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

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