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# 3.15: Triple-Angle Formulas and Linear Combinations

Difficulty Level: At Grade Created by: CK-12
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Practice Triple-Angle Formulas and Linear Combinations
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In other Concepts you've dealt with double angle formulas. This was useful for finding the value of an angle that was double your well known value. Now consider the idea of a "triple angle formula". If someone gave you a problem like this:

sin135\begin{align*}\sin 135^\circ\end{align*}

Could you compute its value?

Keep reading, and at the end of this Concept you'll know how to simplify equations such as this using the triple angle formula.

### Watch This

Deriving a Triple Angle Formula

### Guidance

Double angle formulas are great for computing the value of a trig function in certain cases. However, sometimes different multiples than two times and angle are desired. For example, it might be desirable to have three times the value of an angle to use as the argument of a trig function.

By combining the sum formula and the double angle formula, formulas for triple angles and more can be found.

Here, we take an equation which takes a linear combination of sine and cosine and converts it into a simpler cosine function.

Acosx+Bsinx=Ccos(xD)\begin{align*}A \cos x + B \sin x = C \cos(x - D)\end{align*}, where C=A2+B2\begin{align*}C = \sqrt{A^2 + B^2}\end{align*}, cosD=AC\begin{align*}\cos D = \frac{A}{C} \end{align*} and sinD=BC\begin{align*}\sin D = \frac{B}{C}\end{align*}.

You can also use the TI-83 to solve trigonometric equations. It is sometimes easier than solving the equation algebraically. Just be careful with the directions and make sure your final answer is in the form that is called for. You calculator cannot put radians in terms of π\begin{align*}\pi\end{align*}.

#### Example A

Find the formula for sin3x\begin{align*}\sin 3x \end{align*}

Solution: Use both the double angle formula and the sum formula.

sin3x=sin(2x+x)=sin(2x)cosx+cos(2x)sinx=(2sinxcosx)cosx+(cos2xsin2x)sinx=2sinxcos2x+cos2xsinxsin3x=3sinxcos2xsin3x=3sinx(1sin2x)sin3x=3sinx4sin3x

#### Example B

Transform 3cos2x4sin2x\begin{align*}3 \cos 2x - 4 \sin 2x\end{align*} into the form Ccos(2xD)\begin{align*}C \cos(2x - D)\end{align*}

Solution: A=3\begin{align*}A = 3\end{align*} and B=4\begin{align*}B = -4\end{align*}, so C=32+(4)2=5\begin{align*}C = \sqrt{3^2 + (-4)^2} = 5 \end{align*}. Therefore cosD=35\begin{align*} \cos D = \frac{3}{5}\end{align*} and sinD=45\begin{align*} \sin D = - \frac{4}{5}\end{align*} which makes the reference angle is 53.1\begin{align*}-53.1^\circ\end{align*} or 0.927\begin{align*}-0.927\end{align*} radians. since cosine is positive and sine is negative, the angle must be a fourth quadrant angle. D\begin{align*}D\end{align*} must therefore be 306.9\begin{align*}306.9^ \circ\end{align*} or 5.36 radians. The final answer is 3cos2x4sin2x=5cos(2x5.36)\begin{align*}3 \cos 2x - 4 \sin 2x = 5 \cos (2x - 5.36)\end{align*}.

#### Example C

Solve sinx=2cosx\begin{align*}\sin x = 2 \cos x\end{align*} such that 0x2π\begin{align*}0 \le x \le 2\pi\end{align*} using a graphing calculator.

Solution: In y=\begin{align*}y =\end{align*}, graph y1=sinx\begin{align*}y1 = \sin x\end{align*} and y2=2cosx\begin{align*}y2 = 2 \cos x\end{align*}.

Next, use CALC to find the intersection points of the graphs.

### Guided Practice

1. Transform  5cosx5sinx\begin{align*}\ 5 \cos x - 5 \sin x\end{align*} to the form Ccos(xD)\begin{align*}C \cos(x - D)\end{align*}

2. Transform 15cos3x8sin3x\begin{align*} - 15 \cos 3x - 8 \sin 3x \end{align*} to the form Ccos(xD)\begin{align*}C \cos(x - D)\end{align*}

3. Derive a formula for tan4x\begin{align*}\tan 4x\end{align*}.

Solutions:

1. If 5cosx5sinx\begin{align*}5 \cos x - 5 \sin x\end{align*}, then A=5\begin{align*}A = 5\end{align*} and B=5\begin{align*}B = -5\end{align*}. By the Pythagorean Theorem, C=52\begin{align*}C = 5 \sqrt{2}\end{align*} and cosD=552=12=22\begin{align*}\cos D = \frac{5}{5 \sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\end{align*}. So, because B\begin{align*}B\end{align*} is negative, D\begin{align*}D\end{align*} is in Quadrant IV. Therefore, D=7π4\begin{align*}D = \frac{7 \pi}{4}\end{align*}. Our final answer is 52cos(x7π4)\begin{align*}5 \sqrt{2} \cos \left (x - \frac{7 \pi}{4} \right)\end{align*}.

