The questions no l and p.
please help me as soon as possible

l = 2cosΘ p = see attachment

Step-by-step explanation:

√2+√2+2cos 4Θ

√2+√2(1 + cos 4Θ)

√2 + √2(2 cos² 2Θ)

√2 + √4 cos² 2Θ                      cos 2Θ = 2 cos²Θ - 1

√2 + 2cos 2Θ

√2(1 + cos 2Θ)

√4cos²Θ

2cosΘ

Step-by-step explanation:

I'm only going to do the first one, because the tangent problem is going to give us both night mares.

Start with cos(4*theta), There are various places on the internet which solves this, so I won't bother. Basically it comes down to using cos(2*theta).

cos(4theta) = 8*cos^4(theta) - 8*cos^2 (theta) + 1

2*cos(4*theta) = 16*cos^4(theta) - 16cos^2(theta) + 2

2 + 2cos(4*theta) = 16cos^4(theta) - 16cos^2(theta) + 4

(2 + 2cos(4*theta) = (4 cos^2(theta) - 2 )^2

sqrt(2 + 2cos(4theta) = 4cos^2(theta) - 2

sqrt(2 + 4cos^2(theta) -2 ) = sqrt( 4 cos^2(theta) = 2 cos(theta) = RHS

answer: (1,3) and (-2,0) i think.