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Question Number 76633 by mathmax by abdo last updated on 28/Dec/19
$${calculate}\:{cos}^{\mathrm{6}} \left(\frac{\pi}{\mathrm{8}}\right)+{cos}^{\mathrm{6}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+{cos}^{\mathrm{6}} \left(\frac{\mathrm{5}\pi}{\mathrm{8}}\right)+{cos}^{\mathrm{6}} \left(\frac{\mathrm{7}\pi}{\mathrm{8}}\right) \\ $$
Answered by MJS last updated on 28/Dec/19
$$\mathrm{cos}\:\frac{\pi}{\mathrm{8}}\:=\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\:=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{8}}\:=−\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{8}}\:=−\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{2}×\left(\left(\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\right)^{\mathrm{6}} +\left(\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\right)^{\mathrm{6}} \right)=\frac{\mathrm{5}}{\mathrm{4}} \\ $$
Commented by mathmax by abdo last updated on 29/Dec/19
$${thank}\:{you}\:{sir}. \\ $$
Answered by $@ty@m123 last updated on 29/Dec/19
![= cos^6 ((π/8))+cos^6 (((3π)/8))+cos^6 (((3π)/8))+cos^6 ((π/8)) = 2{cos^6 ((π/8))+cos^6 (((3π)/8))} = 2{cos^6 ((π/8))+sin^6 ((π/8))} = 2{cos^2 ((π/8))+sin^2 ((π/8))}{cos^4 ((π/8))+sin^4 ((π/8))−cos^2 ((π/8)).sin^2 ((π/8))} = 2{cos^4 ((π/8))+sin^4 ((π/8))−(1/4)×4cos^2 ((π/8)).sin^2 ((π/8))} = 2[{cos^2 ((π/8))+sin^2 ((π/8))}^2 −2cos^2 ((π/8)).sin^2 ((π/8))−(1/4)×(sin (π/4))^2 ] = 2[1−(1/2)×sin^2 (π/4)−(1/4)×(1/2)] = 2[1−(1/2)×(1/2)−(1/8)] =2(1−(1/4)−(1/8)) =2×((8−2−1)/8) =(5/4)](Q76659.png)
$$=\:{cos}^{\mathrm{6}} \left(\frac{\pi}{\mathrm{8}}\right)+{cos}^{\mathrm{6}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+{cos}^{\mathrm{6}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+{cos}^{\mathrm{6}} \left(\frac{\pi}{\mathrm{8}}\right) \\ $$$$=\:\mathrm{2}\left\{{cos}^{\mathrm{6}} \left(\frac{\pi}{\mathrm{8}}\right)+{cos}^{\mathrm{6}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\right\} \\ $$$$=\:\mathrm{2}\left\{{cos}^{\mathrm{6}} \left(\frac{\pi}{\mathrm{8}}\right)+{sin}^{\mathrm{6}} \left(\frac{\pi}{\mathrm{8}}\right)\right\} \\ $$$$=\:\mathrm{2}\left\{{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)+{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)\right\}\left\{{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+{sin}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)−{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right).{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)\right\} \\ $$$$=\:\mathrm{2}\left\{{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+{sin}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)−\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{4}{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right).{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)\right\} \\ $$$$=\:\mathrm{2}\left[\left\{{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)+{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)\right\}^{\mathrm{2}} −\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right).{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)−\frac{\mathrm{1}}{\mathrm{4}}×\left(\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\right)^{\mathrm{2}} \right] \\ $$$$=\:\mathrm{2}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{sin}^{\mathrm{2}} \:\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$$=\:\mathrm{2}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{8}}\right] \\ $$$$=\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}\right) \\ $$$$=\mathrm{2}×\frac{\mathrm{8}−\mathrm{2}−\mathrm{1}}{\mathrm{8}} \\ $$$$=\frac{\mathrm{5}}{\mathrm{4}} \\ $$
Commented by abdomathmax last updated on 29/Dec/19
$${thankx}\:\:{sir}. \\ $$