2. If 15cos3x8sin3x\begin{align*}-15 \cos 3x - 8 \sin 3x\end{align*}, then A=15\begin{align*}A = -15\end{align*} and B=8\begin{align*}B = -8\end{align*}. By the Pythagorean Theorem, C=17\begin{align*}C = 17\end{align*}. Because A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} are both negative, D\begin{align*}D\end{align*} is in Quadrant III, which means D=cos1(1517)=0.49+π=3.63\begin{align*}D = \cos^{-1} \left (\frac{15}{17} \right) = 0.49+\pi = 3.63\end{align*} rad. Our final answer is 17cos3(x3.63)\begin{align*}17 \cos 3(x - 3.63)\end{align*}.

3.

tan4x=tan(2x+2x)=tan2x+tan2x1tan2xtan2x=2tan2x1tan22x=22tanx1tan2x1(2tanx1tan2x)2=4tanx1tan2x÷(1tan2x)24tan2x(1tan2x)2=4tanx1tan2x÷12tan2x+tan4x4tan2x(1tan2x)2=4tanx1tan2x(1tan2x)216tan2x+tan4x=4tanx4tan3x16tan2x+tan4x

### Concept Problem Solution

Using the triple angle formula we learned in this Concept for the sine function, we can break the angle down into three times a well known angle:

sin3x=3sinx4sin3x\begin{align*}\sin 3x = 3 \sin x - 4 \sin^3 x \end{align*}

we can solve this problem.

sin(3×45)=3sin454sin345=3224(22)3=322(4(2)2/38)=32222=22

### Explore More

Transform each expression to the form Ccos(xD)\begin{align*}C \cos(x - D)\end{align*}.

1. 3cosx2sinx\begin{align*}3\cos x - 2 \sin x\end{align*}
2. 2cosxsinx\begin{align*}2\cos x - \sin x\end{align*}
3. 4cosx+5sinx\begin{align*}-4\cos x + 5 \sin x\end{align*}
4. 7cosx6sinx\begin{align*}7\cos x - 6 \sin x\end{align*}
5. 11cosx+9sinx\begin{align*}11\cos x + 9 \sin x\end{align*}
6. 14cosx+2sinx\begin{align*}14\cos x + 2 \sin x\end{align*}
7. 2cosx4sinx\begin{align*}-2\cos x - 4 \sin x\end{align*}

Derive a formula for each expression.

1. sin4x\begin{align*}\sin 4x\end{align*}
2. cos6x\begin{align*}\cos 6x\end{align*}
3. cos4x\begin{align*}\cos 4x\end{align*}
4. csc2x\begin{align*}\csc 2x\end{align*}
5. cot2x\begin{align*}\cot 2x\end{align*}

Find all solutions to each equation in the interval [0,2π)\begin{align*}[0,2\pi)\end{align*}.

1. cosx+cos3x=0\begin{align*}\cos x +\cos 3x=0\end{align*}
2. sin2x=cos3x\begin{align*}\sin 2x =\cos 3x\end{align*}
3. cos2x+cos4x=0\begin{align*}\cos 2x +\cos 4x=0\end{align*}

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 3.15.

### Vocabulary Language: English

Linear Combination

Linear Combination

A linear combination is a set of terms that are added or subtracted from each other with a multiplicative constant in front of each term.
Triple Angle Identity

Triple Angle Identity

A triple angle identity (also referred to as a triple angle formula) relates a trigonometric function of three times an argument to a set of trigonometric functions, each containing the original argument. Examples include: the Triple Angle Formula for Sine $\text{sin} (3\theta) = 3 \text{sin} \theta - 4 \text{sin}^3 \theta$, the Triple Angle Formula for Cosine $\text{cos} (3 \theta) = -3 \text{ cos} \theta + 4 \text{ cos}^3 \theta$, and the Triple Angle Formula for Tangent $\text{tan} (3 \theta) = \frac{3 \text{ tan} \theta - \text{ tan}^3 \theta}{1 - 3 \text{ tan}^2 \theta}$.

## Date Created:

Sep 26, 2012

Feb 26, 2015
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