Question

Find the maximum value of the function y = \dfrac{\sin x\cos x}{1+\sin x+\cos x}

Collected in the board: Trigonometry

Steven Zheng posted 1 month ago


Answer

(1+\sin x+\cos x)(1-\sin x-\cos x)

=1-(\sin x+\cos x)^2

=-2\sin x\cos x

Therefore,

y = \dfrac{\sin x\cos x}{1+\sin x+\cos x}

=\sin x+\cos x-1

=\sqrt{2} \cdotp (\dfrac{\sqrt{2} }{2}\sin x+ \dfrac{\sqrt{2} }{2}\cos x)-1

=\sqrt{2}(\sin x \cos 45\degree +\cos x\sin 45\degree )-1

=\sqrt{2}\sin (x-45\degree )-1

So the maximum value of the function is \sqrt{2}-1

Steven Zheng posted 1 month ago

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