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3.9: Applications of Sum and Difference Formulas

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You are quite likely familiar with the values of trig functions for a variety of angles. Angles such as 30^\circ, 60^\circ, and 90^\circ are common. However, if you were asked to find the value of a trig function for a more rarely used angle, could you do so? Or what if you were asked to find the value of a trig function for a sum of angles? For example, if you were asked to find \sin \left( \frac{3\pi}{2} + \frac{\pi}{4} \right) could you?

Read on, and in this section, you'll get practice with simplifying trig functions of angles using the sum and difference formulas.

Watch This

James Sousa Example: Simplify a Trig Expression Using the Sum and Difference Identities

Guidance

Quite frequently one of the main obstacles to solving a problem in trigonometry is the inability to transform the problem into a form that makes it easier to solve. Sum and difference formulas can be very valuable in helping with this.

Here we'll get some extra practice putting the sum and difference formulas to good use. If you haven't gone through them yet, you might want to review the Concepts on the Sum and Difference Formulas for sine, cosine, and tangent.

Example A

Verify the identity \frac{\cos (x-y)}{\sin x \sin y} = \cot x \cot y + 1

\cot x \cot y + 1 & = \frac{\cos(x-y)}{\sin x \sin y} \\& = \frac{\cos x \cos y}{\sin x \sin y} + \frac{\sin x \sin y}{\sin x \sin y} && \text{Expand using the cosine difference formula}. \\& = \frac{\cos x \cos y}{\sin x \sin y} + 1 \\\cot x \cot y +1 & = \cot x \cot y +1 && \text{cotangent equals cosine over sine}

Example B

Solve 3 \sin (x-\pi)=3 in the interval [0, 2\pi).

Solution: First, get \sin(x - \pi) by itself, by dividing both sides by 3.

\frac{3 \sin (x- \pi)}{3} & = \frac{3}{3} \\\sin (x - \pi) & = 1

Now, expand the left side using the sine difference formula.

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

The \sin x = -1 when x is \frac{3\pi}{2}.

Example C

Find all the solutions for 2 \cos^2 \left (x+ \frac{\pi}{2} \right ) = 1 in the interval [0, 2\pi).

Solution: Get the \cos^2 \left (x+ \frac{\pi}{2} \right ) by itself and then take the square root.

2 \cos^2 \left (x+ \frac{\pi}{2} \right ) & = 1 \\\cos^2 \left (x+ \frac{\pi}{2} \right ) & = \frac{1}{2} \\\cos \left (x+ \frac{\pi}{2} \right ) & = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}

Now, use the cosine sum formula to expand and solve.

\cos x \cos \frac{\pi}{2} - \sin x \sin \frac{\pi}{2} & = \frac{\sqrt{2}}{2} \\\cos x(0) - \sin x (1) & = \frac{\sqrt{2}}{2} \\- \sin x & = \frac{\sqrt{2}}{2} \\\sin x & = - \frac{\sqrt{2}}{2}

The \sin x = - \frac{\sqrt{2}}{2} is in Quadrants III and IV, so x = \frac{5 \pi}{4} and \frac{7 \pi}{4}.

Vocabulary

Difference Formula: A difference formula is a formula to help simplify a trigonometric function of the difference of two angles, such as \sin(a-b).

Sum Formula: A sum formula is a formula to help simplify a trigonometric function of the sum of two angles, such as \sin(a+b).

Guided Practice

1. Find all solutions to 2 \cos^2 \left (x+ \frac{\pi}{2} \right ) = 1, when x is between [0, 2\pi).

2. Solve for all values of x between [0, 2\pi) for 2 \tan^2 \left (x+ \frac{\pi}{6} \right ) + 1 = 7.

3. Find all solutions to \sin \left (x+ \frac{\pi}{6} \right ) = \sin \left (x- \frac{\pi}{4} \right ), when x is between [0, 2\pi).

Solutions:

1. To find all the solutions, between [0, 2\pi), we need to expand using the sum formula and isolate the \cos x.

2 \cos^2 \left (x + \frac{\pi}{2} \right ) & = 1 \\\cos^2 \left (x + \frac{\pi}{2} \right ) & = \frac{1}{2} \\\cos \left (x + \frac{\pi}{2} \right ) & = \pm\sqrt{\frac{1}{2}} = \pm\frac{\sqrt{2}}{2} \\\cos x \cos \frac{\pi}{2} - \sin x \sin \frac{\pi}{2} & = \pm\frac{\sqrt{2}}{2} \\\cos x \cdot 0 - \sin x \cdot 1 & = \pm\frac{\sqrt{2}}{2} \\- \sin x & = \pm\frac{\sqrt{2}}{2} \\\sin x & = \pm\frac{\sqrt{2}}{2}

This is true when x = \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, or \frac{7 \pi}{4}

2. First, solve for \tan (x+ \frac{\pi}{6}).

2 \tan^2 \left (x+ \frac{\pi}{6} \right ) +1 & = 7 \\2 \tan^2 \left (x+ \frac{\pi}{6} \right ) & = 6 \\\tan^2 \left (x+ \frac{\pi}{6} \right ) & = 3 \\\tan \left (x+ \frac{\pi}{6} \right ) & = \pm\sqrt{3}

Now, use the tangent sum formula to expand for when \tan (x+ \frac{\pi}{6}) = \sqrt{3}.

