1. Problem 4.1: Prove without using a calculator that $\sin 44^\circ + \sin 16^\circ = \sin 76^\circ$. Use the sum-to-product identity $\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$. Take $A=44^\circ$ and $B=16^\circ$. Then $\sin 44^\circ + \sin 16^\circ = 2\sin\frac{44^\circ+16^\circ}{2}\cos\frac{44^\circ-16^\circ}{2}$. Simplify the angles: $\frac{44^\circ+16^\circ}{2}=30^\circ$ and $\frac{44^\circ-16^\circ}{2}=14^\circ$. So the sum equals $2\sin 30^\circ \cos 14^\circ$. Since $\sin 30^\circ = \frac{1}{2}$ this equals $\cos 14^\circ$. Finally $\cos 14^\circ = \sin 76^\circ$ because $\cos \theta = \sin(90^\circ-\theta)$. Therefore $\sin 44^\circ + \sin 16^\circ = \sin 76^\circ$. 2. Problem 4.2: Simplify without using a calculator $\displaystyle \frac{2\tan 15^\circ}{1-\tan^2 15^\circ}$. Use the double-angle identity $\tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta}$. With $\theta=15^\circ$ the expression equals $\tan 30^\circ$. And $\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$. So $\displaystyle \frac{2\tan 15^\circ}{1-\tan^2 15^\circ}=\frac{\sqrt{3}}{3}$. 3. Problem 4.3: Find the value of $\cos 210^\circ$. Write $210^\circ=180^\circ+30^\circ$ and use $\cos(180^\circ+\theta)=-\cos\theta$. Thus $\cos 210^\circ=-\cos 30^\circ$. Since $\cos 30^\circ=\frac{\sqrt{3}}{2}$ we get $\cos 210^\circ=-\frac{\sqrt{3}}{2}$. 4. Problem 4.4: Solve $2\cos x-1=0$. Solve for $\cos x$ to get $\cos x=\frac{1}{2}$. In degrees the solutions are $x=60^\circ+360^\circ n$ and $x=300^\circ+360^\circ n$ for any integer $n$. In radians the solutions are $x=\pm\frac{\pi}{3}+2\pi n$ for any integer $n$. 5. Problem 4.5: Find the general solution of $\displaystyle \frac{\tan x}{\sin x}+\frac{3}{\sin x}-2\tan x=6$. Assume $\sin x\neq 0$ because the equation has terms with $\displaystyle\frac{1}{\sin x}$. Write everything in terms of $\sin x$ and $\cos x$ using $\tan x=\frac{\sin x}{\cos x}$, so $\frac{\tan x}{\sin x}=\frac{1}{\cos x}$. The equation becomes $\frac{1}{\cos x}+\frac{3}{\sin x}-2\frac{\sin x}{\cos x}=6$. Multiply both sides by $\sin x\cos x$ to clear denominators, giving $\sin x+3\cos x-2\sin^2 x=6\sin x\cos x$. Rearrange to $0=(6\sin x-3)\cos x+2\sin^2 x-\sin x$. Factor $2\sin^2 x-\sin x=-\sin x(\,2\sin x-1\,)$ and note $6\sin x-3=3(2\sin x-1)$, so move terms to obtain $3(2\sin x-1)\cos x-\sin x(2\sin x-1)=0$. Factor out $(2\sin x-1)$ to get $(2\sin x-1)\bigl(3\cos x-\sin x\bigr)=0$. Thus either $2\sin x-1=0$ or $3\cos x-\sin x=0$. Case 1: $2\sin x-1=0$ gives $\sin x=\tfrac{1}{2}$, so $x=\frac{\pi}{6}+2\pi n$ or $x=\frac{5\pi}{6}+2\pi n$ for integer $n$. Case 2: $3\cos x-\sin x=0$ gives $\tan x=3\cos x/\cos x?\,$ better rearrange to $\sin x=3\cos x$, so $\tan x=3$ with a sign check: from $3\cos x-\sin x=0$ we get $\sin x=3\cos x$ so $\tan x=3$. Therefore $\tan x=3$ so $x=\arctan 3+\pi n$ for integer $n$. Combine with the domain restriction $\sin x\neq 0$ which is satisfied by the listed solutions, so the general solutions are $x=\frac{\pi}{6}+2\pi n$, $x=\frac{5\pi}{6}+2\pi n$, or $x=\arctan 3+\pi n$ for any integer $n$. 6. Problem 4.6: A right triangle has angle at Q equal to $38.7^\circ$, the adjacent side to that angle is $100\text{ m}$, and the opposite side is the building height $h$. Find $h$ to the nearest metre. Use $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$ so $\tan 38.7^\circ=\frac{h}{100}$. Hence $h=100\tan 38.7^\circ$. Compute $\tan 38.7^\circ\approx 0.8012$ so $h\approx 100\times 0.8012\approx 80.12$ metres. Rounded to the nearest metre $h\approx 80$ metres. Final answers summary: 4.1 $\sin 44^\circ+\sin 16^\circ=\sin 76^\circ$. 4.2 $\displaystyle \frac{2\tan 15^\circ}{1-\tan^2 15^\circ}=\frac{\sqrt{3}}{3}$. 4.3 $\cos 210^\circ=-\frac{\sqrt{3}}{2}$. 4.4 $x=60^\circ+360^\circ n$ or $x=300^\circ+360^\circ n$ for integer $n$. 4.5 $x=\frac{\pi}{6}+2\pi n$, $x=\frac{5\pi}{6}+2\pi n$, or $x=\arctan 3+\pi n$ for integer $n$. 4.6 $h\approx 80$ metres.