If $$ heta \in [0,2\pi$$], find the maximum value of
$$f = |\sin^2 heta {\sin^3(2 heta )\cdotp \sin^3 (4 heta ) \dots\sin^3 (2^{n-1} }\sin (2^n heta )| $$

Question

If $$ 	heta \in [0,2\pi$$], find the maximum value of 
$$f = |\sin^2 	heta {\sin^3(2	heta )\cdotp \sin^3 (4	heta  )  \dots\sin^3 (2^{n-1}  }\sin (2^n	heta )| $$

Uniteasy Admin · Accepted Answer

$$(\sin^2 	heta \sin 2	heta )^2 $$
$$=\sin^4\cdotp 4\sin^2 	heta (1-\sin^2	heta )  $$
$$=\dfrac{4}{3}\sin^2	heta \cdotp\sin^2	heta \cdotp\sin^2	heta \cdotp(3-3\sin^2	heta )  $$
$$\leq \dfrac{4}{3} \cdotp \Big(  \dfrac{3}{4}\Big)  ^4 =\dfrac{27}{64} $$
Therefore,
$$|\sin^2 	heta \sin 2	heta |\leq \dfrac{3\sqrt{3} }{8}  $$
Similarly,

Question

Answer

This answer is set private or in draft.