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}.