Prove that \(\displaystyle{\frac{{{\sin{{5}}}{x}}}{{{\sin{{x}}}}}}\in{\left(-{\frac{{{5}}}{{{4}}}},{5}\right)}\) for any \(\displaystyle{x}\in{\mathbb{{{R}}}}{k}\pi\)

Alexis Alexander

Alexis Alexander

Answered question

2022-03-31

Prove that sin5xsinx(54,5) for any xRkπ where kZ.

Answer & Explanation

armejantm925

armejantm925

Beginner2022-04-01Added 20 answers

For sinx0,
sin5xsinx=5cos4x10cos2x(1cos2x)+(1cos2x)2
=16cos4x12cos2x+1
=(4cos2x32)2+194
Since anything squared is 0, we have:
0(4cos2x32)2
194(4cos2x32)2+194
5416cos4x12cos2x+1
Also as cosx1
16cos4x12cos2x+11612+1

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