Identities with Angle Sums - Trigonometry

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Question

Find the exact value of the expression:

$\sin $60^{\circ}$\cos $30^{\circ}$ + \cos $60^{\circ}$ \sin $30^{\circ}$$

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Answer

There are two ways to solve this problem. If one recognizes the identity

$\sin (a+b)= \sin a\cos b + \cos a\sin b$ ,

the answer is as simple as:

$\sin $60^{\circ}$\cos $30^{\circ}$ + \cos $60^{\circ}$\sin $30^{\circ}$ = \sin $(60^{\circ}$$+30^{\circ}$) = \sin $90^{\circ}$ = 1$

If one misses the identity, or wishes to be more thorough, you can simplify:

$\sin $60^{\circ}$\cos $30^{\circ}$ + \cos $60^{\circ}$\sin $30^{\circ}$$

$=\bigg(\bigg($\frac{\sqrt{3}$}{2}\bigg)\cdot\bigg($\frac{\sqrt{3}$}{2}\bigg)\bigg)+\bigg(\bigg($\frac{1}{2}$\bigg)\cdot\bigg($\frac{1}{2}$\bigg)\bigg)=\frac{3}{4}$ + $\frac{1}{4}$=1$

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Question

Given $\phi = \pi$ , what is $\tan\left ( \theta + \phi \right )$ ?

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Answer

We need to use the formula

$\tan\left ( \theta \pm \phi\right ) = $\frac{\tan(\theta) \pm \tan(\phi)}{1 \mp \tan(\theta)\tan(\phi)}$$

Substituting $\phi = \pi$ , and $\tan (\pi) = 0$ ,

$\tan\left ( \theta + \pi\right ) = $\frac{\tan(\theta) + \tan(\phi)}{1 - \tan(\theta)\tan(\phi)}$ = $\frac{\tan(\theta) + 0}{1 - \tan(\theta)\left( 0\right)}$ = \tan(\theta)$

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Question

Find the exact value of $\cos $105^{\circ}$$ using .

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Answer

Our basic sum formula for cosine is:

$\cos(a+b)=\cos a\cos b - \sin a\sin b$

Substituting the relevant angles gives us:

$\cos $105^{\circ}$=\cos $60^{\circ}$\cos $45^{\circ}$ - \sin $60^{\circ}$\sin $45^{\circ}$$

Now substitute in the exact values for each function, simplifying to keep radicals out of the denominator:

$\cos $105^{\circ}$ = $\frac{1}{2}$\cdot $\frac{\sqrt{2}$}{2}-$\frac{\sqrt{3}$}{2}\cdot $\frac{\sqrt {2}$}{2}$

Multiply and subtract to obtain:

$$\frac{\sqrt{2}$-$\sqrt{6}$}{4}$

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Question

Find the exact value of the expression:

$\cos\bigg($\frac{5\pi }{6}$\bigg)\cos\bigg($\frac{2\pi}{3}$\bigg)+\sin\bigg($\frac{5\pi}{6}$\bigg)\sin\bigg($\frac{2\pi}{3}$\bigg)$

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Answer

The formula for the cosine of the difference of two angles is

$\cos (a-b) = \cos a\cos b + \sin a\sin b$

Substituting, we find that

$a= $\frac{5\pi}{6}$$

and

$b= $\frac{2\pi}{3}$$

Therefore, what we are really looking for is

$\cos (a-b)= \cos \bigg($\frac{5\pi}{6}$-$\frac{2\pi}{3}$\bigg) = \cos\bigg($\frac{\pi}{6}$\bigg) = $\frac{\sqrt{3}$}{2}$

Thus,

$\cos \bigg($\frac{5\pi}{6}$\bigg)\cos\bigg($\frac{2\pi}{3}$\bigg) + \sin\bigg ($\frac{5\pi}{6}$\bigg)\sin\bigg($\frac{2\pi}{3}$\bigg) = $\frac{\sqrt{3}$}{2}$

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Question

Find the exact value of $\tan $75^{\circ}$$ using $\tan $30^{\circ}$$ and $\tan $45^{\circ}$$ .

