Sum and Difference of Sines and Cosines - Trigonometry

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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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Question

Solve for the following using the formula for the differences of two cosines. Do not simplify.

Answer

We begin by considering the formula for the differences of two cosines.

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

We will let and . Proceed by plugging these values into the formula.

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

$cos(2+x) - cos(2 -x) = -2sin\frac{4}{2}sin\frac{2x}{2}$

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Question

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

Answer

This is a known trigonometry identity and has been proven to be true. It is often helpful to solve for the quantity within a cosine function when there are unknowns or if the quantity needs to be simplified

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Question

Solve for the following using the correct identity:

$sin(\frac{\pi}{4} + \frac{\pi}{6})$

Answer

To solve this problem we must use the identity

$sin(\alpha + \beta) = sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta)$ . We will let $\alpha = \frac{\pi}{4}$ and $\beta = \frac{\pi}{6}$ .

$sin(\frac{\pi}{4} + \frac{\pi}{6}) = sin(\frac{\pi}{4})cos(\frac{\pi}{6}) + cos(\frac{\pi}{4})sin(\frac{\pi}{6})$

$sin(\frac{\pi}{4} + \frac{\pi}{6}) = (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2})(\frac{1}{2})$

$sin(\frac{\pi}{4} + \frac{\pi}{6}) =\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{2}$

$sin(\frac{\pi}{4} + \frac{\pi}{6}) =\frac{\sqrt{6} + 2\sqrt{2}}{4}$

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Question

True or False: To solve for a problem in the form of $cos(\alpha - \beta)$ , I use the identity $cos(\alpha) - cos(\beta) = -2sin\frac{\alpha + \beta}{2}sin\frac{\alpha - \beta}{2}$ .

Answer

This answer is false. $cos(\alpha - \beta)$ is not the same as $cos(\alpha) - cos(\beta)$ .

For example, say $\alpha = \frac{2\pi}{3}$ and $\beta = \frac {\pi}{3}$

$cos(\frac{2\pi}{3} - \frac{\pi}{3}) = cos(\frac{\pi}{3}) = \frac{1}{2}$

$cos(\frac{2\pi}{3}) - cos(\frac{\pi}{3} = -\frac{1}{2} - \frac{1}{2} = -1$

And so the correct identity to use for this is

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

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Question

Solve for the following using the correct identity:

Answer

The correct identity to use for this kind of problem is

$cos(\alpha + \beta) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$ . We will let $\alpha = 45 = \frac{\pi}{4}$ and $\beta = 30 = \frac{\pi}{6}$ .

$cos(45 + 30) = cos(\frac{\pi}{4})cos(\frac{\pi}{6}) - sin(\frac{\pi}{4})sin(\frac{\pi}{6})$

$cos(45 + 30) = (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) - (\frac{\sqrt{2}}{2})(\frac{1}{2})$

$cos(45 + 30) = \frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}$

$cos(45 + 30) = \frac{\sqrt{6} - 2\sqrt{2}}{4}$

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Question

Which of the following is the correct to complete the following identity: $sin(\alpha - \beta) =$ ___?

Answer

This is a known trigonometry identity and has been proven to be true. It is often helpful to solve for the quantity within a cosine function when there are unknowns or if the quantity needs to be simplified

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