Sum and Difference of Sines and Cosines - Trigonometry

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

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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

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

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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

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)$

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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}$$

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

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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}$$

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

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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

Solve for the following using the correct identity:

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

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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}$

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}$$ .

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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

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

Solve for the following using the correct identity:

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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}$

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

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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

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

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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

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)$

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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}$$

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

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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}$$

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

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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

Solve for the following using the correct identity:

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

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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}$

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}$$ .

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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

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

Solve for the following using the correct identity:

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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}$

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

Tap to see back →

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

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)$

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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}$$

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

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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}$$

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

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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