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14.4: Sum and Difference Formulas

Created by: CK-12

Objective

To use and derive the sum and difference formulas.

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Using your calculator, find the value of each trig function below. Round your answer to 4 decimal places.

1. $\sin 75^\circ$

2. $\cos 15^\circ$

3. $\tan 105^\circ$

4. $\cos 255^\circ$

Finding Exact Trig Values using Sum and Difference Formulas

Objective

Here you'll use the sum and difference formulas to find exact values of angles other than the critical angles.

Guidance

You know that $\sin 30^\circ=\frac{1}{2}, \cos 135^\circ=-\frac{\sqrt{2}}{2}, \tan 300 ^\circ = -\sqrt{3},$ etc... from the special right triangles. In this concept, we will learn how to find the exact values of the trig functions for angles other than these multiples of $30^\circ, 45^\circ,$ and $60^\circ$. Using the Sum and Difference Formulas, we can find these exact trig values.

Sum and Difference Formulas

$\sin(a\pm b) &=\sin a \cos b \pm \cos a \sin b \\\cos(a\pm b) &=\cos a \cos b \mp \sin a \sin b \\\tan(a \pm b) &=\frac{\tan a \pm \tan b}{1 \mp \tan a \tan b}$

Example A

Find the exact value of $\sin 75^\circ$.

Solution: This is an example of where we can use the sine sum formula from above, $\sin(a+b)=\sin a \cos b+\cos a \sin b$, where $a = 45^\circ$ and $b = 30^\circ$.

$\sin 75^\circ &=\sin(45^\circ + 30 ^\circ) \\&= \sin 45^\circ \cos 30^\circ +\cos 45^\circ \sin 30 ^\circ \\&= \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot \frac{1}{2} \\&= \frac{\sqrt{6}+\sqrt{2}}{4}$

In general, $\sin (a+b)\ne \sin a+\sin b$ and similar statements can be made for the other sum and difference formulas.

Example B

Find the exact value of $\cos \frac{11 \pi}{12}$.

Solution: For this example, we could use either the sum or difference cosine formula, $\frac{11\pi}{12}=\frac{2\pi}{3}+\frac{\pi}{4}$ or $\frac{11\pi}{12}=\frac{7\pi}{6}-\frac{\pi}{4}$. Let’s use the sum formula.

$\cos \frac{11\pi}{12} &=\cos \left(\frac{2\pi}{3}+\frac{\pi}{4}\right) \\&=\cos \frac{2\pi}{3}\cos \frac{\pi}{4}-\sin\frac{2\pi}{3}\sin \frac{\pi}{4} \\&= -\frac{1}{2}\cdot \frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} \\&= -\frac{\sqrt{2}+\sqrt{6}}{4}$

Example C

Find the exact value of $\tan \left(-\frac{\pi}{12}\right)$.

Solution: This angle is the difference between $\frac{\pi}{4}$ and $\frac{\pi}{3}$.

$\tan \left(\frac{\pi}{4}-\frac{\pi}{3}\right) &=\frac{\tan \frac{\pi}{4}-\tan \frac{\pi}{3}}{1+\tan \frac{\pi}{4} \tan \frac{\pi}{3}} \\&=\frac{1-\sqrt{3}}{1+\sqrt{3}}$

This angle is also the same as $\frac{23 \pi}{12}$. You could have also used this value and done $\tan\left(\frac{\pi}{4}+\frac{5 \pi}{3}\right)$ and arrived at the same answer.

