Trigonometry : Sum, Difference, and Product Identities

Study concepts, example questions & explanations for Trigonometry

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Example Questions

Example Question #3 : Product Of Sines And Cosines

Use the product of cosines to evaluate \(\displaystyle cos(\frac{\pi}{6})cos(\frac{5\pi}{3})\)

Possible Answers:

\(\displaystyle 4\pi\)

\(\displaystyle cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\sqrt{3}\)

\(\displaystyle cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{\sqrt{3}}{4}\)

\(\displaystyle cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{3\pi}{4}\)

Correct answer:

\(\displaystyle cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{\sqrt{3}}{4}\)

Explanation:

We are using the identity \(\displaystyle cos(\alpha)cos(\beta)=\frac{1}{2}[cos(\alpha + \beta) + cos(\alpha - \beta)]\).  We will let \(\displaystyle \alpha = \frac{\pi}{6}\) and \(\displaystyle \beta = \frac{5\pi}{3}\).

 

\(\displaystyle cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) = \frac{1}{2}[cos(\frac{\pi}{6} + \frac{5\pi}{3}) + cos(\frac{\pi}{6} - \frac{5\pi}{3})]\)

\(\displaystyle cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{1}{2}[cos(\frac{11\pi}{6}) + cos(\frac{-9\pi}{6})]\)

\(\displaystyle cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{1}{2}[\frac{\sqrt{3}}{2} +0]\)

\(\displaystyle cos(\frac{\pi}{6})cos(\frac{5\pi}{3}) =\frac{\sqrt{3}}{4}\)

 

 

Example Question #4 : Product Of Sines And Cosines

Use the product of sines to evaluate \(\displaystyle sin(\frac{3x}{4})sin(\frac{6x}{2})\) where \(\displaystyle x = \frac{pi}{3}\)

Possible Answers:

\(\displaystyle sin(\frac{3x}{4})sin(\frac{6x}{2}) =\frac{\sqrt{3}}{2}\)

\(\displaystyle sin(\frac{3x}{4})sin(\frac{6x}{2}) =0\)

\(\displaystyle sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{\sqrt{2}}{2}\)

\(\displaystyle sin(\frac{3x}{4})sin(\frac{6x}{2}) =\sqrt{2}\)

Correct answer:

\(\displaystyle sin(\frac{3x}{4})sin(\frac{6x}{2}) =0\)

Explanation:

The formula for the product of sines is \(\displaystyle sin(\alpha)sin(\beta) = \frac{1}{2}[cos(\alpha-\beta) - cos(\alpha + \beta)\).  We will let \(\displaystyle \alpha = \frac{3x}{4}\) and \(\displaystyle \beta = \frac{6x}{2}\).

 

\(\displaystyle sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[cos(\frac{3x}{4} - \frac{6x}{2}) - cos(\frac{3x}{4} + \frac{6x}{2})]\)

\(\displaystyle sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[cos(\frac{-9x}{4}) - cos(\frac{15x}{4})]\)

\(\displaystyle sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[cos(\frac{-9}{4}\frac{\pi}{3}) - cos(\frac{15}{4}\frac{\pi}{3})]\)

\(\displaystyle sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[cos(\frac{-9\pi}{12}) - cos(\frac{15\pi}{12})]\)

\(\displaystyle sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[cos(\frac{-3\pi}{4}) - cos(\frac{5\pi}{4})]\)

\(\displaystyle sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[\frac{-\sqrt{2}}{2} - \frac{-\sqrt{2}}{2}]\)

\(\displaystyle sin(\frac{3x}{4})sin(\frac{6x}{2}) = \frac{1}{2}[0]\)

\(\displaystyle sin(\frac{3x}{4})sin(\frac{6x}{2}) = 0\)

 

 

Example Question #5 : Product Of Sines And Cosines

True or False: All of the product-to-sum identities can be obtained from the sum-to-product identities

Possible Answers:

False

True 

Correct answer:

True 

Explanation:

All of these identities are able to be obtained by the sum-to-product identities by either adding or subtracting two of the sum identities and canceling terms.  Through some algebra and manipulation, you are able to derive each product identity.

Example Question #6 : Product Of Sines And Cosines

Use the product of sine and cosine to evaluate \(\displaystyle sin(\frac{\pi}{4})cos(\pi)\).

Possible Answers:

\(\displaystyle sin(\frac{\pi}{4]cos(\pi) = -\sqrt{2}\)

\(\displaystyle sin(\frac{\pi}{4})cos(\pi) = \frac{-\sqrt{2}}{2}\)

\(\displaystyle sin(\frac{\pi}{4]cos(\pi) = \frac{\sqrt{2}}{2}\)

\(\displaystyle sin(\frac{\pi}{4]cos(\pi) = \sqrt{2}\)

Correct answer:

\(\displaystyle sin(\frac{\pi}{4})cos(\pi) = \frac{-\sqrt{2}}{2}\)

Explanation:

The identity that we will need to utilize to solve this problem is \(\displaystyle sin(\alpha)cos(\beta) = \frac{1}{2}[sin(\alpha + \beta) + sin(\alpha - \beta)]\).  We will let \(\displaystyle \alpha = \frac{\pi}{4}\) and  \(\displaystyle \beta = \pi\).

