Given $\sin x=3/4$ and $\pi/2<x<\pi$, find $\sin(x+\pi/6)$.
I have been trying to calculate this question. My current working out is as follows:
$\sin\theta = 3/4$
$\pi/2 < \theta < \pi$
$$\begin{align}\sin(\theta+\pi/6)&=\sin\theta.\cos\pi/6 +\cos\theta.\sin\pi/6\\&=3/4.\sqrt{3}/2+\cos\theta.1/2\\&=(3.\sqrt{3}+4\cos\theta)/8\end{align}$$
The answer is: $(3.\sqrt{3}-\sqrt{7})/8$
However I do not know how to get to it.
Thanks in advance.
$\endgroup$1 Answer
$\begingroup$You are on the right track.
We have $$\sin\left(x+\frac\pi6\right)=\sin x\cos\frac\pi6+\sin\frac\pi6\cos x=\frac34\cdot\frac{\sqrt3}2+\frac12\cos x$$ and since $$\sin x=\frac34=\frac3{\sqrt{3^2+(\sqrt7)^2}}$$ we know that $x>\pi/2$ so it means that by Pythagoras, $$\cos x=\frac{-\sqrt7}{\sqrt{3^2+(\sqrt7)^2}}=-\frac{\sqrt7}4$$ so the answer is $$\frac{3\sqrt3}8-\frac{\sqrt7}8$$
$\endgroup$ 5
