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Trigonometry : Product of Sines and Cosines

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Example Questions

Example Question #1 : Product Of Sines And Cosines

Which of the following completes the identity $cos(\alpha)cos(\beta) =$

Possible Answers:

$\frac{1}{2}[sin(\alpha +\beta)+cos(\alpha - \beta)]$

$\frac{1}{2}[cos(\alpha +\beta)+sin(\alpha - \beta)]$

$\frac{1}{2}[cos(\alpha +\beta)+cos(\alpha - \beta)]$

$\frac{1}{2}[sin(\alpha +\beta)+sin(\alpha - \beta)]$

Correct answer:

$\frac{1}{2}[cos(\alpha +\beta)+cos(\alpha - \beta)]$

Explanation:

This formula is able to be derived directly from the identities for the sum and difference of cosines, .

$cos(\alpha + \beta) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$

$+cos(\alpha - \beta) = cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)$

$cos(\alpha + \beta) + cos(\alpha - \beta) = 2cos(\alpha)cos(\beta)$

$\implies cos(\alpha)cos(\beta)=\frac{1}{2}[cos(\alpha + \beta) + cos(\alpha - \beta)]$

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Example Question #1 : Product Of Sines And Cosines

Derive the product of sines from the identities for the sum and differences of trigonometric functions.

Possible Answers:

$cos(\alpha + \beta) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$

$+cos(\alpha - \beta) = cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)$

$cos(\alpha + \beta) + cos(\alpha - \beta) = 2cos(\alpha)cos(\beta)$

$\implies cos(\alpha)cos(\beta)=\frac{1}{2}[cos(\alpha + \beta) + cos(\alpha - \beta)]$

$cos(\alpha + \beta) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$

$-cos(\alpha - \beta) = cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)$

$cos(\alpha + \beta) - cos(\alpha- \beta) = -2sin(\alpha)sin(\beta)$

$\implies sin(\alpha)sin(\beta) = \frac{1}{2}[cos(\alpha - \beta) - cos(\alpha + \beta)]$

$sin(\alpha+\beta) = sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta)$

$-sin(\alpha - \beta) = sin(\alpha)cos(\beta) - cos(\alpha)sin(\beta)$

$sin(\alpha + \beta) + sin(\alpha - \beta) = 2sin(\alpha)cos(\beta)$

$\implies sin(\alpha)cos(\beta) = \frac{1}{2}[sin(\alpha + \beta) + sin(\alpha - \beta)]$

Correct answer:

$cos(\alpha + \beta) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$

$-cos(\alpha - \beta) = cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)$

$cos(\alpha + \beta) - cos(\alpha- \beta) = -2sin(\alpha)sin(\beta)$

$\implies sin(\alpha)sin(\beta) = \frac{1}{2}[cos(\alpha - \beta) - cos(\alpha + \beta)]$

Explanation:

First, we must know the formula for the product of sines so that we know what we are searching for. The formula for this identity is $sin(\alpha)sin(\beta) = \frac{1}{2}[cos(\alpha -\beta) - cos(\alpha + \beta)]$ . Using the known identities of the sum/difference of cosines, we are able to derive the product of sines in this way. Sometimes it is helpful to be able to expand the product of trigonometric functions as sums. It can either simplify a problem or allow you to visualize the function in a different way.

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Example Question #1 : Product Of Sines And Cosines

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

Possible Answers:

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

$4\pi$

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

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

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 #1 : 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:

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

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

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

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

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 #1 : 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:

False

True

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 #2 : Product Of Sines And Cosines

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

Possible Answers:

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

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

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

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

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 #2 : 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) =cos(6x) + cos(2x)

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

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

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 #2 : Product Of Sines And Cosines

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

Possible Answers:

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

sin(45)sin(30) = 2

sin(45)sin(30) = 4

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

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 : Product of Sines and Cosines

Study concepts, example questions & explanations for Trigonometry

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Example Question #1 : Product Of Sines And Cosines

Example Question #1 : Product Of Sines And Cosines

Example Question #1 : Product Of Sines And Cosines

Example Question #1 : Product Of Sines And Cosines

Example Question #1 : Product Of Sines And Cosines

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