Question

Equation: sin^2XcosX

Show that at the maximum, cosX=(1/3)^1/2

I've gotten to 2cos^2X-sin^2X=0, but need to get to cosX=(1/3)^1/2 from there. I believe it requires trig identities. Any help?

Thanks in advance

Answer 1

this does not make sense, the max of cos(x) = 1

Answer 2

write the question as it was asked

Answer 3

OKay, the equation was (sinX)^2cosX.

Prove that at the maximum point of the graph, cosX=(1/3)^1/2

Answer 4

Find the maximum point on the graph of \(\sin^2x\cos x\), then plug that point into \(\cos x\).

Answer 5

\[\begin{align*}f(x)=\sin^2x\cos x~~\Rightarrow~~f'(x)=2\sin x\cos^2x-\sin^3x&=0\\
\sin x\left(2\cos^2x-\sin^2x\right)&=0
\end{align*}\]
From the second factor, you have
\[2\cos^2x=\sin^2x~~\iff~~\tan^2x=2~~\iff~~\tan x=\pm\sqrt2\]

Answer 6

Fantastic, thanks for the help!