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Trigonometry : Sum, Difference, and Product Identities

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Example Question #3 : Product Of Sines And Cosines

Use the product of cosines to evaluate $cos(\frac{\pi}{6})cos(\frac{5\pi}{3})$

Possible Answers:

$4\pi$

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{\sqrt{3}}{4}$

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\sqrt{3}$

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{3\pi}{4}$

Correct answer:

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{\sqrt{3}}{4}$

Explanation:

We are using the identity $cos(\alpha)cos(\beta)=\frac{1}{2}[cos(\alpha + \beta) + cos(\alpha - \beta)]$ . We will let $\alpha = \frac{\pi}{6}$ and $\beta = \frac{5\pi}{3}$ .

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) = \frac{1}{2}[cos(\frac{\pi}{6} + \frac{5\pi}{3}) + cos(\frac{\pi}{6} - \frac{5\pi}{3})]$

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{1}{2}[cos(\frac{11\pi}{6}) + cos(\frac{-9\pi}{6})]$

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{1}{2}[\frac{\sqrt{3}}{2} +0]$

$cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{\sqrt{3}}{4}$

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Example Question #4 : Product Of Sines And Cosines

Use the product of sines to evaluate $sin(\frac{3x}{4})sin(\frac{6x}{2})$ where $x = \frac{pi}{3}$

Possible Answers:

$sin(\frac{3x}{4})sin(\frac{6x}{2}) =\frac{\sqrt{3}}{2}$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) =0$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{\sqrt{2}}{2}$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) =\sqrt{2}$

Correct answer:

$sin(\frac{3x}{4})sin(\frac{6x}{2}) =0$

Explanation:

The formula for the product of sines is $sin(\alpha)sin(\beta) = \frac{1}{2}[cos(\alpha-\beta) - cos(\alpha + \beta)$ . We will let $\alpha = \frac{3x}{4}$ and $\beta = \frac{6x}{2}$ .

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[cos(\frac{3x}{4} - \frac{6x}{2}) - cos(\frac{3x}{4} + \frac{6x}{2})]$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[cos(\frac{-9x}{4}) - cos(\frac{15x}{4})]$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[cos(\frac{-9}{4}\frac{\pi}{3}) - cos(\frac{15}{4}\frac{\pi}{3})]$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[cos(\frac{-9\pi}{12}) - cos(\frac{15\pi}{12})]$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[cos(\frac{-3\pi}{4}) - cos(\frac{5\pi}{4})]$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[\frac{-\sqrt{2}}{2} - \frac{-\sqrt{2}}{2}]$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[0]$

$sin(\frac{3x}{4})sin(\frac{6x}{2}) = 0$

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Example Question #5 : Product Of Sines And Cosines

True or False: All of the product-to-sum identities can be obtained from the sum-to-product identities

Possible Answers:

True

False

Correct answer:

True

Explanation:

All of these identities are able to be obtained by the sum-to-product identities by either adding or subtracting two of the sum identities and canceling terms. Through some algebra and manipulation, you are able to derive each product identity.

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Example Question #6 : Product Of Sines And Cosines

Use the product of sine and cosine to evaluate $sin(\frac{\pi}{4})cos(\pi)$ .

Possible Answers:

$sin(\frac{\pi}{4})cos(\pi) = \frac{-\sqrt{2}}{2}$

$sin(\frac{\pi}{4]cos(\pi) = -\sqrt{2}$

$sin(\frac{\pi}{4]cos(\pi) = \sqrt{2}$

$sin(\frac{\pi}{4]cos(\pi) = \frac{\sqrt{2}}{2}$

Correct answer:

$sin(\frac{\pi}{4})cos(\pi) = \frac{-\sqrt{2}}{2}$

Explanation:

The identity that we will need to utilize to solve this problem is $sin(\alpha)cos(\beta) = \frac{1}{2}[sin(\alpha + \beta) + sin(\alpha - \beta)]$ . We will let $\alpha = \frac{\pi}{4}$ and $\beta = \pi$ .

