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Fundamental Trigonometric Identity Trigonometry Trigonometry Series

Solving Trig Equations Using Identities Exercise 7

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 \(\cot{x} = \sin\left(\dfrac{\pi}{2}-x\right).\)

Solution: \(x = \frac{\pi}{2}+ \pi n, n \in \mathbb{Z}\)

Solution Method: I can choose to start with either the LHS or the RHS for this equation, because I’m going to need to apply an identity to both.

For the RHS, I’ll apply the ratio identity for cotangent, \(\cot{x} = \frac{\cos{x}}{\sin{x}}\). \[\begin{aligned} \sin\left(\frac{\pi}{2}-x\right) &= \cot{x} \\ \sin\left(\frac{\pi}{2}-x\right) &= \frac{\cos{x}}{\sin{x}} \end{aligned}\]

So I now have terms of sine and cosine, but the sine terms have different arguments, \((\frac{\pi}{2}-x) \text{ and } x\). So I’ll use a cofunction identity to get \(\sin(\frac{\pi}{2}-x)\) in terms of a trig function with argument \(x\).

Cofunction Identities for Sine/Cosine

\[\begin{aligned} \cos{x}&=\sin{\left(\frac{\pi}{2}-x\right)}\\ \sin{x}&=\cos{\left(\frac{\pi}{2}-x\right)} \end{aligned}\]

So substituting \(\sin{\left(\frac{\pi}{2}-x\right)}=\cos{x}\), \[\begin{aligned} \sin\left(\frac{\pi}{2}-x\right) &= \frac{\cos{x}}{\sin{x}} \\ \cos{x} &= \frac{\cos{x}}{\sin{x}} \\ \sin{x}\cos{x} &= \cos{x} \end{aligned}\] Now it may be tempting to divide out a \(\cos{x}\) here but if we do that, we’ll actually lose a solution to the equation, the case where \(\cos{x}=0\). So instead, we’ll get everything on one side, equal to zero, and factor out a \(\cos{x}\). \[\begin{aligned} 0 &= \cos{x}-\sin{x}\cos{x} \\ 0 &= \cos{x}(1-\sin{x}) \end{aligned}\] So now I’ll set both these factors equal to 0 and solve for x.

\[\begin{aligned} \cos{x}=0 \end{aligned}\] The quadrantal angles satisfying cosine of 0 on the unit circle are \(\frac{\pi}{2}, \frac{3\pi}{2}\). But we were asked for all solutions. Trig functions are periodic, they repeat themselves every rotation. So every full rotation from \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) is another value satisfying the equation. Since every rotation is \(2\pi\) in measure, we can represent all the values satisfying the equation as \(x=\frac{\pi}{2} + 2\pi n, n\in \mathbb{Z}\) and \(x=\frac{3\pi}{2} + 2\pi n, n\in \mathbb{Z}\). 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).

\[\begin{aligned} 1-\sin{x} = 0 \\ 1 = \sin{x} \end{aligned}\] 1 is another quadrantal value and the angle satisfying \(\sin{x}=1\) on the unit circle is \(\frac{\pi}{2}\). This value is already included as a solution of \(\cos{x}=0\), so we don’t need to do anything else.

Now, we could leave these two equations as they are, but I want to point something out. Notice that \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) are a \(\pi\) apart. So \(\frac{\pi}{2} + \pi = \frac{3\pi}{2}, + \pi = \frac{5\pi}{2}...\) So we can actually represent all our solutions in one equation: \(x = \frac{\pi}{2} + \pi n, n \in \mathbb{Z}.\)

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