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Question Number 90700 by jagoll last updated on 25/Apr/20

sin^2 (((7π)/8))+sin^2 (((3π)/8))+sin^2 (((5π)/8))+sin^2 ((π/8)) ?

$$\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{8}}\right)+\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{5}\pi}{\mathrm{8}}\right)+\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)\:? \\ $$

Commented by john santu last updated on 25/Apr/20

sin^2 (((7π)/8)) = sin^2 (π−((7π)/8)) = sin^2 ((π/8)) sin^2 (((5π)/8)) = sin^2 (π−((5π)/8)) = sin^2 (((3π)/8)) ⇒ 2 { sin^2 (((3π)/8)) + sin^2 ((π/8))} = 2 { sin^2 (((3π)/8)) + cos^2 ((π/2)−(π/8))} = 2 { sin^2 (((3π)/8)) + cos^2 (((3π)/8))} = 2 .

$$\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{8}}\right)\:=\:\mathrm{sin}\:^{\mathrm{2}} \left(\pi−\frac{\mathrm{7}\pi}{\mathrm{8}}\right)\:=\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right) \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{5}\pi}{\mathrm{8}}\right)\:=\:\mathrm{sin}\:^{\mathrm{2}} \left(\pi−\frac{\mathrm{5}\pi}{\mathrm{8}}\right)\:=\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\ $$$$\Rightarrow\:\mathrm{2}\:\left\{\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\:+\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)\right\} \\ $$$$=\:\mathrm{2}\:\left\{\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\:+\:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{8}}\right)\right\} \\ $$$$=\:\mathrm{2}\:\left\{\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\:+\:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\right\}\: \\ $$$$=\:\mathrm{2}\:.\: \\ $$

Commented by jagoll last updated on 25/Apr/20

thank you

$${thank}\:{you} \\ $$

Commented by peter frank last updated on 26/Apr/20

thanks

$${thanks} \\ $$

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