Solving Trig Equations Using Identities Exercise 6

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).