Sum, Difference, and Product Identities - Trigonometry

Card 0 of 20

Question

True or false:

$\sin x + \sin y = \sin (x + y)$ .

Answer

The sum of sines is given by the formula $\sin x + \sin y = \2\sin ((x + y)/2) * \cos((x - y)/2)$ .

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Question

True or false: $\sin x - \sin y = \sin (x - y)$ .

Answer

The difference of sines is given by the formula $\sin x - \sin y = \2\sin ((x - y)/2) * \cos((x + y)/2)$ .

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Question

True or false: $\cos x + \cos y = \cos (x + y)$ .

Answer

The sum of cosines is given by the formula $\cos x + \cos y = 2\cos ((x + y)/2) * \cos((x - y)/2)$ .

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Question

True or false: $\cos x - \cos y = \cos (x - y)$ .

Answer

The difference of cosines is given by the formula $\cos x - \cos y = 2\sin ((x + y)/2) * \sin((y - x)/2)$ .

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Question

Which of the following correctly demonstrates the compound angle formula?

Answer

The compound angle formula for sines states that $\sin(x +- y) = sinxcosy +- cosxsiny$ .

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Question

Which of the following correctly demonstrates the compound angle formula?

Answer

The compound angle formula for cosines states that $\cos(x +- y) = sinxsiny -+ cosxcosy$ .

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Question

Simplify by applying the compound angle formula:

Answer

Using the compound angle formula, we can rewrite each half of the non-coefficient terms in the given expression. Given that $\sin (x - y) = sinxcosy - cosxsiny$ and $\sin (x + y) = sinxcosy + cosxsiny$ , substitution yields the following:

This is the formula for the product of sine and cosine, $sinxcosy = (1/2)(\sin(x - y) + \sin(x + y))$ .

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Question

Simplify by applying the compound angle formula:

Answer

Using the compound angle formula, we can rewrite each half of the non-coefficient terms in the given expression. Given that $\cos(x - y) = sinxsiny + cosxcosy$ and $\cos(x + y) = sinxsiny - cosxcosy$ , substitution yields the following:

$(1/2)(\cos(x - y) + \cos(x + y))$

This is the formula for the product of two cosines, $cosxcosy = (1/2)(\cos(x - y) + \cos(x + y))$ .

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Question

Using $\sin(90 - x) = \cos x$ and the formula for the sum of two sines, rewrite the sum of cosine and sine:

$\cos x + \sin x$

Answer

Substitute $\sin(90 - x)$ for $\cos x$ :

$\cos x + \sin x$

$\sin(90 - x) + \sin x$

Apply the formula for the sum of two sines, $\sin x + \sin y = 2\sin ((x + y)/2) * \cos((x - y)/2)$ :

$\sin(90 - x) + \sin x$

$2\sin((90 - x + x)/2) * \cos((90 - x - x)/2)$

$2\sin(45) * \cos(45 - x)$

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Question

Using $\sin(90 - x) = \cos x$ and the formula for the difference of two sines, rewrite the difference of cosine and sine:

$\cos x - \sin x$

Answer

Substitute $\sin(90 - x)$ for $\cos x$ :

$\cos x - \sin x$

$\sin(90 - x) - \sin x$

Apply the formula for the difference of two sines, $\sin x - \sin y = 2\sin((x - y)/2) * \cos((x + y)/2)$ .

$\sin(90 - x) - \sin x$

$2\sin((90 - x - x)/2) * \cos((90 - x + x)/2)$

$2\sin(45 - x) * \cos(45)$

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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

Use the product of sines to evaluate .

Answer

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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Question

What is the correct formula for the sum of two sines: ?

Answer

This is a known trigonometry identity. Whenever you are adding two sine functions, you can plug and into the formula to solve for this sum

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Question

Solve for the following given that $x = 2\pi$ . Use the formula for the sum of two sines.

$sin(\frac{\pi}{3} +x) + sin(\frac{\pi}{3} - x)$

Answer

We begin by considering our formula for the sum of two sines

$sin(x) + sin(y) = 2sin\frac{x + y}{2}cos\frac{x-y}{2}$

We will let $x = \frac{\pi}{3} + x$ and $y = \frac{\pi}{3} - x$ and plug these values into our formula.

$sin(\frac{\pi}{3} +x) + sin(\frac{\pi}{3} - x) = 2sin\frac{(\frac{\pi}{3} + x) + (\frac{\pi}{3} - x)}{2}cos\frac{(\frac{\pi}{3} + x) - (\frac{\pi}{3} - x)}{2}$

$sin(\frac{\pi}{3} + x) + sin(\frac{\pi}{3} - x) = 2sin\frac{\frac{2\pi}{3}}{2}cos\frac{2x}{2}$

$sin(\frac{\pi}{3} + x) + sin(\frac{\pi}{3} - x) = 2sin\frac{\pi}{3}cosx$

$sin(\frac{\pi}{3} + x) + sin(\frac{\pi}{3} - x) = 2(\frac{\sqrt{3}}{2})(1)$

$sin(\frac{\pi}{3} + x) + sin(\frac{\pi}{3} - x) = \sqrt{3}$

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