Please help me with this question of trigonometry
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2 may be...hope it helps
Answered by
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Answer:
cos⁴π/8+cos⁴3π/8+cos⁴5π/8+cos⁴7π/8
=cos⁴π/8+cos⁴3π/8+{cos(π/2+π/8)}⁴+{cos(π/2+3π/8)}⁴
=cos⁴π/8++cos⁴3π/8+(-sinπ/8)⁴+(-sin3π/8)⁴
=sin⁴π/8+cos⁴π/8+sin⁴3π/8+cos⁴3π/8
={(sin²π/8)²+(cos²π/8)²}+{(sin²3π/8)²+(cos²3π/8)²}
={(sin²π/8+cos²π/8)²-2sin²π/8cos²π/8}+{(sin²3π/8+cos²3π/8)²- 2sin²3π/8cos²3π/8}
=1-2sin²π/8cos²π/8+1-2sin²3π/8cos²3π²/8
=(1/2){4-(2sinπ/8cosπ/8)²-(2sin3π/8cos3π/8)²}
=(1/2)[4-(sinπ/4)²-(sin3π/4)²] [∵ sin2A=2sinAcosA]
=(1/2)[4-(1/√2)²-(cosπ/4)²] [∵sin3π/4=sin{(π/2×1)+π/4}=cosπ/4]
=(1/2)[4-1/2-(1/√2)²]
=(1/2)[4-(1/2+1/2)]
=(1/2)(4-1)
=3/2
HENCE PROVED
✌
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