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Answer

The sum identity for tangent states that

$\tan (a+b) = $\frac{\tan a + \tan b}{1-\tan a\tan b}$$

Substituting known values for $\tan $30^{\circ}$$ and $\tan $45^{\circ}$$ , we have

$\tan $(30^{\circ}$ + $45^{\circ}$) = $\frac{1+\frac{1}{\sqrt{3}$}}{1-$\frac{1}{\sqrt{3}$}}$

For ease, multiply all terms by $\sqrt{3}$ to get $$\frac{\sqrt{3}$+1}{$\sqrt{3}$-1}$ .

At this point, multiply both halves of the fraction by the conjugate of the denominator:

$\bigg($\frac{\sqrt{3}$+1}{$\sqrt{3}$-1}\bigg)\cdot\bigg($\frac{\sqrt{3}$+1}{$\sqrt{3}$+1}\bigg) = $\frac{3+2\sqrt{3}$+1}{3-1}$

Finally, simplify.

$\frac{3+2\sqrt{3}$+1}{3-1} = $\frac{4+2\sqrt{3}$}{2} = 2+$\sqrt{3}$

So, $\tan $75^{\circ}$ = 2+$\sqrt{3}$$ .

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Question

Suppose we have two angles, and , such that:

$sin\ A = 0.10$

$cos\ B = 0.10$

Furthermore, suppose that angle is located in the first quadrant and angle is located in the fourth.

What is the measure of:

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Answer

We can calculate some missing values using the pythagorean identities.

$sin\ B = {0.99}^$\frac{1}{2}$ = -0.995$

(Note the negative sign, because is in the fourth quadrant, where the sine of the angle is always negative).

$cos\ A = $\sqrt{1 - $sin^2A$}$ = $\sqrt{1 - 0.01}$ = $\sqrt{0.99}$ = 0.995$

Note the positive value, since is in the first quadrant, where cosine is positive.

Now using the rules for double angles:

$cos\ 2B = 2\ $cos^2$ B - 1 = 2 * $0.1^2$ - 1 = -0.98$

$sin\ 2B = 2\ cos\ B\ sinB = 2 \ * 0.1 \ * -0.995 = -0.199$

And then the angle subtraction formula:

$sin(A - 2B) = sin\ A \ cos \ 2B\ - cos\ A \ sin 2B$

$= (0.1 * -0.98) \ - (0.995 * -0.199)$

$= \textbf{0.10}$

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Question

Calculate $sin(105^\circ)$ .

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Answer

Recall the formula for the sine of the sum of two angles:

Here, we can evaluate by noticing that and applying the above formula to the sines and cosines of these two angles.

$=($\frac{\sqrt3}{2}$)($\frac{\sqrt2}{2}$)+($\frac{\sqrt2}{2}$)($\frac{1}{2}$)$

$=\frac{\sqrt6+\sqrt2}{4}$$

Hence,

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Question

What is the value of $\cos(540)$ , using the sum formula.

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Answer

The formula for

$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$ .

We can expand

,

where and .

Substituting these values into the equation, we get

$\cos(180+360)=\cos(180)\cos(360)-\sin(180)\sin(360)$ .

The final answer is -1, using what we know about the unit circle values.

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Question

Simplify the given expression.

