Sigh Sine Cos Cosine

Algebra Level 5

cos x ( sin x + sin 2 x + 3 4 ) \large \cos x \left(\sin x + \sqrt{\sin ^2 x + \frac{3}{4}}\right)

Find the maximum value of the expression above, where x R x \in \mathbb R .


The answer is 1.323.

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2 solutions

Thanks for the advice from Harsh Shrivastava , the solution is as follows:

By Cauchy-Schwarz inequality:

( sin x cos x + cos x sin 2 x + 3 4 ) 2 ( sin 2 x + cos 2 x ) ( cos 2 x + sin 2 x + 3 4 ) = ( 1 ) ( 1 + 3 4 ) = 7 4 cos x ( sin x + sin 2 x + 3 4 ) 7 2 1.323 \begin{aligned} \left(\sin x \cos x + \cos x \sqrt{\sin^2 x +\frac 34}\right)^2 & \le \left(\sin^2 x + \cos^2 x \right) \left(\cos^2 x + \sin^2 x + \frac 34 \right) = (1)\left(1+\frac 34\right) = \frac 74 \\ \implies \cos x \left(\sin x + \sqrt{\sin^2 x +\frac 34}\right) & \le \frac {\sqrt 7}2 \approx \boxed{1.323} \end{aligned}

Thanks for posting a solution!

Harsh Shrivastava - 4 years, 1 month ago

Hint : When all fails, Cauchy Schwartz hails~!

(Though we can solve this problem without Cauchy Schwartz also but that would be too lengthy/)

Nice statement.Although I solved by a quadratic m a x ( 1 / 4 ) ( x 2 + 11 / 2 x 9 / 16 ) max√(1/4)(-x^2+11/2x-9/16) .This wasn't much lengthy though.Max value is ( 7 / 2 ) √(7/2) .my app was set S i n x + ( s i n 2 x + 3 / 4 ) = y Sinx+√(sin^2x+3/4)=y yields ( y 2 3 / 4 ) / 2 y = s i n x (y^2-3/4)/2y=sinx .Now we need to max y c o s x ycosx .so square it and express c o s cos in terms of s i n sin gives us to maximise a quadratic.

Spandan Senapati - 4 years, 1 month ago

Quite creative.

Tapas Mazumdar - 4 years, 1 month ago

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