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Precalculus : Sum and Difference Identities For Sine

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Example Questions

Example Question #1 : Trigonometric Identities

Find $\sin\bigg(\frac{\pi }{3} + \frac{\pi }{6}\bigg)$ using the sum identity.

Possible Answers:

$\frac{\pi }{2}$

1.5

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

Correct answer:

Explanation:

Using the sum formula for sine,

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

where,

$a=\frac{\pi}{3}$ , $b=\frac{\pi}{6}$

yeilds:

$\sin \bigg(\frac{\pi}{3}+\sin\frac{\pi}{6}\bigg)=\sin\bigg(\frac{\pi }{3}\bigg)\cdot\cos\bigg(\frac{\pi }{6}\bigg) +\cos \bigg(\frac{\pi}{3}\bigg)\cdot\sin\bigg(\frac{\pi }{6}\bigg)$

$=\frac{\sqrt{3}}2{}\cdot\frac{\sqrt{3}}2{}+\frac{1}{2}\cdot\frac{1}{2}$

$=\frac{3}{4}+\frac{1}{4}=1$ .

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Example Question #1 : Sum And Difference Identities

Calculate $\sin(75)$ .

Possible Answers:

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

$\frac{1}2{}$

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

Correct answer:

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

Explanation:

Notice that $\sin(75)$ is equivalent to $\sin(30+45)$ . With this conversion, the sum formula can be applied using,

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

where

a=30 , b=45 .

Therefore the result is as follows:

$\sin(a+b)=\sin(30)\cdot \cos(45)+\cos(30)\cdot\sin(45)$

$=\frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}2{}$

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

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Example Question #3 : Trigonometric Identities

Find the exact value for: $4\sin (15^{\circ})$

Possible Answers:

$\frac{\sqrt6}{4}-\frac{\sqrt2}{4}$

$\sqrt6-\sqrt2$

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

$\sqrt2-\frac{\sqrt2}{3}$

Correct answer:

$\sqrt6-\sqrt2$

Explanation:

In order to solve this question, it is necessary to know the sine difference identity.

sin(A-B)=sinAcosB-cosAsinB

The values of and must be a special angle, and their difference must be 15 degrees.

A possibility of their values that match the criteria are:

A=45

B=30

Substitute the values into the formula and solve.

sin(45-30)=(sin45)(cos30)-(cos45)(sin30)

$sin(15)=(\frac{\sqrt2}{2})(\frac{\sqrt3}{2})-(\frac{\sqrt2}{2})(\frac{1}{2})= \frac{\sqrt6}{4}-\frac{\sqrt2}{4}$

Evaluate 4sin(15) .

$=4(\frac{\sqrt6}{4}-\frac{\sqrt2}{4})= \sqrt6-\sqrt2$

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Example Question #1 : Trigonometric Identities

Find the exact value of: $\sin(105^{\circ})$

Possible Answers:

$\frac{\sqrt6-\sqrt2}{4}$

$\frac{\sqrt2+\sqrt6}{2}$

$\frac{\sqrt2+\sqrt5}{4}$

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

$\frac{\sqrt2+3\sqrt2}{4}$

Correct answer:

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

Explanation:

In order to find the exact value of sin(105) , the sum identity of sine must be used. Write the formula.

sin(A+B)=sinAcosB+cosAsinB

The only possibilites of and are 45 and 60 degrees interchangably. Substitute these values into the equation and evaluate.

sin(45+60)=sin45cos60+cos45sin60

$sin(105)=(\frac{\sqrt2}{2})(\frac{1}{2})+(\frac{\sqrt2}{2})(\frac{\sqrt3}{2})=\frac{\sqrt2+\sqrt6}{4}$

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Example Question #5 : Trigonometric Identities

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\sin (\alpha + \beta)$ .

Possible Answers:

$\frac {\sqrt{6} - \sqrt{15}}{10}$

$\frac {2\sqrt{6} - \sqrt{15}}{20}$

$\frac {2\sqrt{6} + \sqrt{15}}{20}$

$\frac {1+ 6\sqrt{10}}{20}$

$\frac {1- 6\sqrt{10}}{20}$

Correct answer:

$\frac {1+ 6\sqrt{10}}{20}$

Explanation:

Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that y = 1 and r = 5 and therefore:

$x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\cos \alpha = \frac {2\sqrt{6}}{5}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that x = 1 and r = 4 and therefore:

$y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}$ .

So $\sin \beta = \frac {\sqrt{15}}{4}$ .

