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tags:
Fundamental Trigonometric Identity Sine Trigonometry Trigonometry Series

Using Identities to Find Exact Trig Values Exercise 2

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: Given \(\sin{\theta}=\dfrac{4}{5},\) and \(\dfrac{\pi}{2}<\theta<\dfrac{3\pi}{2},\) find \(\sin\left(\dfrac{\pi}{3}+\theta\right).\)

Solution: \(-\dfrac{3\sqrt{3}}{10} + \dfrac{2}{5}\)
 

Solution Method: A lot of things to unpack here in this question. First thing we see is we are given the sine ratio of some angle \(\theta\). Then we see that angle \(\theta\) is between \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\). Finally, the question asks us to find \(\sin(\frac{\pi}{3}+\theta)\). So actually, we don’t need to calculate what \(\theta\) is to solve this question.

We’re asked to find sine of a sum of angles, so we’re going to use the sum identity for sine. \[\sin{(A+B)}= \sin{A}\cos{B}+\cos{A}\sin{B}\] I don’t know what \(\theta\) is but, I have the sine ratio of \(\theta\) and a domain restriction on it. Using those two facts, I can find the other trig ratios of \(\theta\), including \(\cos{\theta}\), which is what I need to use this formula.

To find \(\cos{\theta}\) given \(\sin{\theta}\), I construct a reference triangle. First, I need to know which quadrant this triangle is going in. \(\frac{\pi}{2}<\theta<\frac{3\pi}{2}\) restricts us to Q2 and Q3. Now, sine is positive in Q2, but negative in Q3, so since \(\sin{\theta}\) is equal to positive \(\frac{4}{5}\), we know we have to be in Q2.

Sine is opposite over hypotenuse so this side opposite \(\theta\) is length 4 and the hypotenuse is length 5. To find the length of this adjacent side we use pythagorean theorem. \[\begin{aligned} x^2+4^2&=5^2 \\ x^2 + 16 &= 25 \\ x^2 &= 9 \\ x &= 3 \end{aligned}\] (Of course, we also could have recognized this is a 3-4-5 triangle, and gotten there a little faster).

So the length of the adjacent side is 3, BUT we have to notice what quadrant we are in for the sign. We are in Q2, so this side is actually -3. Then, \(\cos{\theta}\) is adjacent over hypotenuse so \(\cos{\theta}=\frac{-3}{5}\)

Finally, we can rewrite \(\sin(\frac{\pi}{3}+\theta)\) using the identity as \[\sin\left(\frac{\pi}{3}+\theta\right) = \sin{\frac{\pi}{3}}\cos{\theta}+\cos{\frac{\pi}{3}}\sin{\theta}\] Then we substitute the values of \(\sin{\theta}\) and \(\cos{\theta}\), and \(\frac{\pi}{3}\) is a special angle so I know its value from the unit circle. \[\begin{aligned} \sin\left(\frac{\pi}{3}+\theta\right) &= \left(\frac{\sqrt{3}}{2}\right) \left(-\frac{3}{5}\right) + \left(\frac{1}{2}\right) \left(\frac{4}{5}\right) \\ &= -\frac{3\sqrt{3}}{10} + \frac{2}{5} \\ \end{aligned}\]

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