Answer:
Below.
Step-by-step explanation:
cos2a = 2 cos^2 a - 1
cos^4a = 8cos^4 a - 8cos^2 a + 1
Right side of the identity =
1 /8(3 + 4cos2a + cos4a)
= 1/8( 3 + 4(2 cos^2 a - 1) + 8cos^4 a - 8cos^2 a+1)
= 1/8 (3 + 8 cos^2 a - 4 + 8 cos^4 a - 8cos^2 a + 1)
= 1/8 (8cos^4 a)
= cos^4 a = the left side.
I realised that I should have derived the identity for cos 4a as well , which i have done in the picture.
The given identity, cos⁴a = 1/8(3 + 4cos2a + cos4a) is proved.
To prove a given identity, say f(a) = f(b), we proceed to simplify any of the expressions among the L.H.S. f(a) and the R.H.S. f(b), to get the other term. If we get the simplified form as the other term, our identity holds and is proved.
We know cos2A = 2cos²A - 1.
Going by the same formula, we derive the value of cos4A.
cos4A = 2cos²2A - 1
or, cos4A = 2(2cos²A - 1)² - 1, putting value of cos2A
or, cos4A = 2(4cos⁴A - 4cos²A + 1) - 1, (expanding (2cos²A - 1)² using the formula of (a-b)² = a² -2ab + b²)
or, cos4A = 8cos⁴A - 8cos²A + 2 - 1, (expanding)
or, cos4A = 8cos⁴A - 8cos²A + 1, (simplifying)
∴ cos4A = 8cos⁴A - 8cos²A + 1.
Given identity to us is, cos⁴a = 1/8(3 + 4cos2a + cos4a).
To prove the identity, we proceed with the R.H.S.:
= 1/8(3 + 4cos2a + cos4a)
We put values of cos2A and cos4A in this expression, to get
= 1/8(3 + 4(2cos²a - 1) + (8cos⁴a - 8cos²a + 1))
Expanding the equation, we get
= 1/8(3 + 8cos²a - 4 + 8cos⁴a - 8cos²a + 1)
Simplifying, we get
= 1/8(8cos⁴a) = cos⁴a = L.H.S.
Hence, the given identity is proved.
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