Guided Practice

Find the exact values of:

1. $\cos 15^\circ$

2. $\tan 255^\circ$

1. $\cos 15^\circ &=\cos(45^\circ - 30^\circ) \\&= \cos 45^\circ \cos 30^\circ + \sin 45 ^\circ \sin 30 ^\circ \\&= \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot \frac{1}{2} \\&= \frac{\sqrt{6}+\sqrt{2}}{4}$

2. $\tan (210^\circ + 45^\circ) &=\frac{\tan 210^\circ+\tan 45^\circ}{1-\tan 210^\circ \tan 45^\circ} \\&= \frac{\frac{\sqrt{3}}{3}+1}{1-\frac{\sqrt{3}}{3}}=\frac{\frac{\sqrt{3}+3}{3}}{\frac{3-\sqrt{3}}{3}}=\frac{\sqrt{3}+3}{3-\sqrt{3}}$

Vocabulary

Sum and Difference Formulas
$\sin(a\pm b) &=\sin a \cos b \pm \cos a \sin b \\\cos (a \pm b) &=\cos a \cos b \mp \sin a \sin b \\\tan (a \pm b) &=\frac{\tan a \pm \tan b}{1 \mp \tan a \tan b}$

Problem Set

Find the exact value of the following trig functions.

1. $\sin 15^\circ$
2. $\cos \frac{5\pi}{12}$
3. $\tan 345^\circ$
4. $\cos (-255^\circ)$
5. $\sin \frac{13 \pi}{12}$
6. $\sin \frac{17\pi}{12}$
7. $\cos 15^\circ$
8. $\tan (-15^\circ)$
9. $\sin 345^\circ$
10. Now, use $\sin 15^\circ$ from #1, and find $\sin 345^\circ$. Do you arrive at the same answer? Why or why not?
11. Using $\cos 15^\circ$ from #7, find $\cos 165^\circ$. What is another way you could find $\cos 165^\circ$?
12. Describe any patterns you see between the sine, cosine, and tangent of these “new” angles.
13. Using your calculator, find the $\sin 142^\circ$. Now, use the sum formula and your calculator to find the $\sin 142^\circ$ using $83^\circ$ and $59^\circ$.
14. Use the sine difference formula to find $\sin 142^\circ$ with any two angles you choose. Do you arrive at the same answer? Why or why not?
15. Challenge Using $\sin (a+b)=\sin a \cos b +\cos a \sin b$ and $\cos (a+b)=\cos a \cos b - \sin a \sin b$, show that $\tan (a+b)=\frac{\tan a + \tan b}{1-\tan a \tan b}$.

Simplifying Trig Expressions using Sum and Difference Formulas

Objective

Here you'll use the sum and difference formulas to simplify expressions.

Guidance

We can also use the sum and difference formulas to simplify trigonometric expressions.

Example A

The $\sin a = -\frac{3}{5}$ and $\cos b =\frac{12}{13}$. $a$ is in the $3^{rd}$ quadrant and $b$ is in the $1^{st}$. Find $\sin(a+b)$.

Solution: First, we need to find $\cos a$ and $\sin b$. Using the Pythagorean Theorem, missing lengths are 4 and 5, respectively. So, $\cos a=-\frac{4}{5}$ because it is in the $3^{rd}$ quadrant and $\sin b = \frac{5}{13}$. Now, use the appropriate formulas.

$\sin (a+b) &=\sin a \cos b + \cos a \sin b \\&= -\frac{3}{5}\cdot \frac{12}{13}+-\frac{4}{5}\cdot \frac{5}{13} \\&= -\frac{56}{65}$

Example B

Using the information from Example A, find $\tan (a+b)$.

Solution: From the cosine and sine of $a$ and $b$, we know that $\tan a=\frac{3}{4}$ and $\tan b=\frac{5}{12}$.

$\tan (a+b) &=\frac{\tan a +\tan b}{1-\tan a \tan b} \\&= \frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot\frac{5}{12}} \\&= \frac{\frac{14}{12}}{\frac{11}{16}}=\frac{56}{33}$

Example C

Simplify $\cos (\pi - x)$.

Solution: Expand this using the difference formula and then simplify.

