Question

Find the minimum value of the trig function

y = \dfrac{\sin 3x \sin^3x +\cos 3x \cos^3x}{\cos^2 2x} +\sin 2x

Collected in the board: Trigonometry

Steven Zheng posted 1 month ago


Answer

Using the product to sum identity

y = \dfrac{\sin 3x \sin^3x +\cos 3x \cos^3x}{\cos^2 2x} +\sin 2x

= \dfrac{(\sin 3x\sin x) \sin^2x +(\cos 3x\cos x) \cos^2x}{\cos^2 2x} +\sin 2x

=\dfrac{1}{2} \cdotp \dfrac{(\cos 2x-\cos 4x)\sin^2x+(\cos 2x+\cos 4x) \cos^2x}{\cos^2 2x} +\sin 2x

=\dfrac{1}{2} \cdotp \dfrac{\cos 2x(\sin^2 x+\cos^2 x)+(\cos^2 x-\sin^2 x)\cos 4x}{\cos^2 2x} +\sin 2x

=\dfrac{1}{2} \cdotp \dfrac{\cos 2x+\cos 2x\cos 4x}{\cos^2 2x} +\sin 2x

=\dfrac{1}{2} \cdotp \dfrac{\cos 2x(1+\cos 4x)}{\cos^2 2x} +\sin 2x

= \dfrac{\cos^3 2x}{\cos^2 2x} +\sin 2x

=\cos^2 2x+\sin 2x

=\sqrt{2}\sin (2x+\dfrac{\pi}{4} )

So the minimum value of y is -\sqrt{2}


Steven Zheng posted 1 month ago

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