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3.15: 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:

\sin 135^\circ

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.

A \cos x + B \sin x = C \cos(x - D) , where C = \sqrt{A^2 + B^2} , \cos D = \frac{A}{C} and \sin D = \frac{B}{C} .

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 \pi .

Example A

Find the formula for \sin 3x

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

\sin 3x &= \sin(2x + x) \\&= \sin(2x) \cos x + \cos(2x)\sin x \\&= (2 \sin x \cos x) \cos x + (\cos^2 x - \sin^2 x) \sin x \\&= 2 \sin x \cos^2 x + \cos^2 x \sin x - \sin^3 x \\&= 3 \sin x \cos^2 x - \sin^3 x \\&= 3 \sin x(1 - \sin^2 x) - \sin^3 x \\&= 3 \sin x - 4 \sin^3 x

Example B

Transform 3 \cos 2x - 4 \sin 2x into the form C \cos(2x - D)

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

Example C

Solve \sin x = 2 \cos x such that 0 \le x \le 2\pi using a graphing calculator.

Solution: In y = , graph y1 = \sin x and y2 = 2 \cos x .

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

Vocabulary

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: A triple angle identity relates to a trigonometric function of three times an argument to a set of trigonometric functions, each containing the original argument.

Guided Practice

1. Transform \  5 \cos x - 5 \sin x to the form C \cos(x - D)

2. Transform  - 15 \cos 3x - 8 \sin 3x to the form C \cos(x - D)

3. Derive a formula for \tan 4x .

Solutions:

1. If 5 \cos x - 5 \sin x , then A = 5 and B = -5 . By the Pythagorean Theorem, C = 5 \sqrt{2} and \cos D = \frac{5}{5 \sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} . So, because B is negative, D is in Quadrant IV. Therefore, D = \frac{7 \pi}{4} . Our final answer is 5 \sqrt{2} \cos \left (x - \frac{7 \pi}{4} \right) .

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

3.

\tan  4x &= \tan (2x + 2x)\\&= \frac {\tan 2x + \tan 2x}{1 - \tan 2x \tan 2x}\\&= \frac {2 \tan 2x}{1 - \tan^2 2x}\\&= \frac{2 \cdot \frac{2 \tan x}{1 - \tan^2x}}{1 - \left ( \frac{2 \tan x}{1 - \tan^2x} \right )^2}\\&= \frac{4 \tan x}{1 - \tan^2 x} \div \frac{(1 - \tan ^2 x)^2 - 4 \tan^2 x}{(1 - \tan^2 x)^2}\\&= \frac{4 \tan x}{1 - \tan^2 x} \div \frac{1 - 2 \tan^2 x + \tan^4 x- 4 \tan^2 x}{(1 -  \tan^2 x)^2}\\&= \frac{4 \tan x}{1 - \tan^2 x} \cdot \frac{(1 - \tan^2 x)^2}{1 - 6 \tan^2 x   + \tan^4 x}\\&= \frac{4 \tan x - 4 \tan^3 x}{1 - 6 \tan^2 x + \tan^4 x}

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:

\sin 3x = 3 \sin x - 4 \sin^3 x

we can solve this problem.

\sin(3 \times 45^\circ) = 3 \sin 45^\circ - 4 \sin^3 45^\circ\\=3 \frac{\sqrt{2}}{2} - 4 \left(\frac{\sqrt{2}}{2} \right)^3\\=3 \frac{\sqrt{2}}{2} - \left( \frac{4(2)^{2/3}}{8} \right)\\=\frac{3\sqrt{2} - 2\sqrt{2}}{2}\\=\frac{\sqrt{2}}{2}\\

Explore More

Transform each expression to the form C \cos(x - D) .

  1. 3\cos x - 2 \sin x
  2. 2\cos x -  \sin x
  3. -4\cos x + 5 \sin x
  4. 7\cos x - 6 \sin x
  5. 11\cos x + 9 \sin x
  6. 14\cos x + 2 \sin x
  7. -2\cos x - 4 \sin x

Derive a formula for each expression.

  1. \sin 4x
  2. \cos 6x
  3. \cos 4x
  4. \csc 2x
  5. \cot 2x

Find all solutions to each equation in the interval [0,2\pi) .

  1. \cos x +\cos 3x=0
  2. \sin 2x =\cos 3x
  3. \cos 2x +\cos 4x=0

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Sep 26, 2012

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Oct 28, 2014
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