cos4 θ + sin4 θ – 2 sin2 θ cos2 θ = (2 cos2 θ – 1)2
aum1052:
how to do that
hey
Can you pls check this question too
Brilliant option will come ,you have to click that
pls check the question
Cos^4 θ + sin^4 θ – 2 sin^2 θ cos^2 θ = (2 cos^2 θ – 1)^2
you may find this easier
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Answered by
7
Answer:
Hey Mate.!
Step-by-step explanation:
L.H.S. = sin4 θ – cos4 θ
= (sin2 θ)2 – (cos2 θ)2
= (sin2 θ – cos2 θ)(sin2 θ + cos2 θ)
= (sin2 θ – cos2 θ) × 1
= sin2 θ – cos2 θ = R.H.S.
= sin2 θ – (1 – sin2 θ)
= sin2 θ – 1 + sin2 θ
= 2 sin2 θ – 1 = R.H.S.
= 2(1 – cos2 θ) – 1
= 2 – 2 cos2 θ – 1
= 1 – 2 cos2 θ = R.H.S.
but can you pls first check the question
I think there is some mistake
@legend welcome dear
I think it should be sin^4Q-cos^4 =1-2cos^2
cos^4 θ + sin^4 θ – 2 sin^2 θ cos^2 θ = (2 cos^2 θ – 1)^2
this is it
okay I will try
pls give fast
hello are you answering
Answered by
2
L.H.S. = sin4 θ – cos4 θ
= (sin2 θ)2 – (cos2 θ)2
= (sin2 θ – cos2 θ)(sin2 θ + cos2 θ)
= (sin2 θ – cos2 θ) × 1
= sin2 θ – cos2 θ = R.H.S.
= sin2 θ – (1 – sin2 θ)
= sin2 θ – 1 + sin2 θ
= 2 sin2 θ – 1 = R.H.S.
= 2(1 – cos2 θ) – 1
= 2 – 2 cos2 θ – 1
= 1 – 2 cos2 θ = R.H.S.
Mark ❤️
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