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Trigonometry : Product of Sines and Cosines

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例子问题tions

例子问题tion #1 :Product Of Sines And Cosines

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

Possible Answers:

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

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

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

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

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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例子问题tion #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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例子问题tion #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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例子问题tion #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:

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

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

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

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

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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例子问题tion #5 :Product Of Sines And Cosines

真的或假的:所有的product-to-sum identities can be obtained from the sum-to-product identities

Possible Answers:

True

False

Correct answer:

True

Explanation:

所有的这些身份能够得到的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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例子问题tion #2 :Product Of Sines And Cosines

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

我们将n的身份eed 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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例子问题tion #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) =cos(6x) + cos(2x)

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

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

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

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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例子问题tion #8 :Product Of Sines And Cosines

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

Possible Answers:

sin(45)sin(30) = 4

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

sin(45)sin(30) = 2

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

Correct answer:

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

Explanation:

我们将n的身份eed 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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例子问题tions

例子问题tion #1 :Product Of Sines And Cosines

例子问题tion #1 :Product Of Sines And Cosines

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