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\operatorname { sin } ^ { 2 } \frac { \pi } { 8 } + \operatorname { sin } ^ { 2 } \frac { 3 \pi } { 8 } + \operatorname { sin } ^ { 2 } \frac { 5 \pi } { 8 } + \operatorname { sin } ^ { 2 } \frac { 7 \pi } { 8 } = 2 |
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Answer» Sin²π/8+sin²3π/8+sin²5π/8+sin²7π/8=sin²π/8+sin²3π/8+sin²(π/2+π/8)+sin²(π/2+3π/8)=sin²π/8+sin²3π/8+cos²π/8+cos²3π/8=(sin²π/8+cos²π/8)+(sin²3π/8+cos²3π/8) [∵, sin²Ф+cos²Ф=1]=1+1=2 Ans. |
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