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Applications of Sum and Difference Formulas

Sine, cosine, and tangent sum and difference formulas.

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Applications of Sum and Difference Formulas

You are quite likely familiar with the values of trig functions for a variety of angles. Angles such as \begin{align*}30^\circ\end{align*}, \begin{align*}60^\circ\end{align*}, and \begin{align*}90^\circ\end{align*} 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 \begin{align*}\sin \left( \frac{3\pi}{2} + \frac{\pi}{4} \right)\end{align*} could you?

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

Sum and Difference Formulas

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 sections on the Sum and Difference Formulas for sine, cosine, and tangent.

Solve using the Sum Formula

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

\begin{align*}\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}\end{align*}

Solve using the Difference Formula

Solve \begin{align*}3 \sin (x-\pi)=3\end{align*} in the interval \begin{align*}[0, 2\pi)\end{align*}.

First, get \begin{align*}\sin(x - \pi)\end{align*} by itself, by dividing both sides by \begin{align*}3\end{align*}.

\begin{align*}\frac{3 \sin (x- \pi)}{3} & = \frac{3}{3} \\ \sin (x - \pi) & = 1\end{align*}

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

\begin{align*}\sin x \cos \pi - \cos x \sin \pi & = 1 \\ \sin x (-1) - \cos x (0) & = 1 \\ - \sin x & = 1 \\ \sin x & = - 1 \end{align*}

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

Solve using the Sum Formula

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

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

\begin{align*}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}\end{align*}

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

\begin{align*}\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} \end{align*}

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

Examples

Example 1

Earlier, you were asked to find \begin{align*}\sin \left( \frac{3\pi}{2} + \frac{\pi}{4} \right)\end{align*}, use the sine sum formula:

\begin{align*} \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}\\ \end{align*}

Example 2

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

To find all the solutions, between \begin{align*}[0, 2\pi)\end{align*}, we need to expand using the sum formula and isolate the \begin{align*}\cos x\end{align*}.

\begin{align*}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} \end{align*}

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

Example 3

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

First, solve for \begin{align*}\tan (x+ \frac{\pi}{6})\end{align*}.

\begin{align*}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}\end{align*}

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

\begin{align*}\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}\end{align*}

This is true when \begin{align*}x = \frac{\pi}{6}\end{align*} or \begin{align*}\frac{7 \pi}{6}\end{align*}.

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

\begin{align*}\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}\\\end{align*}

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

\begin{align*}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}\end{align*}

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

Example 4

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

To solve, expand each side:

\begin{align*}\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\end{align*}

Set the two sides equal to each other:

\begin{align*}\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}\end{align*}

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

Review

Prove each identity.

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

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

Solve each equation on the interval \begin{align*}[0,2\pi)\end{align*}.

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

To see the Review answers, open this PDF file and look for section 3.9.

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Color Highlighted Text Notes

Vocabulary Language: English

TermDefinition
Difference Formula Trigonometric function difference formulas exist for each of the primary trigonometric functions. For example, the cosine difference formula is $cos(A - B) = cosA cosB + sinA sinB$.
Sum Formula A sum formula is a formula to help simplify a trigonometric function of the sum of two angles, such as $\sin(a+b)$.