thanks alot Lets express cos5 t in terms of cosines and sine

thanks alot

Let\'s express cos^5 t in terms of cosines and sines of multiples of t. For this we need recall the Binomial theorem for n = 5, namely that (a + b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5 and the expression cos t = e^it + e^-it/2. We group together like terms and use the facts that cost = e^int + e^-int/2 and sin nt = -e^int-e^int/2i to get that cos^5 t = a cos 5t + b cos 3t + c cos where a =, b =, c =. Let\'s express sin^4t in terms of cosines and sines of multiples of t. For this we need the Binomial theorem for n = 4, along with the expression sint = e^it -e^-it/2i. Don\'t forget that important i in the denominator! We must group together like terms and use the facts that cos nt = e^int + e^-int/2 and sin nt = e^int_e^-int/2i to get that sin^4 t = a cos 4t + b cos 2t + c where a =,b = and c =.

Solution

Part 1

Cos⁵t = [(e^{i t} + e^{- i t})/2]⁵

^{On expanding and solving we get}

1/16 * Cos5t + 5/16 * Cos3t + 5/8 * Cos t

a = 1/16 , b = 5/16 ,c = 5/8

Part 2

Using Binomial for degree 4

(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

Sin⁴t = [(e^{i t -} e^{- i t})/2i]⁴

On expanding and solving we get

Sin⁴t = 1/8 * Cos4t - 1/2 * Cos2t + 6/16

a = 1/8 , b = - 1/2 , c = 3/8