# Show that (cos x)^n=cos(x+npi/2)?

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The request of this problem is vague, hence, supposing that `(cos x)^(n)` means the n-th derivative of the fuction cos x (since you have tagged calculus as the category of this problem) you should perform mathematical induction to prove the identity.

Starting with the base case, you need to consider n = 1 such that:

`(cos x)' = -sin x`

`cos (x + pi/2) = cos x*cos (pi/2) - sin x*sin (pi/2) =- sin x`

Notice that `(cos x)' = cos (x + pi/2)`

You need to perform the inductive step such that:

If `(cos x)^(k) = cos (x + k*pi/2)` holds, then `(cos x)^(k+1) = cos (x + (k+1)*pi/2)` also holds.

`(cos x)^(k+1) = ((cos x)^(k))'`

Assuming as valid (cos x)^(k) = cos (x + k*pi/2), then, you need to substitute `cos (x + k*pi/2)` for `(cos x)^(k)` such that:

`((cos x)^(k))'= (cos (x + k*pi/2))' = -sin (x + k*pi/2)`

You may write `-sin (x + k*pi/2)` as `cos (x + k*pi/2 + pi/2)` such that:

`cos (x + k*pi/2 + pi/2) = cos (x + pi/2*(k+1))`

Hence, `(cos x)^(k+1) = ((cos x)^(k))' = cos (x + pi/2*(k+1))`

**Hence, using the mathematical induction yields that `(cos x)^(n) = cos (x + n*pi/2), ` if `n>=1` .**

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