Complete a Proof Using Sums, Differences, or Products of Sines and Cosines - Trigonometry

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True or false:

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

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

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

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

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

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

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

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

Which of the following correctly demonstrates the compound angle formula?

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

Which of the following correctly demonstrates the compound angle formula?

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

Simplify by applying the compound angle formula:

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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, $sinxcosy = (1/2)(\sin(x - y) + \sin(x + y))$ .

Simplify by applying the compound angle formula:

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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, $cosxcosy = (1/2)(\cos(x - y) + \cos(x + y))$ .

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

$\cos x + \sin x$

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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, $\sin x + \sin y = 2\sin ((x + y)/2) * \cos((x - y)/2)$ :

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

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

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

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

$\cos x - \sin x$

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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, $\sin x - \sin y = 2\sin((x - y)/2) * \cos((x + y)/2)$ .

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

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

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

True or false:

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

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

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

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

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

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

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

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

Which of the following correctly demonstrates the compound angle formula?

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

Which of the following correctly demonstrates the compound angle formula?

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

Simplify by applying the compound angle formula:

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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, $sinxcosy = (1/2)(\sin(x - y) + \sin(x + y))$ .

Simplify by applying the compound angle formula:

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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, $cosxcosy = (1/2)(\cos(x - y) + \cos(x + y))$ .

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

$\cos x + \sin x$

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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, $\sin x + \sin y = 2\sin ((x + y)/2) * \cos((x - y)/2)$ :

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

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

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

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

$\cos x - \sin x$

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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, $\sin x - \sin y = 2\sin((x - y)/2) * \cos((x + y)/2)$ .

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

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

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