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tags:
Fundamental Trigonometric Identity Trigonometry Trigonometry Series

Solving Trig Equations Using Identities Exercise 1

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: Use Even/Odd Identities and Cofunction identities to find all x in \([0, 2\pi]\) satisfying  \(-\frac{1}{2}=\sin\left( -\frac{\pi}{2}+x \right).\)

Solution: \(x=\dfrac{\pi}{3}, \dfrac{5\pi}{3}\)

Solution Method: This is a case where we’ll use multiple trig identities to solve a trigonometric equation. In this case, cofunction and even/odd identities. Cofunction identities are a special case of the sum/difference identites. Even/Odd identities derive from cosine being an even function and sine and tangent being odd functions. You can easily verify that these functions and even and odd respectively by looking at their graphs. Cosine has y-axis symmetry, so it’s even. Sine and tangent have origin symmetry, and are odd.

Even/Odd Identities
\begin{align*}    \cos(-x) &= \cos(x) \\
    \sec(-x) &= \sec(x) \\
    \sin(-x)&=-\sin(x) \\
    \csc(-x)&=-\csc(x) \\
    \tan(-x)&=-\tan(x) \\
    \cot(-x)&=-\cot(x)
    \end{align*}

Comparing my expression to the identities, I notice that \(\sin\left( -\frac{\pi}{2}+x \right)\) is very close to the cofunction identity \(\cos{x}= \sin\left( \frac{\pi}{2}-x \right)\). I just need to distribute out a negative to make them match. \[\begin{aligned} -\frac{1}{2}&=\sin\left( -\frac{\pi}{2}+x \right) \\ -\frac{1}{2}&=\sin\left( -(\frac{\pi}{2}-x) \right) \end{aligned}\] Then by the odd identity of sine, that negative can come out of the argument. \[\begin{aligned} -\frac{1}{2}&=-\sin\left( \frac{\pi}{2}-x\right) \end{aligned}\] So now I can apply the cofunction identity, \[\begin{aligned} -\frac{1}{2} &= -\cos{x} \\ \frac{1}{2} &= \cos{x} \end{aligned}\] So now we’re looking for values of x in the interval \([0,2\pi]\) that satisfy this. That interval is just the unit circle so I need to remember what special angles produce cosine of one-half on the unit circle. First, this is a positive half so I can only have angle values in Q1 and Q4. Cosine corresponds to the x-values on the unit circle and I know that my x-coordinate is one-half for reference angle values of \(\frac{\pi}{3}\). In Q1, that is just \(\frac{\pi}{3}\), in Q4 it is \(\frac{5\pi}{3}\). So the values of x in \([0, 2\pi]\) satisfying the equation are \(\frac{\pi}{3}\) and \(\frac{5\pi}{3}\).

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