Identities of Halved Angles - Trigonometry

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Question

If , then calculate .

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Answer

Because , we can use the half-angle formula for cosines to determine .

In general,

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

for .

For this problem,

$= $\sqrt{\frac{1}{2}$(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,

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Question

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

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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 , we see that , 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.

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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 , then our full angle is . 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 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 ?

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Answer

Write the half angle identity for cosine.

Replace theta with two theta.

$\theta=2\theta$

Therefore:

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Question

What is the amplitude of ?

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Answer

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

In this case, , so therefore...

Consequently, has an amplitude of .

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Question

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

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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?

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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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Question

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

Tap to reveal answer

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 , we see that , 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.

Tap to reveal answer

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 , then our full angle is . 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 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 ?

Tap to reveal answer

Answer

Write the half angle identity for cosine.

Replace theta with two theta.

$\theta=2\theta$

Therefore:

← Didn't Know|Knew It →

Question

What is the amplitude of ?

Tap to reveal answer

Answer

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

In this case, , so therefore...

Consequently, has an amplitude of .

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Question

If , then calculate .

Tap to reveal answer

Answer

Because , we can use the half-angle formula for cosines to determine .

In general,

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

for .

For this problem,

$= $\sqrt{\frac{1}{2}$(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,

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Question

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

Tap to reveal answer

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?

Tap to reveal answer

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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Question

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

Tap to reveal answer

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 , we see that , 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.

Tap to reveal answer

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 , then our full angle is . 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 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 ?

Tap to reveal answer

Answer

Write the half angle identity for cosine.

Replace theta with two theta.

$\theta=2\theta$

Therefore:

← Didn't Know|Knew It →

Question

What is the amplitude of ?

Tap to reveal answer

Answer

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

In this case, , so therefore...

Consequently, has an amplitude of .

← Didn't Know|Knew It →

Question

If , then calculate .

Tap to reveal answer

Answer

Because , we can use the half-angle formula for cosines to determine .

In general,

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

for .

For this problem,

$= $\sqrt{\frac{1}{2}$(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,

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Question

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

Tap to reveal answer

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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