Question

Prove that: Sin⁴A=1/8(3-4Cos2A+Cos4A)​

Answer 1

Answer:

sin⁴A => (sin²A)(sin²A)

Now, cos2A => 1 - 2sin²A

so, sin²A = (1 - cos2A)/2

Then, sin⁴A = [(1 - cos2A)/2][(1 - cos2A)/2]

=> (1/4)[1 - 2cos2A + cos²2A]

And cos2A => 2cos²A - 1

so, cos4A => 2cos²2A - 1

i.e. cos²2A = (cos4A + 1)/2

Then, (1/4)[1 - 2cos2A + ((cos4A + 1)/2)]

=> (1/8)[2 - 4cos2A + cos4A + 1]

Hence, (1/8)[3 - 4cos2A + cos4A]

Step-by-step explanation:

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