# Recitation 2

Doing stuff with triangles!

Exercise. Re-write the expression below with just one sine:
\cos^2(x)\sin^3(x)

\cos^2(x)\sin^3(x)=\cos^2(x)\sin^2(x)\sin(x)=\cos^2(x)(1-\cos^2(x))\sin(x)

Exercise. If \tan(\theta)=\frac{x}{4}, determine an expression for \sin(\theta).

Proof. Since \tan(\theta)=\frac{\sin(x)}{\cos(x)}, we have
4\sin(x)=x\cos(x)
Square both sides, we have
16\sin^2(x)=x^2\cos^2(x)=x^2(1-\sin^2(x)),
which gives us
(16+x^2)\sin^2(x)=x^2.
Hence, \sin(x)=\frac{\pm x}{\sqrt{16+x^2}}.

Exercise. If \sin(\theta)=\frac{3}{5}, what is the value of \sin(2\theta)?

Proof. To calculate \sin(2\theta), use the formula \sin(2\theta)=2\sin(x)\cos(x). We have \sin(2\theta)=2\sin(x)(\pm\sqrt{1-\sin^2(x)})=\pm 2\times\frac{3}{5}\sqrt{1-(\frac{3}{5})^2}=\pm\frac{24}{25}.

Exercise. Evaluate each of the following:

1. \sin\left(\frac{\pi}{6}\right)
2. \cos\left(\frac{\pi}{4}\right)
3. \sec\left(\frac{\pi}{3}\right)

Proof.

1. \sin\left(\frac{\pi}{6}\right)=\frac{1}{2}
2. \cos\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}
3. \sec\left(\frac{\pi}{3}\right)=\frac{1}{\cos\left(\frac{\pi}{3}\right)}=\frac{1}{\frac{1}{2}}=2

Exercise. Determine an exact value of \sin\left(\frac{7\pi}{12}\right)

Proof. \begin{aligned}\sin\left(\frac{7\pi}{12}\right)&=\sin\left(\frac{\pi}{4}+\frac{\pi}{3}\right)\\&=\sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{3}\right)+\cos\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{3}\right)\\&=\frac{\sqrt{2}}{2}\frac{1}{2}+\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}\\&=\frac{\sqrt{2}+\sqrt{6}}{4}\end{aligned}.