Sum and Difference of Sines and Cosines - Trigonometry

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Question

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

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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

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

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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)$

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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.

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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 $cos(\alpha + \beta) =$

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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}$ + $\frac{\pi}{6}$)$

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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}$$ .

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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:

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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

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

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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)$

Tap to reveal answer

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.

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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 $cos(\alpha + \beta) =$

Tap to reveal answer

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}$ + $\frac{\pi}{6}$)$

Tap to reveal answer

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}$$ .

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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:

Tap to reveal answer

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: $sin(\alpha - \beta) =$ ?

Tap to reveal answer

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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

← Didn't Know|Knew It →

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)$

Tap to reveal answer

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.

Tap to reveal answer

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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