$\cos (\pi - x) &=\cos \pi \cos x +\sin \pi \sin x \\&=-1\cdot \cos x +0\cdot \sin x \\&=-\cos x$

Guided Practice

1. Using the information from Example A, find $\cos(a-b)$.

2. Simplify $\tan (x+\pi)$.

1. $\cos(a-b) &=\cos a \cos b + \sin a \sin b \\&=-\frac{4}{5}\cdot \frac{12}{13}+-\frac{3}{5}\cdot\frac{5}{13} \\&=-\frac{63}{65}$

2. $\tan (x+\pi)&=\frac{\tan x +\tan \pi}{1-\tan x \tan \pi} \\&=\frac{\tan x +0}{1-\tan 0} \\&=\tan x$

Problem Set

$\sin a =-\frac{8}{17}, \pi \le a < \frac{3\pi}{2}$ and $\sin b =-\frac{1}{2}, \frac{3\pi}{2}\le b <2\pi$. Find the exact trig values of:

1. $\sin (a+b)$
2. $\cos (a+b)$
3. $\sin (a-b)$
4. $\tan (a+b)$
5. $\cos (a-b)$
6. $\tan (a-b)$

Simplify the following expressions.

1. $\sin (2\pi-x)$
2. $\sin \left(\frac{\pi}{2}+x\right)$
3. $\cos (x+\pi)$
4. $\cos \left(\frac{3\pi}{2}-x\right)$
5. $\tan(x+2\pi)$
6. $\tan(x-\pi)$
7. $\sin \left(\frac{\pi}{6}-x\right)$
8. $\tan \left(\frac{\pi}{4}+x\right)$
9. $\cos \left(x-\frac{\pi}{3}\right)$

Determine if the following trig statements are true or false.

1. $\sin(\pi - x)=\sin (x-\pi)$
2. $\cos(\pi - x)=\cos (x-\pi)$
3. $\tan(\pi - x)=\tan (x-\pi)$

Solving Trig Equations using Sum and Difference Formulas

Objective

Here you'll solve trig equations using the sum and difference formulas.

Guidance

Lastly, we can use the sum and difference formulas to solve trigonometric equations. For this concept, we will only find solutions in the interval $0\le x <2\pi$.

Example A

Solve $\cos (x-\pi)=\frac{\sqrt{2}}{2}$.

Solution: Use the formula to simplify the left-hand side and then solve for $x$.

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

The cosine negative in the $2^{nd}$ and $3^{rd}$ quadrants. $x=\frac{3\pi}{4}$ and $\frac{5\pi}{4}$.

Example B

Solve $\sin \left(x+\frac{\pi}{4}\right)+1=\sin \left(\frac{\pi}{4}-x\right)$.

Solution:

$\sin \left(x+\frac{\pi}{4}\right)+1 &=\sin \left(\frac{\pi}{4}-x\right) \\\sin x \cos \frac{\pi}{4}+\cos x \sin \frac{\pi}{4}+1 &=\sin \frac{\pi}{4}\cos x -\cos \frac{\pi}{4}\sin x \\\sin x \cdot \frac{\sqrt{2}}{2}+\cos x \cdot \frac{\sqrt{2}}{2}+1 &=\frac{\sqrt{2}}{2}.\cos x -\frac{\sqrt{2}}{2} \cdot \sin x \\\sqrt{2} \sin x &=-1 \\\sin x &=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}$

In the interval, $x=\frac{5\pi}{4}$ and $\frac{7\pi}{4}$.

Example C

Solve $2\sin \left(x+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}$.

Solution:

$2\sin\left(x+\frac{\pi}{3}\right) &=\tan \frac{\pi}{3} \\2\left(\sin x \cos \frac{\pi}{3}+\cos x \sin \frac{\pi}{3}\right) &=\sqrt{3} \\2\sin x \cdot \frac{1}{2}+2\cos x \cdot \frac{\sqrt{3}}{2} &=\sqrt{3} \\\sin x +\sqrt{3}\cos x &=\sqrt {3} \\\sin x &=\sqrt{3}(1-\cos x) \\\sin ^2x &=3(1-2\cos x+\cos ^2x) \qquad \text{square both sides} \\1-\cos ^2x &=3-6\cos x + 3\cos ^2 x \qquad \ \ \text{substitute} \ \sin ^2{x} = 1-\cos ^2x \\0 &=4\cos^2x-6\cos x +2 \\0 &=2\cos^2x-3\cos x +1$

At this point, we can factor the equation to be $(2\cos x -1)(\cos x -1)=0$. $\cos x =\frac{1}{2}$, and 1, so $x=0, \frac{\pi}{3}, \frac{5 \pi}{3}$. Be careful with these answers. When we check these solutions it turns out that $\frac{5\pi}{3}$ does not work.

