hey im new to the site but i have some ap calc summer work due tomorrow and i am STUCK with this trig identity. any ideas how to start it off? (cos^4)x = 1/8 (3+4cos2x+cos4x)
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Originally Posted by still_nights hey im new to the site but i have some ap calc summer work due tomorrow and i am STUCK with this trig identity. any ideas how to start it off? (cos^4)x = 1/8 (3+4cos2x+cos4x) Start with the left hand side: $\displaystyle (\cos^2 x)^2 = \left( \frac{\cos (2x) + 1}{2}\right)^2$ using the well known double angle formula $\displaystyle \cos (2x) = 2 \cos^2 (x) - 1$. Now expand. Then use the double angle formula again.
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