√2+√2+√2+ 2 cos 4 theta= 2 cos theta by 2
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Answer:
√7 teheta √27cos theta. 77
Answered by
0
Answer:
According to question,
2+
2+2cos4θ
=2cosθ
squaring on both sides,
=> 2+
2+2cos4θ
=4cos
2
θ
=>
2+2cos4θ
=4cos
2
θ−2
=>
2+2cos4θ
=2(2cos
2
θ−1)
=>
2+2cos4θ
=2(cos2θ)
Again squaring,
=> 2+2cos4θ=4(cos
2
2θ)
=> 2cos4θ=4(cos
2
2θ)−2
=> 2cos4θ=2(2cos
2
2θ−1)
=> 2cos4θ=2(cos4θ)
=> LHS=RHS
Hence proved
Step-by-step explanation:
Step-by-step explanation:please mark me as brainliest please don't forget
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