$\sin(x+y)-sin(x-y)$

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Answer

This problem requires the use of two angle sum/difference identities:

$\sin(x+y)=\sin x\cos y+\cos x\sin y$

$\sin(x-y)=\sin x\cos y-\cos x\sin y$

Using these identities, we get

$(\sin x\cos y+\cos x\sin y)-(\sin x\cos y-\cos x\sin y)$

which simplifies to

$\sin x\cos y+\cos x\sin y-\sin x\cos y+\cos x\sin y$

which equals

$2\cos x\sin y$

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Question

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Answer

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Question

Find the exact value of the expression:

$\sin $60^{\circ}$\cos $30^{\circ}$ + \cos $60^{\circ}$ \sin $30^{\circ}$$

Tap to reveal answer

Answer

There are two ways to solve this problem. If one recognizes the identity

$\sin (a+b)= \sin a\cos b + \cos a\sin b$ ,

the answer is as simple as:

$\sin $60^{\circ}$\cos $30^{\circ}$ + \cos $60^{\circ}$\sin $30^{\circ}$ = \sin $(60^{\circ}$$+30^{\circ}$) = \sin $90^{\circ}$ = 1$

If one misses the identity, or wishes to be more thorough, you can simplify:

$\sin $60^{\circ}$\cos $30^{\circ}$ + \cos $60^{\circ}$\sin $30^{\circ}$$

$=\bigg(\bigg($\frac{\sqrt{3}$}{2}\bigg)\cdot\bigg($\frac{\sqrt{3}$}{2}\bigg)\bigg)+\bigg(\bigg($\frac{1}{2}$\bigg)\cdot\bigg($\frac{1}{2}$\bigg)\bigg)=\frac{3}{4}$ + $\frac{1}{4}$=1$

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Question

Given $\phi = \pi$ , what is $\tan\left ( \theta + \phi \right )$ ?

Tap to reveal answer

Answer

We need to use the formula

$\tan\left ( \theta \pm \phi\right ) = $\frac{\tan(\theta) \pm \tan(\phi)}{1 \mp \tan(\theta)\tan(\phi)}$$

Substituting $\phi = \pi$ , and $\tan (\pi) = 0$ ,

$\tan\left ( \theta + \pi\right ) = $\frac{\tan(\theta) + \tan(\phi)}{1 - \tan(\theta)\tan(\phi)}$ = $\frac{\tan(\theta) + 0}{1 - \tan(\theta)\left( 0\right)}$ = \tan(\theta)$

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Question

Find the exact value of $\cos $105^{\circ}$$ using .

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Answer

Our basic sum formula for cosine is:

$\cos(a+b)=\cos a\cos b - \sin a\sin b$

Substituting the relevant angles gives us:

$\cos $105^{\circ}$=\cos $60^{\circ}$\cos $45^{\circ}$ - \sin $60^{\circ}$\sin $45^{\circ}$$

Now substitute in the exact values for each function, simplifying to keep radicals out of the denominator:

$\cos $105^{\circ}$ = $\frac{1}{2}$\cdot $\frac{\sqrt{2}$}{2}-$\frac{\sqrt{3}$}{2}\cdot $\frac{\sqrt {2}$}{2}$

Multiply and subtract to obtain:

$$\frac{\sqrt{2}$-$\sqrt{6}$}{4}$

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Question

Find the exact value of the expression:

$\cos\bigg($\frac{5\pi }{6}$\bigg)\cos\bigg($\frac{2\pi}{3}$\bigg)+\sin\bigg($\frac{5\pi}{6}$\bigg)\sin\bigg($\frac{2\pi}{3}$\bigg)$

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Answer

The formula for the cosine of the difference of two angles is

$\cos (a-b) = \cos a\cos b + \sin a\sin b$

Substituting, we find that

$a= $\frac{5\pi}{6}$$

and

$b= $\frac{2\pi}{3}$$

Therefore, what we are really looking for is

$\cos (a-b)= \cos \bigg($\frac{5\pi}{6}$-$\frac{2\pi}{3}$\bigg) = \cos\bigg($\frac{\pi}{6}$\bigg) = $\frac{\sqrt{3}$}{2}$

Thus,

$\cos \bigg($\frac{5\pi}{6}$\bigg)\cos\bigg($\frac{2\pi}{3}$\bigg) + \sin\bigg ($\frac{5\pi}{6}$\bigg)\sin\bigg($\frac{2\pi}{3}$\bigg) = $\frac{\sqrt{3}$}{2}$

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Question

Find the exact value of $\tan $75^{\circ}$$ using $\tan $30^{\circ}$$ and $\tan $45^{\circ}$$ .

