6. Expand the left side:
[tex](\cos x+\cos y)^2+(\sin x-\sin y)^2[/tex]
[tex]=(\cos^2x+2\cos x\cos y+\cos^2y)+(\sin^2x-2\sin x\sin y+\sin^2y)[/tex]
Recall the Pythagorean identity,
[tex]\cos^2x+\sin^2x=1[/tex]
which leaves us with
[tex](\cos x+\cos y)^2+(\sin x-\sin y)^2=2+2\cos x\cos y-2\sin x\sin y[/tex]
Recall the angle sum formula for cosine:
[tex]\cos(x+y)=\cos x\cos y-\sin x\sin y[/tex]
which gets us to where we need to be:
[tex](\cos x+\cos y)^2+(\sin x-\sin y)^2=2+2\cos(x+y)[/tex]
and we're done.
7. A. The formulas you need are
[tex]\sin(x\pm y)=\sin x\cos y\mp\sin y\cos x[/tex]
This gives
[tex]\sin\left(x+\dfrac\pi4\right)-\sin\left(x-\dfrac\pi4\right)=1[/tex]
[tex]\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[/tex]
[tex]2\sin\dfrac\pi4\cos x=1[/tex]
[tex]\dfrac2{\sqrt2}\cos x=1[/tex]
7. B. Solving for [tex]\cos x[/tex] gives
[tex]\cos x=\dfrac1{\sqrt2}[/tex]
which holds for
[tex]x=\dfrac\pi4+2n\pi\text{ and }x=-\dfrac\pi4+2n\pi[/tex]
where [tex]n[/tex] is any integer.
7. C. Factorize as
[tex]\tan^2x-3\tan x+2=0[/tex]
[tex](\tan x-1)(\tan x-2)=0[/tex]
From here, we have
[tex]\tan x-1=0\text{ or }\tan x-2=0[/tex]
[tex]\tan x=1\text{ or }\tan x=2[/tex]
which hold for
[tex]x=\dfrac\pi4+n\pi\text{ or }x=\tan^{-1}(2)+n\pi[/tex]
where [tex]n[/tex] is any integer.
