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\[\cos^{n} x \le \cos(nx)\]

Not sure if I got this right, but have you used trial numbers?

to test the inequality ?

I have no clue.

Grade 12 Advanced Functions?

f'(x) - g'(x) > 0 implies f'(x) > g'(x)
right?
i think i found a method to figure it out

I have no clue.

showing (cos(nx))' .>= (cos^n(x))' is sufficient after showing they begin at some point.

ok .. this becomes
\[n \left( -\sin(nx) + \cos^{p - 1}(x) * \sin x \right)\]

youre right, the question uses p for the exponent but i'm used to n to denote an exponent . or k

wow i should seriously review what i type before i post it. lets just say p = n