Product of Sines and Cosines - Trigonometry

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

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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)]$

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

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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.

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

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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}$

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

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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}$) = 0$

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

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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.

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

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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}$

Use the product of cosines to evaluate . Keep your answer in terms of .

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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)]$

Use the product of sines to evaluate .

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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}$

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

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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)]$

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

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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.

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

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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}$

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

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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}$) = 0$

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

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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.

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

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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}$

Use the product of cosines to evaluate . Keep your answer in terms of .

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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)]$

Use the product of sines to evaluate .

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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}$

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

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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)]$

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

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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.

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

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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}$

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

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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}$) = 0$