\frac{\tan x + \tan \frac{\pi}{6}}{1 - \tan x \tan \frac{\pi}{6}} & = \sqrt{3} \\\tan x + \tan \frac{\pi}{6} & = \sqrt{3} \left (1 - \tan x \tan \frac{\pi}{6} \right ) \\\tan x + \frac{\sqrt{3}}{3} & = \sqrt{3} - \sqrt{3} \tan x \cdot \frac{\sqrt{3}}{3} \\\tan x + \frac{\sqrt{3}}{3} & = \sqrt{3} - \tan x \\2 \tan x & = \frac{2 \sqrt{3}}{3} \\\tan x & = \frac{\sqrt{3}}{3}

This is true when x = \frac{\pi}{6} or \frac{7 \pi}{6}.

If the tangent sum formula to expand for when \tan (x+ \frac{\pi}{6}) = -\sqrt{3}, we get no solution as shown.

\frac{\tan x + \tan \frac{\pi}{6}}{1 - \tan x \tan \frac{\pi}{6}} & = -\sqrt{3} \\\tan x + \tan \frac{\pi}{6} & = -\sqrt{3} \left (1 - \tan x \tan \frac{\pi}{6} \right ) \\\tan x + \frac{\sqrt{3}}{3} & = -\sqrt{3} + \sqrt{3} \tan x \cdot \frac{\sqrt{3}}{3} \\\tan x + \frac{\sqrt{3}}{3} & = -\sqrt{3} + \tan x \\\frac{\sqrt{3}}{3} & = -\sqrt{3}\\

Therefore, the tangent sum formula cannot be used in this case. However, since we know that \tan(x+\frac{\pi}{6}) = -\sqrt{3} when x+\frac{\pi}{6} = \frac{5\pi}{6} or \frac{11\pi}{6}, we can solve for x as follows.

x+\frac{\pi}{6}=\frac{5\pi}{6} \\x = \frac{4\pi}{6} \\x = \frac{2\pi}{3} \\ \\x+\frac{\pi}{6}=\frac{11\pi}{6} \\x = \frac{10\pi}{6} \\x = \frac{5\pi}{3}

Therefore, all of the solutions are x=\frac{\pi}{6}, \frac{2\pi}{3}, \frac{7 \pi}{6}, \frac{5\pi}{3}

3. To solve, expand each side:

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

Set the two sides equal to each other:

\frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x & = \frac{\sqrt{2}}{2} \sin x - \frac{\sqrt{2}}{2} \cos x \\\sqrt{3} \sin x + \cos x & = \sqrt{2} \sin x - \sqrt{2} \cos x \\\sqrt{3} \sin x - \sqrt{2} \sin x & = - \cos x - \sqrt{2} \cos x \\\sin x \left (\sqrt{3} - \sqrt{2} \right ) & = \cos x \left (-1 - \sqrt{2} \right ) \\\frac{\sin x}{\cos x} & = \frac{-1- \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\\tan x & = \frac{-1- \sqrt{2}}{\sqrt{3} - \sqrt{2}} \cdot \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} \\& = \frac{- \sqrt{3} - \sqrt{2} + \sqrt{6} - 2}{3-2} \\& = -2 + \sqrt{6} - \sqrt{3} - \sqrt{2}

As a decimal, this is -2.69677, so \tan^{-1}(-2.69677) = x, x = 290.35^\circ and 110.35^\circ.

Concept Problem Solution

To find \sin \left( \frac{3\pi}{2} + \frac{\pi}{4} \right), use the sine sum formula:

\sin (a + b) = \sin(a)\cos(b) + \cos(a)\sin(b)\\\sin \left( \frac{3\pi}{2} + \frac{\pi}{4} \right) = \sin \left( \frac{3\pi}{2} \right) \times \cos \left( \frac{\pi}{4} \right) + \cos \left( \frac{3\pi}{2} \right) \times \sin \left( \frac{\pi}{4} \right)\\= (-1)\left( \frac{\sqrt{2}}{2} \right) + (0)\left( \frac{\sqrt{2}}{2} \right)\\= -\frac{\sqrt{2}}{2}\\

Practice

Prove each identity.

  1. \cos(3x)+\cos(x)=2\cos(2x)\cos(x)
  2. \cos(3x)=\cos^3(x)-3\sin^2(x)\cos(x)
  3. \sin(3x)=3\cos^2(x)\sin(x)-\sin^3(x)
  4. \sin(4x)+\sin(2x)=2\sin(3x)\cos(x)
  5. \tan(5x)\tan(3x)=\frac{\tan^2(4x)-\tan^2(x)}{1-\tan^2(4x)\tan^2(x)}
  6. \cos((\frac{\pi}{2}-x)-y)=\sin(x+y)

Use sum and difference formulas to help you graph each function.

  1. y=\cos(3)\cos(x)+\sin(3)\sin(x)
  2. y=\cos(x)\cos(\frac{\pi}{2})+\sin(x)\sin(\frac{\pi}{2})
  3. y=\sin(x)\cos(\frac{\pi}{2})+\cos(x)\sin(\frac{\pi}{2})
  4. y=\sin(x)\cos(\frac{3\pi}{2})-\cos(3)\sin(\frac{\pi}{2})
  5. y=\cos(4x)\cos(2x)-\sin(4x)\sin(2x)
  6. y=\cos(x)\cos(x)-\sin(x)\sin(x)

Solve each equation on the interval [0,2\pi).

  1. 2\sin(x-\frac{\pi}{2})=1
  2. 4\cos(x-\pi)=4
  3. 2\sin(x-\pi)=\sqrt{2}

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Date Created:

Sep 26, 2012

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Jan 23, 2014
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