Doing stuff with triangles!

**Exercise.** Re-write the expression below with just one sine:

\cos^2(x)\sin^3(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:

- \sin\left(\frac{\pi}{6}\right)
- \cos\left(\frac{\pi}{4}\right)
- \sec\left(\frac{\pi}{3}\right)

*Proof.*

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