Sidhe kijiye ki cosπ/7 .cos2π/7 .cos4π/7=-1/8
Answers
Answer:
-1/8
Step-by-step explanation:
To prove ---->
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Cosπ/7 Cos2π/7 Cos4π/7 =-1/8
Proof--->
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L. H. S.=Cosπ/7 Cos2π/7 Cos4π/7
mutiplying and dividing by 2Sin π/7
=(1/2Sinπ/7) 2Sinπ/7 Cosπ/7 Cos2π/7
Cos4π/7
We have a formula 2SinACosA=Sin2A
Applying it here
=(1/2Sinπ/7) Sin2π/7 Cos2π/7 Cos4π/7
multiplying and dividing by 2
=(1/4Sinπ/7) 2Sin2π/7 Cos2π/7
Cos4π/7
=(1/4Sinπ/7) Sin4π/7 Cos4π/7
Multiplying and dividing by 2
=(1/8 Sinπ/7) 2Sin4π/7 Cos4π/7
=(1/8Sinπ/7) Sin8π/7
=(1/8Sinπ/7) Sin(π + π/7)
Sin(π+π/7) = -Sinπ/7 because (π+π/7) lies in third quadrant in which Sin is negative
=(1/8Sinπ/7) (-Sin π/7)
Sin π/7 cancel out from numerator and denominator
=-(1/8)=R.H.S.
-1/8
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