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Trigonometry : Trigonometry

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6 Diagnostic Tests 155 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Sum And Difference Of Sines And Cosines

What is the correct formula for the sum of two sines: sin(x) + sin(y) ?

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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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Example Question #2 : Sum And Difference Of Sines And Cosines

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

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

Possible Answers:

$\pi$

$\sqrt{3}$

$\frac{1}{2}$

Correct answer:

$\sqrt{3}$

Explanation:

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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Example Question #3 : Sum And Difference Of Sines And Cosines

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

cos(2 + x) - cos(2 - x)

Possible Answers:

-2sin(2)sin(x)

-2cos(2)cos(x)

$\pi$

Correct answer:

-2sin(2)sin(x)

Explanation:

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 x = 2+x and y = 2 - x . Proceed by plugging these values into the formula.

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

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

cos(2+x) - cos(2 -x) = -2sin(2)sin(x)

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Example Question #2 : Sum And Difference Of Sines And Cosines

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

Possible Answers:

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

$sin(\alpha)sin(\beta) - sin(\alpha)sin(\beta)$

$cos(\alpha)cos(\beta) - cos(\alpha)cos(\beta)$

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

Correct answer:

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

Explanation:

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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Example Question #5 : Sum And Difference Of Sines And Cosines

Solve for the following using the correct identity:

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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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Example Question #6 : Sum And Difference Of Sines And Cosines

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

Possible Answers:

True

False

Correct answer:

False

Explanation:

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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Example Question #7 : Sum And Difference Of Sines And Cosines

Solve for the following using the correct identity:

cos(45 + 30)

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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(45)cos(30) - sin(45)sin(30)

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

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

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

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

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Example Question #3 : Sum And Difference Of Sines And Cosines

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

Possible Answers:

$sin(\alpha)sin(\beta) - sin(\alpha)sin(\beta)$

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

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

$cos(\alpha)cos(\beta) - cos(\alpha)cos(\beta)$

Correct answer:

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

Explanation:

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Example Question #1 : Product Of Sines And Cosines

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

Possible Answers:

$\frac{1}{2}[cos(\alpha +\beta)+sin(\alpha - \beta)]$

$\frac{1}{2}[sin(\alpha +\beta)+sin(\alpha - \beta)]$

$\frac{1}{2}[sin(\alpha +\beta)+cos(\alpha - \beta)]$

$\frac{1}{2}[cos(\alpha +\beta)+cos(\alpha - \beta)]$

Correct answer:

$\frac{1}{2}[cos(\alpha +\beta)+cos(\alpha - \beta)]$

Explanation:

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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Example Question #2 : Product Of Sines And Cosines

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

Possible Answers:

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

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

$sin(\alpha + \beta) + sin(\alpha - \beta) = 2sin(\alpha)cos(\beta)$

$\implies sin(\alpha)cos(\beta) = \frac{1}{2}[sin(\alpha + \beta) + sin(\alpha - \beta)]$

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

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

$cos(\alpha + \beta) - cos(\alpha- \beta) = -2sin(\alpha)sin(\beta)$

$\implies sin(\alpha)sin(\beta) = \frac{1}{2}[cos(\alpha - \beta) - cos(\alpha + \beta)]$

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

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

$cos(\alpha + \beta) + cos(\alpha - \beta) = 2cos(\alpha)cos(\beta)$

$\implies cos(\alpha)cos(\beta)=\frac{1}{2}[cos(\alpha + \beta) + cos(\alpha - \beta)]$

Correct answer:

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

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

$cos(\alpha + \beta) - cos(\alpha- \beta) = -2sin(\alpha)sin(\beta)$

$\implies sin(\alpha)sin(\beta) = \frac{1}{2}[cos(\alpha - \beta) - cos(\alpha + \beta)]$

Explanation:

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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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #1 : Sum And Difference Of Sines And Cosines

Example Question #2 : Sum And Difference Of Sines And Cosines

Example Question #3 : Sum And Difference Of Sines And Cosines

Example Question #2 : Sum And Difference Of Sines And Cosines

Example Question #5 : Sum And Difference Of Sines And Cosines

Example Question #6 : Sum And Difference Of Sines And Cosines

Example Question #7 : Sum And Difference Of Sines And Cosines

Example Question #3 : Sum And Difference Of Sines And Cosines

Example Question #1 : Product Of Sines And Cosines

Example Question #2 : Product Of Sines And Cosines

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