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Answer

The sum identity for tangent states that

$\tan (a+b) = $\frac{\tan a + \tan b}{1-\tan a\tan b}$$

Substituting known values for $\tan $30^{\circ}$$ and $\tan $45^{\circ}$$ , we have

$\tan $(30^{\circ}$ + $45^{\circ}$) = $\frac{1+\frac{1}{\sqrt{3}$}}{1-$\frac{1}{\sqrt{3}$}}$

For ease, multiply all terms by $\sqrt{3}$ to get $$\frac{\sqrt{3}$+1}{$\sqrt{3}$-1}$ .

At this point, multiply both halves of the fraction by the conjugate of the denominator:

$\bigg($\frac{\sqrt{3}$+1}{$\sqrt{3}$-1}\bigg)\cdot\bigg($\frac{\sqrt{3}$+1}{$\sqrt{3}$+1}\bigg) = $\frac{3+2\sqrt{3}$+1}{3-1}$

Finally, simplify.

$\frac{3+2\sqrt{3}$+1}{3-1} = $\frac{4+2\sqrt{3}$}{2} = 2+$\sqrt{3}$

So, $\tan $75^{\circ}$ = 2+$\sqrt{3}$$ .

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Question

Suppose we have two angles, and , such that:

$sin\ A = 0.10$

$cos\ B = 0.10$

Furthermore, suppose that angle is located in the first quadrant and angle is located in the fourth.

What is the measure of:

Tap to reveal answer

Answer

We can calculate some missing values using the pythagorean identities.

$sin\ B = {0.99}^$\frac{1}{2}$ = -0.995$

(Note the negative sign, because is in the fourth quadrant, where the sine of the angle is always negative).

$cos\ A = $\sqrt{1 - $sin^2A$}$ = $\sqrt{1 - 0.01}$ = $\sqrt{0.99}$ = 0.995$

Note the positive value, since is in the first quadrant, where cosine is positive.

Now using the rules for double angles:

$cos\ 2B = 2\ $cos^2$ B - 1 = 2 * $0.1^2$ - 1 = -0.98$

$sin\ 2B = 2\ cos\ B\ sinB = 2 \ * 0.1 \ * -0.995 = -0.199$

And then the angle subtraction formula:

$sin(A - 2B) = sin\ A \ cos \ 2B\ - cos\ A \ sin 2B$

$= (0.1 * -0.98) \ - (0.995 * -0.199)$

$= \textbf{0.10}$

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Question

Calculate $sin(105^\circ)$ .

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Answer

Recall the formula for the sine of the sum of two angles:

Here, we can evaluate by noticing that and applying the above formula to the sines and cosines of these two angles.

$=($\frac{\sqrt3}{2}$)($\frac{\sqrt2}{2}$)+($\frac{\sqrt2}{2}$)($\frac{1}{2}$)$

$=\frac{\sqrt6+\sqrt2}{4}$$

Hence,

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Question

What is the value of $\cos(540)$ , using the sum formula.

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Answer

The formula for

$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$ .

We can expand

,

where and .

Substituting these values into the equation, we get

$\cos(180+360)=\cos(180)\cos(360)-\sin(180)\sin(360)$ .

The final answer is -1, using what we know about the unit circle values.

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Question

Simplify the given expression.

$\sin(x+y)-sin(x-y)$

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Answer

This problem requires the use of two angle sum/difference identities:

$\sin(x+y)=\sin x\cos y+\cos x\sin y$

$\sin(x-y)=\sin x\cos y-\cos x\sin y$

Using these identities, we get

$(\sin x\cos y+\cos x\sin y)-(\sin x\cos y-\cos x\sin y)$

which simplifies to

$\sin x\cos y+\cos x\sin y-\sin x\cos y+\cos x\sin y$

which equals

$2\cos x\sin y$

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