sin^2 pi_12 +sin^2 3pi_12 +sin^2 5pi_12 +sin^2 7pi_12 + sin^2 9pi_12 +sin^2 11pi_12 please help meeeee nowww

Đáp án:

 3

Giải thích các bước giải:

$\sin ^{2}\frac{\pi }{12}+\sin ^{2}\frac{3\pi }{12}+\sin ^{2}\frac{5\pi }{12}+\sin ^{2}\frac{7\pi }{12}+\sin ^{2}\frac{9\pi }{12}+\sin ^{2}\frac{11\pi }{12}$

=$\sin ^{2}(\pi-\frac{11\pi }{12})+\sin ^{2}(\pi-\frac{9\pi }{12})+\sin ^{2}(\pi-\frac{7\pi }{12}+\sin ^{2}\frac{7\pi }{12})+\sin ^{2}\frac{9\pi }{12}+\sin ^{2}\frac{11\pi }{12}$

=$2.\sin ^{2}\frac{7\pi }{12}+2\sin ^{2}\frac{9\pi }{12}+2\sin ^{2}\frac{11\pi }{12}$

=$2.(\sin ^{2}\frac{7\pi }{12}+\sin ^{2}\frac{11\pi }{12})+2\sin ^{2}\frac{9\pi }{12}$

=$2((\sin \frac{7\pi }{12}+\sin \frac{11\pi }{12})^{2}-2\sin \frac{7\pi }{12}\sin \frac{11\pi }{12})+2\sin ^{2}\frac{9\pi }{12}$

=$2.((2\sin \frac{\pi (\frac{7}{12}+\frac{11}{12})}{2}\cos (\pi .\frac{\frac{11}{12}-\frac{7}{12}}{2} ))^{2}-(\cos \frac{4\pi }{12}-\cos \frac{8\pi }{12}))+2\sin ^{2}\frac{3\pi }{4}$

=$2((\sqrt{\frac{3}{2}})^{2}-\frac{1}{2}))+2.(\frac{1}{\sqrt{2}})^{2}$

=3