第13章 13.14

(1)

$\sin(x+y)\cdot\sin(x-y)=(\sin x\cdot\cos y+\cos x\cdot\sin y)(\sin x\cdot\cos y-\cos x\cdot\sin y)$

$=\sin^2x\cdot\cos^2y-\cos^2x\cdot\sin^2y=\sin^2x(1-\sin^2y)-(1-\sin^2x)\sin^2y=\sin^2x-\sin^2y$

(2)

$\cos(x+y)\cdot\cos(x-y)=(\cos x\cdot\cos y-\sin x\cdot\sin y)(\cos x\cdot\cos y+\sin x\cdot\sin y)$

$=\cos^2x\cdot\cos^2y-\sin^2x\cdot\sin^2y=\cos^2x(1-\sin^2y)-(1-\cos^2x)\sin^2y=\cos^2x-\sin^2y$

(3)

$\sin(x+y)+\cos(x-y)=\sin x\cdot\cos y+\cos x\cdot\sin y+\cos x\cdot\cos y+\sin x\cdot\sin y$

$=(\cos x+\sin x)(\cos y+\sin y)$

(4)

$\frac{\sin(x+y)}{\cos(x-y)}=\frac{\sin x\cdot\cos y+\cos x\cdot\sin y}{\cos x\cdot\cos y+\sin x\cdot\sin y}=\dfrac{\dfrac{\sin x\cdot\cos y}{\cos x\cdot\cos y}+\dfrac{\cos x\cdot\sin y}{\cos x\cdot\cos y}}{\dfrac{\cos x\cdot\cos y}{\cos x\cdot\cos y}+\dfrac{\sin x\cdot\sin y}{\cos x\cdot\cos y}}=\frac{\tan x+\tan y}{1+\tan x\cdot\tan y}$

(5)

$\sin\left(x-\frac{\pi}{6}\right)+\cos\left(x-\frac{\pi}{3}\right)=\sin x\cdot\cos\frac{\pi}{6}-\cos x\cdot\sin\frac{\pi}{6}+\cos x\cdot\cos\frac{\pi}{3}+\sin x\cdot\sin\frac{\pi}{3}$

$=\frac{\sqrt3}{2}\sin x-\frac12\cos x+\frac12\cos x+\frac{\sqrt3}{2}\sin x=\sqrt3\sin x$

(6)

$\tan\left(x+\frac{\pi}{4}\right)-\tan\left(x-\frac{3\pi}{4}\right)=\frac{\tan x+\tan\dfrac{\pi}{4}}{1-\tan x\cdot\tan\dfrac{\pi}{4}}-\frac{\tan x-\tan\dfrac{3}{4}\pi}{1+\tan x\cdot\tan\dfrac{3}{4}\pi}=\frac{\tan x+1}{1-\tan x}-\frac{\tan x+1}{1-\tan x}=0$