$sin(\frac{\pi}{4})cos(\pi) = \frac{1}{2}[sin(\frac{\pi}{4} + \pi) +sin(\frac{\pi}{4} - \pi)]$

$sin(\frac{\pi}{4})cos(\pi) = \frac{1}{2}[sin(\frac{5\pi}{4} + sin(\frac{-3\pi}{4}]$

$sin(\frac{\pi}{4})cos(\pi) = \frac{1}{2}[\frac{-\sqrt{2}}{2} - \frac{\sqrt{2}}{2}]$

$sin(\frac{\pi}{4})cos(\pi) = \frac{1}{2}[\frac{-2\sqrt{2}}{2}]$

$sin(\frac{\pi}{4})cos(\pi) = \frac{1}{2}[-\sqrt{2}]$

$sin(\frac{\pi}{4})cos(\pi) = \frac{-\sqrt{2}}{2}$

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Example Question #7 : Product Of Sines And Cosines

Use the product of cosines to evaluate cos(4x)cos(2x) . Keep your answer in terms of .

Possible Answers:

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

$cos(4x)cos(2x) =\frac{1}{2}[cos(6x) + cos(2x)]$

cos(4x)cos(2x) =cos(6x) + cos(2x)

cos(4x)cos(2x) =cos(3x) + cos(x)

Correct answer:

$cos(4x)cos(2x) =\frac{1}{2}[cos(6x) + cos(2x)]$

Explanation:

The identity we will be using is $cos(\alpha)cos(\beta) = \frac{1}{2}[cos(\alpha + \beta) + cos(\alpha - \beta)]$ . We will let $\alpha = 4x$ and $\beta = 2x$ . $cos(4x)cos(2x) = \frac{1}{2}[cos(4x + 2x) + cos(4x - 2x)]$

$cos(4x)cos(2x) =\frac{1}{2}[cos(6x) + cos(2x)]$

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Example Question #8 : Product Of Sines And Cosines

Use the product of sines to evaluate sin(45)sin(30) .

Possible Answers:

sin(45)sin(30) = 4

$sin(45)sin(30) = \sqrt{2}$

sin(45)sin(30) = 2

$sin(45)sin(30) = \frac{\sqrt{2}}{4}$

Correct answer:

$sin(45)sin(30) = \frac{\sqrt{2}}{4}$

Explanation:

The identity that we will need to use is $sin(\alpha)sin(\beta) = \frac{1}{2}[cos(\alpha - \beta) - cos(\alpha + \beta)]$ . We will let $\alpha = 45 = \frac{\pi}{4}$ and $\beta = 30 = \frac{\pi}{6}$ .

$sin(45)sin(30) = \frac{1}{2}[cos(45 - 30) - cos(45 + 30)]$

$sin(45)sin(30) = \frac{1}{2}[cos(\frac{\pi}{4} - \frac{\pi}{6}) - cos(\frac{\pi}{4} + \frac{\pi}{6})]$

$sin(45)sin(30) = \frac{1}{2}[cos(\frac{\pi}{12}) - cos(\frac{5\pi}{12})]$

$sin(45)sin(30) = \frac{1}{2}[\frac{\sqrt{2}}{4}(\sqrt{3} +1) - \frac{\sqrt{2}}{4}(\sqrt{3} -1)]$

$sin(45)sin(30) = \frac{1}{2}[\frac{\sqrt{6}+\sqrt{2}}{4} - \frac{\sqrt{6}-\sqrt{2}}{4}]$

$sin(45)sin(30) = \frac{1}{2}[\frac{2\sqrt{2}}{4}]$

$sin(45)sin(30) = \frac{\sqrt{2}}{4}$

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Trigonometry : Sum, Difference, and Product Identities

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #3 : Product Of Sines And Cosines

Example Question #4 : Product Of Sines And Cosines

Example Question #5 : Product Of Sines And Cosines

Example Question #6 : Product Of Sines And Cosines

Example Question #7 : Product Of Sines And Cosines

Example Question #8 : Product Of Sines And Cosines

All Trigonometry Resources

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