Question

I need assistance in this multipart question. This question is quite difficult for me and would gladly appreciate it if someone would help me solve this. (50) point question and brainliest

Answer 1

6. Expand the left side:

$(\cos x+\cos y)^2+(\sin x-\sin y)^2$

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

Recall the Pythagorean identity,

$\cos^2x+\sin^2x=1$

which leaves us with

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

Recall the angle sum formula for cosine:

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

which gets us to where we need to be:

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

and we're done.

7. A. The formulas you need are

$\sin(x\pm y)=\sin x\cos y\mp\sin y\cos x$

This gives

$\sin\left(x+\dfrac\pi4\right)-\sin\left(x-\dfrac\pi4\right)=1$

$\left(\sin x\cos\dfrac\pi4+\sin\dfrac\pi4\cos x\right)-\left(\sin x\cos\dfrac\pi4-\sin\dfrac\pi4\cos x\right)=1$

$2\sin\dfrac\pi4\cos x=1$

$\dfrac2{\sqrt2}\cos x=1$

7. B. Solving for $\cos x$ gives

$\cos x=\dfrac1{\sqrt2}$

which holds for

$x=\dfrac\pi4+2n\pi\text{ and }x=-\dfrac\pi4+2n\pi$

where $n$ is any integer.

7. C. Factorize as

$\tan^2x-3\tan x+2=0$

$(\tan x-1)(\tan x-2)=0$

From here, we have

$\tan x-1=0\text{ or }\tan x-2=0$

$\tan x=1\text{ or }\tan x=2$

which hold for

$x=\dfrac\pi4+n\pi\text{ or }x=\tan^{-1}(2)+n\pi$

where $n$ is any integer.