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Question Number 21009 by xxyy@gmail.com last updated on 10/Sep/17

Answered by Tinkutara last updated on 10/Sep/17

2cos^2 (π/(16))×2cos^2 ((3π)/(16))×2cos^2 ((5π)/(16))×2cos^2 ((7π)/(16)) =(4cos (π/(16))cos ((3π)/(16))cos ((5π)/(16))cos ((7π)/(16)))^2 =(4cos (π/(16))cos ((3π)/(16))sin ((3π)/(16))sin (π/(16)))^2 =(sin (π/8)sin ((3π)/8))^2 =(1/4)(2sin (π/8)cos (π/8))^2 =(1/4)sin^2 (π/4)=(1/8)

$$\mathrm{2cos}^{\mathrm{2}} \:\frac{\pi}{\mathrm{16}}×\mathrm{2cos}^{\mathrm{2}} \:\frac{\mathrm{3}\pi}{\mathrm{16}}×\mathrm{2cos}^{\mathrm{2}} \:\frac{\mathrm{5}\pi}{\mathrm{16}}×\mathrm{2cos}^{\mathrm{2}} \:\frac{\mathrm{7}\pi}{\mathrm{16}} \\ $$$$=\left(\mathrm{4cos}\:\frac{\pi}{\mathrm{16}}\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{16}}\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{16}}\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{16}}\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{4cos}\:\frac{\pi}{\mathrm{16}}\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{16}}\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{16}}\mathrm{sin}\:\frac{\pi}{\mathrm{16}}\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{sin}\:\frac{\pi}{\mathrm{8}}\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2sin}\:\frac{\pi}{\mathrm{8}}\mathrm{cos}\:\frac{\pi}{\mathrm{8}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}^{\mathrm{2}} \:\frac{\pi}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$

Commented by xxyy@gmail.com last updated on 10/Sep/17

thank very much sir!

$$\mathrm{thank}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir}! \\ $$

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