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tags:
Cosine Fundamental Trigonometric Identity Trigonometry Trigonometry Series

Solving Trig Equations Using Identities Exercise 6

Author: Hannah Solomon

The following problem is solved in this video. It is recommended that you try to solve the problem before watching the video. You can click "Reveal Answer" to see the answer to the problem.

Problem: Find all solutions of \(\cos\left(x+\frac{\pi}{4}\right)-\cos\left(x-\frac{\pi}{4}\right)=1.\)

Solution: \(x=\dfrac{5\pi}{4} + 2\pi n, n\in\mathbb{Z}\) and \(x=\dfrac{7\pi}{4} +2\pi n, n\in \mathbb{Z}\)

Solution Method: To solve this equation, we need to utilize the sum and difference identities for cosine.

Sum and Difference Identities \[\begin{align*} \cos{(A+B)}&=\cos{A}\cos{B}-\sin{A}\sin{B} \\ \cos{(A-B)}&=\cos{A}\cos{B}+\sin{A}\sin{B} \\ \end{align*}\]

With these identities, we can rewrite the equation as \[\begin{aligned} \cos\left(x+\frac{\pi}{4}\right)&-\cos\left(x-\frac{\pi}{4}\right)&=1 \\ \left( \cos{x}\cos{\frac{\pi}{4}} -\sin{x}\sin{\frac{\pi}{4}}\right) &- \left( \cos{x}\cos{\frac{\pi}{4}} +\sin{x}\sin{\frac{\pi}{4}}\right) &= 1 \\ \cancel{\cos{x}\cos{\frac{\pi}{4}}} -\sin{x}\sin{\frac{\pi}{4}} &- \cancel{\cos{x}\cos{\frac{\pi}{4}}} -\sin{x}\sin{\frac{\pi}{4}} &= 1 \\ &-2\sin{x}\sin{\frac{\pi}{4}} &= 1 \end{aligned}\] Since \(\frac{\pi}{4}\) is a special angle value I know that \(\sin{\frac{\pi}{4}}=\frac{\sqrt{2}}{2}\). So I substitute that in, \[\begin{aligned} -2\sin{x}\sin{\frac{\pi}{4}} &= 1 \\ -2\sin{x}\left(\frac{\sqrt{2}}{2}\right) &= 1 \\ -\sqrt{2}\sin{x} &= 1 \\ \sin{x} &= \frac{1}{-\sqrt{2}} \\ \sin{x} &= -\frac{\sqrt{2}}{2} \end{aligned}\] The reference angle producing a sine of \(\frac{\sqrt{2}}{2}\) is \(\frac{\pi}{4}\). Then (ASTC) sine is negative in Q3 and Q4. So the values between 0 and \(2\pi\) satisfying the equation are \(\frac{5\pi}{4}\) in Q3 and \(\frac{7\pi}{4}\) in Q4. To find all x satisfying the equation we have to consider every full rotation from \(\frac{5\pi}{4}\) and \(\frac{7\pi}{4}\) is also a solution. So we can write the full solution as \(x=\frac{5\pi}{4} + 2\pi n, n\in \mathbb{Z}\) and \(x=\frac{7\pi}{4} + 2\pi n, n\in \mathbb{Z}\). The \(2\pi\) represents a full rotation back to where we started and \(n\) is an integer (...-2,-1,0,1,2...) which represents how many rotations we’ve made (forwards or backwards).

 

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