sin^4 A-cos^4 A=2sin^2A-1
prove the following identity
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Answer:
Step-by-step explanation:
Given:
I−sin
4
A−cos
4
A;II−2sin
2
A−1;III=1−2cos
2
A;IV−sin
2
A−cos
2
A
Taking I term
=sin
4
A−cos
4
A=(sin
2
A)
2
−(cos
2
A)
2
=(sin
2
A−cos
2
A)(sin
2
A+cos
2
A)[∵(a
2
−b
2
)=(a+b)(a−b)]
=(sin
2
A−cos
2
A)(1)[∵cos
2
θ+sin
2
θ=1]=(sin
2
A−cos
2
A)...(1)→IV term
From Eq (1)
=[sin
2
A−(1−sin
2
A)]=sin
2
A−1+sin
2
A=2sin
2
A−1→II term
again from Eq (1)
=[(1−cos
2
A)−cos
2
A][∵cos
2
θ+sin
2
θ=1]=1−2cos
2
A→III term
Hence, I=II=III=IV
Hence proved
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