cos pi/7 +cos 3pi/7 +cos 5pi/7=1/2
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Answered by
92
Step-by-step explanation:
cos(π/7) + cos(3π/7) + cos(5π/7)
Multuiply and divide by 2sin(π/7)
= 1/2sin(π/7) [ 2sin(π/7)cos(π/7) + 2sin(π/7)cos(3π/7) + 2sin(π/7)cos(5π/7) ]
Use the following
sin(2A) = 2sinAcosA
2sinAcosB = sin(A+B) + sin(A-B)
= 1/2sin(π/7) [ sin(2π/7) + sin(π/7+3π/7) + sin(π/7-3π/7) + sin(π/7+5π/7) + sin(π/7-5π/7) ]
= 1/2sin(π/7) [ sin(2π/7) + sin(4π/7) + sin(-2π/7) + sin(6π/7) + sin(-4π/7) ]
= 1/2sin(π/7) [ sin(2π/7) + sin(4π/7) - sin(2π/7) + sin(6π/7) - sin(4π/7) ]
= 1/2sin(π/7) [sin(6π/7)]
= 1/2sin(π/7) [sin(π - π/7)]
= [sin(π/7)]/2sin(π/7)
= 1/2
Answered by
2
Answer:
1 by 2
Step-by-step explanation:
the solution in above is correct check it from the above one ☺️☺️
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