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Precalculus : Sum and Difference Identities For Cosine

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Example Questions

Example Question #13 : Trigonometric Identities

Evaluate the exact value of:

2cos(75)

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

In order to solve 2cos(75) , two special angles will need to be used to solve for the exact values.

The angles chosen are A=30 and B=45 degrees, since:

A+B=30+45 =75

Write the formula for the cosine additive identity.

cos(A+B)= cosAcosB-sinAsinB

Substitute the known variables.

$cos(30)cos(45)-sin(30)sin(45)=\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{\sqrt2}{2}\right)$

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

$2cos(75)=2\left(\frac{\sqrt6-\sqrt2}{4}\right)=\frac{\sqrt6-\sqrt2}{2}$

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Example Question #14 : Trigonometric Identities

Find the exact value of cos(15)-cos(75) .

Possible Answers:

$\frac{\sqrt2}{2}$

$\sqrt2$

$\sqrt6+\sqrt2$

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

Correct answer:

$\frac{\sqrt2}{2}$

Explanation:

To solve cos(15)-cos(75) , we will need to use both the sum and difference identities for cosine.

Write the formula for these identities.

$cos(A\pm B)= cos(A)cos(B)\pm sin(A)sin(B)$

To solve for cos(15) and cos(75) , find two special angles whose difference and sum equals to the angle 15 and 75, respectively. The two special angles are 45 and 30.

A=45

B=30

Substitute the special angles in the formula.

$cos(45\pm 30)= cos(45)cos(30)\pm sin(45)sin(30)$

Evaluate both conditions.

cos(75)=cos(45+30)=cos(45)cos(30)+sin(45)sin(30)

$=\frac{\sqrt2}{2}\left(\frac{\sqrt3}{2}\right)+\frac{\sqrt2}{2}\left(\frac{1}{2}\right)=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}$

cos(15)=cos(45-30)=cos(45)cos(30)-sin(45)sin(30)

$=\frac{\sqrt2}{2}\left(\frac{\sqrt3}{2}\right)-\frac{\sqrt2}{2}\left(\frac{1}{2}\right)=\frac{\sqrt6}{4}-\frac{\sqrt2}{4}$

Solve for cos(15)-cos(75) .

$=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}-\left[\frac{\sqrt6}{4}-\frac{\sqrt2}{4}\right]$

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

$= \frac{2\sqrt2}{4}$

$= \frac{\sqrt2}{2}$

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Example Question #15 : 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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 cosine sum formula, we then see:

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

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Example Question #16 : 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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 cosine difference formula, we see:
$\\ \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\\ \\* = \frac {2 \sqrt {6}}{5} \cdot \frac {1}{4} + \frac {1}{5} \cdot \frac {\sqrt {15}}{4} = \frac {2 \sqrt{6} + \sqrt {15}}{20}$

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Precalculus : Sum and Difference Identities For Cosine

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #13 : Trigonometric Identities

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