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Question

The value of sin2π8+sin23π8+sin25π8+sin27π8 is

A
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B
1
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C
2
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D
3
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Solution

The correct option is C 2
sin2π8+sin23π8+sin25π8+sin27π8
=(sin2π8+sin27π8)+(sin23π8+sin25π8)
=(sin2π8+sin2(ππ8))+(sin23π8+sin2(π3π8))
=(sin2π8+sin2π8)+(sin23π8+sin23π8)=2(sin2π8+sin2(π2π8))=2(sin2π8+cos2π8)=2

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