 

\(\displaystyle sin(\frac{\pi}{4})cos(\pi) = \frac{1}{2}[sin(\frac{\pi}{4} + \pi) +sin(\frac{\pi}{4} - \pi)]\)

\(\displaystyle sin(\frac{\pi}{4})cos(\pi) = \frac{1}{2}[sin(\frac{5\pi}{4} + sin(\frac{-3\pi}{4}]\)

\(\displaystyle sin(\frac{\pi}{4})cos(\pi) = \frac{1}{2}[\frac{-\sqrt{2}}{2} - \frac{\sqrt{2}}{2}]\)

\(\displaystyle sin(\frac{\pi}{4})cos(\pi) = \frac{1}{2}[\frac{-2\sqrt{2}}{2}]\)

\(\displaystyle sin(\frac{\pi}{4})cos(\pi) = \frac{1}{2}[-\sqrt{2}]\)

\(\displaystyle sin(\frac{\pi}{4})cos(\pi) = \frac{-\sqrt{2}}{2}\)

 

 

Example Question #1 : Product Of Sines And Cosines

Use the product of cosines to evaluate \(\displaystyle cos(4x)cos(2x)\).  Keep your answer in terms of \(\displaystyle x\).

Possible Answers:

\(\displaystyle cos(4x)cos(2x) =cos(6x) + cos(2x)\)

\(\displaystyle cos(4x)cos(2x) =cos(3x) + cos(x)\)

\(\displaystyle cos(4x)cos(2x) =\frac{1}{2}[cos(8x)]\)

\(\displaystyle cos(4x)cos(2x) =\frac{1}{2}[cos(6x) + cos(2x)]\)

Correct answer:

\(\displaystyle cos(4x)cos(2x) =\frac{1}{2}[cos(6x) + cos(2x)]\)

Explanation:

The identity we will be using is \(\displaystyle cos(\alpha)cos(\beta) = \frac{1}{2}[cos(\alpha + \beta) + cos(\alpha - \beta)]\).  We will let \(\displaystyle \alpha = 4x\) and \(\displaystyle \beta = 2x\).\(\displaystyle cos(4x)cos(2x) = \frac{1}{2}[cos(4x + 2x) + cos(4x - 2x)]\)

\(\displaystyle cos(4x)cos(2x) =\frac{1}{2}[cos(6x) + cos(2x)]\)

Example Question #1 : Product Of Sines And Cosines

Use the product of sines to evaluate \(\displaystyle sin(45)sin(30)\).

Possible Answers:

\(\displaystyle sin(45)sin(30) = \frac{\sqrt{2}}{4}\)

\(\displaystyle sin(45)sin(30) = 4\)

\(\displaystyle sin(45)sin(30) = \sqrt{2}\)

\(\displaystyle sin(45)sin(30) = 2\)

Correct answer:

\(\displaystyle sin(45)sin(30) = \frac{\sqrt{2}}{4}\)

Explanation:

The identity that we will need to use is \(\displaystyle sin(\alpha)sin(\beta) = \frac{1}{2}[cos(\alpha - \beta) - cos(\alpha + \beta)]\).  We will let \(\displaystyle \alpha = 45 = \frac{\pi}{4}\) and \(\displaystyle \beta = 30 = \frac{\pi}{6}\).

 

\(\displaystyle sin(45)sin(30) = \frac{1}{2}[cos(45 - 30) - cos(45 + 30)]\)

\(\displaystyle sin(45)sin(30) = \frac{1}{2}[cos(\frac{\pi}{4} - \frac{\pi}{6}) - cos(\frac{\pi}{4} + \frac{\pi}{6})]\)

\(\displaystyle sin(45)sin(30) = \frac{1}{2}[cos(\frac{\pi}{12}) - cos(\frac{5\pi}{12})]\)

\(\displaystyle sin(45)sin(30) = \frac{1}{2}[\frac{\sqrt{2}}{4}(\sqrt{3} +1) - \frac{\sqrt{2}}{4}(\sqrt{3} -1)]\)

\(\displaystyle sin(45)sin(30) = \frac{1}{2}[\frac{\sqrt{6}+\sqrt{2}}{4} - \frac{\sqrt{6}-\sqrt{2}}{4}]\)

\(\displaystyle sin(45)sin(30) = \frac{1}{2}[\frac{2\sqrt{2}}{4}]\)

\(\displaystyle sin(45)sin(30) = \frac{\sqrt{2}}{4}\)

 

 

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