \left ( \frac{2sin^2x}{2cos^2x} \right ) = \sec^2x - \tan^2x = 1$

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Trigonometry : Identities of Doubled Angles

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #1 : Identities Of Doubled Angles

Example Question #1 : Identities Of Doubled Angles

Example Question #1 : Identities Of Doubled Angles

Example Question #1 : Identities Of Doubled Angles

Example Question #5 : Identities Of Doubled Angles

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Example Question #1 : Identities Of Doubled Angles

Example Question #1 : Identities Of Doubled Angles

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