Identities of Halved Angles - Trigonometry

Card 0 of 7

Question

Find $\sin 2\theta$ if $\cos \theta = \frac{4}{5}$ and $\frac{3\pi}{2}<\theta<2\pi$ .

Answer

The double-angle identity for sine is written as

$\sin 2\theta= 2\sin\theta \cos\theta$

and we know that

$\cos\theta = \frac{4}{5}$

Using $x^{2}+y^{2}=r^{2}$ , we see that $(4)^{2} + (y)^{2} = (5)^{2}$ , which gives us

$y=\pm3$

Since we know $\theta$ is between $\frac{3\pi}{2}$ and $2\pi$ , sin $\theta$ is negative, so . Thus,

$\sin \theta = -\frac{3}{5}$ .

Finally, substituting into our double-angle identity, we get

$\sin 2\theta = 2 \sin\theta \cos\theta = 2 \cdot\bigg(-\frac{3}{5}\bigg)\cdot\frac{4}{5} = -\frac{24}{25}$

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Question

Find the exact value of $\sin 15^{\circ}$ using an appropriate half-angle identity.

Answer

The half-angle identity for sine is:

$\sin \bigg(\frac{x}{2}\bigg) = \pm\sqrt\frac{1-\cos x}{2}$

If our half-angle is $15^{\circ}$ , then our full angle is $30^{\circ}$ . Thus,

$\sin 15^{\circ}= \pm\sqrt\frac{1-\cos30^{\circ}}{2}$

The exact value of $\cos 30^{\circ}$ is expressed as $\frac{\sqrt{3}}{2}$ , so we have

$\sin 15^{\circ} = \pm\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}$

Simplify under the outer radical and we get

$\sin 15^{\circ} = \pm\sqrt\frac{2-\sqrt{3}}{4}$

Now simplify the denominator and get

$\sin 15^{\circ} = \pm\frac{\sqrt{2-\sqrt{3}}}{2}$

Since $15^{\circ}$ is in the first quadrant, we know sin is positive. So,

$\sin 15^{\circ} = \frac{\sqrt{2-\sqrt{3}}}{2}$

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Question

Which of the following best represents $2cos^2(2\theta)$ ?

Answer

Write the half angle identity for cosine.

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

Replace theta with two theta.

$\theta=2\theta$

Therefore:

$2cos^2(2\theta)= 2\left[\frac{1+cos(2 \times 2\theta)}{2}\right] = 1+cos(4\theta)$

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Question

What is the amplitude of ?

Answer

The key here is to use the half-angle identity for to convert it and make it much easier to work with.

$acos^2(x) = \frac{a}{2}[1 + cos(x)]$

In this case, , so therefore...

$8cos^2(x) = \frac{8}{2}[1 + cos(x)] = 4 + 4cos(x)$

Consequently, has an amplitude of .

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Question

If $cos(45^{\circ})=\frac{\sqrt{2}}{2}$ , then calculate $cos(22.5^{\circ})$ .

Answer

Because $22.5^{\circ}=\frac{45^{\circ}}{2}$ , we can use the half-angle formula for cosines to determine $cos(22.5^{\circ})$ .

In general,

$cos(\frac{\Theta}{2})= \sqrt{\frac{1}{2}(1+cos\Theta)}$

for $0^{\circ}\leq\Theta<360^{\circ}$ .

For this problem,

$cos(22.5^{\circ})= \sqrt{\frac{1}{2}(1+cos(45^{\circ}))}$

$= \sqrt{\frac{1}{2}(1+\frac{\sqrt{2}}{2})}$

$=\sqrt{\frac{1}{2}(\frac{2+\sqrt{2}}{2})}$

$=\sqrt{\frac{1}{4}(2+\sqrt{2})}$

$=\frac{1}{2}\sqrt{(2+\sqrt{2})}$

Hence,

$cos(22.5^{\circ})=\frac{1}{2}\sqrt{2+\sqrt{2}}$

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Question

What is $\cos(\frac{5\pi}{12})$ ?

Answer

Let $x=\frac{5\pi}{6}$ ; then

$\cos(\frac{5\pi}{12})=\cos(\frac{x}{2})$ .

We'll use the half-angle formula to evaluate this expression.

$\cos(\frac{x}{2})=\pm \sqrt{\frac{1+\cos x}{2}}$

Now we'll substitute $\frac{5\pi}{6}$ for .

$\begin{align} \cos(\frac{5\pi}{12})=\cos(\frac{\frac{5\pi}{6}}{2})&=\pm\sqrt{\frac{1+\cos(\frac{5\pi}{6})}{2}}\ &=\pm\sqrt{\frac{1+\frac{-\sqrt{3}}{2}}{2}}\ &=\pm\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\ &=\pm\sqrt{\frac{2-\sqrt{3}}{2}}\ &=\pm\left(\frac{\sqrt{2-\sqrt{3}}}{2}\right) \end{align}$

$\frac{5\pi}{6}$ is in the first quadrant, so $\cos(\frac{5\pi}{6})$ is positive. So

$\cos(\frac{5\pi}{12})=\frac{\sqrt{2-\sqrt{3}}}{2}$ .

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Question

What is $\tan(15)$ , given that $\sin(30)$ and $\cos(30)$ are well defined values?

Answer

Using the half angle formula for tangent,

$\tan(\frac{\theta}{2})=\frac{\sin(\theta)}{(1+\cos(\theta))}$ ,

we plug in 30 for $\theta$ .

We also know from the unit circle that $\sin(30)$ is and $\cos(30)$ is $\sqrt3/2$ .

Plug all values into the equation, and you will get the correct answer.

$\ \tan(\frac{30}{2})=\frac{\sin(30)}{(1+\cos(30))} \ \ \ \tan(15)=\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}$

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