$2\sin\left(\frac{5\pi}{3}+\frac{\pi}{3}\right) &=\tan \frac{\pi}{3} \\2\sin 2\pi &=\sqrt{3} \\0 &\ne \sqrt{3}$

Therefore, $\frac{5\pi}{3}$ is an extraneous solution.

Guided Practice

Solve the following equations in the interval $0\le x<2\pi$.

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

2. $\sin \left(\frac{\pi}{6}-x\right)+1 = \sin \left(x+\frac{\pi}{6}\right)$

3. $\cos \left(\frac{\pi}{2}+x\right)=\tan \frac{\pi}{4}$

1. $\cos (2\pi-x) &=\frac{1}{2} \\\cos 2\pi \cos x +\sin 2 \pi \sin x &=\frac{1}{2} \\\cos x &=\frac{1}{2} \\x &=\frac{\pi}{3} \ and \ \frac{5\pi}{3}$

2. $\sin \left(\frac{\pi}{6}-x\right)+1 &=\sin \left(x+\frac{\pi}{6}\right)\\\sin \frac{\pi}{6}\cos x-\cos \frac{\pi}{6}\sin x +1 &=\sin x \cos \frac{\pi}{6}+\cos x \sin \frac{\pi}{6} \\\frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x+1 &=\frac{\sqrt{3}}{2}\sin x +\frac{1}{2} \cos x \\1 &=\sqrt{3}\sin x \\\frac{1}{\sqrt{3}} &=\sin x \\x &=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)=0.6155 \ and \ 2.5261 \ \text{rad}$

3. $\cos \left(\frac{\pi}{2}+x\right) &=\tan \frac{\pi}{4} \\\cos \frac{\pi}{2} \cos x - \sin \frac{\pi}{2} \sin x &=1 \\-\sin x &=1 \\\sin x &=-1 \\x &=\frac{3\pi}{2}$

Problem Set

Solve the following trig equations in the interval $0\le x < 2\pi$.

1. $\sin (x-\pi)=-\frac{\sqrt{2}}{2}$
2. $\cos(2\pi +x)=-1$
3. $\tan \left(x+\frac{\pi}{4}\right)=1$
4. $\sin \left(\frac{\pi}{2}-x\right)=\frac{1}{2}$
5. $\sin \left(x+\frac{3\pi}{4}\right)+\sin \left(x-\frac{3\pi}{4}\right)=1$
6. $\sin \left(x+\frac{\pi}{6}\right)=-\sin \left(x-\frac{\pi}{6}\right)$
7. $\cos \left(x+\frac{\pi}{6}\right)=\cos \left(x-\frac{\pi}{6}\right)+1$
8. $\cos \left(x+\frac{\pi}{3}\right)+\cos \left(x-\frac{\pi}{3}\right)=1$
9. $\tan(x+\pi)+2\sin (x+\pi)=0$
10. $\tan (x+\pi)+\cos \left(x+\frac{\pi}{2}\right)=0$
11. $\tan \left(x+\frac{\pi}{6}\right)=\tan \left(x+\frac{\pi}{4}\right)$
12. $\sin \left(x-\frac{5\pi}{3}\right)-2\sin \left(x-\frac{2\pi}{3}\right)=0$
13. $4\sin (x+\pi)-2=2\cos\left(x+\frac{\pi}{2}\right)$
14. $1+2\cos(x-\pi)+\cos x =0$
15. Real Life Application The height, $h$ (in feet), of two people in different seats on a Ferris wheel can be modeled by $h_1=50\cos 6t+46$ and $h_2=50\cos 6\left(t-\frac{\pi}{3}\right)+46$ where $t$ is the time (in minutes). When are the two people at the same height?

Date Created:

Apr 23, 2013

Aug 21, 2014
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