Trigonometry - Identities of Doubled Angles

Simplify the function below:

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

$\csc^2 x -2$

$1 - 2\sec^2 x$

$\tan 2x$

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

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:

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

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

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

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

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

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

Using trigonometric identities, determine whether the following is valid:

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

True

False

Uncertain

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

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

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.

Using trigonometric identities determine whether the following is true:

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

True

False

Uncertain

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

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

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.

Using a double angle formula, find the value of

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

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

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

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

$\frac{1}{2}$

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

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.

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

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

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

$\frac{1}{9}$

$-\frac{4}{9}$

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

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

$\frac{5}{9}$

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.

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

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

$tan\ X = 0.83$

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

Explanation

We can exploit the following trigonometric identity:

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:

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

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

$\pi$

$2\pi$

Explanation

The key here is to double-angle identity for to simplify the function.

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 or is $\frac{2\pi}{b}$ . Consequently,

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

Expand the following expression using double-angle identities.

$\sin 4x$

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

$\sin^4 x$

$2\sin 2x \cos x$

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

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:

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

Simplify the function below:

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

$\tan^2x$

$\tanx$ $\tan x$

$\cos x$

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:

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

Identities of Doubled Angles

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