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# Need Help on Trig Identity

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Thanks in advance if someone is there.

Veteran  Mar 18, 2017
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#1
+1

Verify the following identity:
3/8 - (cos(2 x))/(2) + (cos(4 x))/(8) = sin(x)^4

Put 3/8 - 1/2 cos(2 x) + 1/8 cos(4 x) over the common denominator 8: 3/8 - 1/2 cos(2 x) + 1/8 cos(4 x) = (3 - 4 cos(2 x) + cos(4 x))/8:
(3 - 4 cos(2 x) + cos(4 x))/8 = ^?sin(x)^4

Multiply both sides by 8:
3 - 4 cos(2 x) + cos(4 x) = ^?8 sin(x)^4

cos(2 x) = 1 - 2 sin(x)^2:
3 - 41 - 2 sin(x)^2 + cos(4 x) = ^?8 sin(x)^4

-4 (1 - 2 sin(x)^2) = 8 sin(x)^2 - 4:
3 + 8 sin(x)^2 - 4 + cos(4 x) = ^?8 sin(x)^4

cos(4 x) = 1 - 2 sin(2 x)^2:
3 - 4 + 8 sin(x)^2 + 1 - 2 sin(2 x)^2 = ^?8 sin(x)^4

sin(2 x) = 2 sin(x) cos(x):
3 - 4 + 8 sin(x)^2 + 1 - 2 2 cos(x) sin(x)^2 = ^?8 sin(x)^4

Multiply each exponent in 2 sin(x) cos(x) by 2:
3 - 4 + 8 sin(x)^2 + 1 - 24 cos(x)^2 sin(x)^2 = ^?8 sin(x)^4

cos(x)^2 = 1 - sin(x)^2:
3 - 4 + 8 sin(x)^2 + 1 - 2×4 1 - sin(x)^2 sin(x)^2 = ^?8 sin(x)^4

4 (1 - sin(x)^2) sin(x)^2 = 4 sin(x)^2 - 4 sin(x)^4:
3 - 4 + 8 sin(x)^2 + 1 - 24 sin(x)^2 - 4 sin(x)^4 = ^?8 sin(x)^4

3 - 4 + 8 sin(x)^2 + 1 - 2 (4 sin(x)^2 - 4 sin(x)^4) = 8 sin(x)^4:
8 sin(x)^4 = ^?8 sin(x)^4

The left hand side and right hand side are identical:
Answer: |(identity has been verified)

Guest Mar 18, 2017
#2
+26366
+1

Guest#1 beat me to it, but here's my answer anyway:

Alan  Mar 18, 2017
#3
+8836
+1

Thanks in advance if someone is there.

Omi67  Mar 18, 2017
#4
+8836
0
Omi67  Mar 18, 2017

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