Using the sine sum formula, we see:

$\\ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \\* = \frac {1}{5} \cdot \frac {1}{4} + \frac {2 \sqrt {6}}{5} \cdot \frac {\sqrt {15}}{4} = \frac {1 + 6 \sqrt {10}}{20}$

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Example Question #6 : Trigonometric Identities

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\sin (\alpha - \beta)$ .

Possible Answers:

$\frac {1- 6\sqrt{10}}{20}$

$\frac {2\sqrt{6} - \sqrt{15}}{20}$

$\frac {\sqrt{6} - \sqrt{15}}{10}$

$\frac {1+ 6\sqrt{10}}{20}$

$\frac {2\sqrt{6} + \sqrt{15}}{20}$

Correct answer:

$\frac {1- 6\sqrt{10}}{20}$

Explanation:

Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that y = 1 and r = 5 and therefore:

$x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\cos \alpha = \frac {2\sqrt{6}}{5}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that x = 1 and r = 4 and therefore:

$y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}$ .

So $\sin \beta = \frac {\sqrt{15}}{4}$ .

Using the sine difference formula, we see:

$\\ \sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\ \\* = \frac {1}{5} \cdot \frac {1}{4} - \frac {2 \sqrt {6}}{5} \cdot \frac {\sqrt {15}}{4} = \frac {1 - 6 \sqrt {10}}{20}$

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Example Question #11 : Trigonometric Identities

Evaluate

$\sin \left(\frac{5 \pi}{12}\right)$ .

Possible Answers:

$\frac{-1 - \sqrt 3 }{\sqrt 2 }$

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

$\frac{1+\sqrt3}{2 \sqrt 2 }$

$\frac{-1+\sqrt3}{2 \sqrt 2 }$

$\frac{1}{\sqrt 2 }$

Correct answer:

$\frac{1+\sqrt3}{2 \sqrt 2 }$

Explanation:

$\frac{5 \pi }{12}$ is equivalent to $\frac{2 \pi }{12 } + \frac{3 \pi }{12 }$ or more simplified $\frac{\pi}{6 } + \frac{\pi }{4}$ .

We can use the sum identity to evaluate this sine:

$\sin \left(\frac{\pi }{6} + \frac{\pi }{4 } \right) = \sin\left (\frac{\pi }{6} \right) \cos \left(\frac { \pi }{4} \right) + \cos \left(\frac{\pi }{6}\right) \sin \left( \frac{ \pi }{4} \right)$

From the unit circle, we can determine these measures:

$\frac{1}{2} \cdot \frac{1}{\sqrt 2 } + \frac{ \sqrt 3 }{2} \cdot \frac{1}{ \sqrt 2 } = \frac{1}{2 \sqrt 2 }+ \frac{\sqrt 3 }{2 \sqrt 2 } = \frac{ 1 + \sqrt 3 }{ 2 \sqrt 2 }$

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Example Question #11 : Trigonometric Identities

Evaluate

$\sin(285 ^ o)$ .

Possible Answers:

$\frac { 1 - \sqrt 3 }{ 2 \sqrt 2 }$

$\frac{1+ \sqrt 3 }{ 2 \sqrt 2 }$

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

$\frac{-1 - \sqrt {3 }}{2 \sqrt 2 }$

$\frac{\sqrt 3 - 1 }{2 \sqrt 2 }$

Correct answer:

$\frac{-1 - \sqrt {3 }}{2 \sqrt 2 }$

Explanation:

The angle 285 ^o = 225^ o + 60 ^ o or 45^o + 240 ^ o .

Using the first one:

$\sin(60^o + 225^o ) = \sin (60^o) \cos (225^o) + \cos (60^o) \cos(225^o )$

We can find these values in the unit circle:

$\frac{\sqrt 3 }{2 } \cdot \frac{-1}{\sqrt 2 } + \frac{1}{2} \cdot \frac{-1}{\sqrt 2 } = \frac{-\sqrt3 }{2 \sqrt 2 } + \frac{-1}{2\sqrt2} = \frac{-\sqrt 3 - 1 }{2 \sqrt 2 }$

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Precalculus : Sum and Difference Identities For Sine

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Trigonometric Identities

Example Question #1 : Sum And Difference Identities

Example Question #3 : Trigonometric Identities

Example Question #1 : Trigonometric Identities

Example Question #5 : Trigonometric Identities

Example Question #6 : Trigonometric Identities

Example Question #11 : Trigonometric Identities

Example Question #11 : Trigonometric Identities

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