SOLUTION: If cos^2 x + cos x = 1. so, sin^12 x + 3 sin^10 x + 3 sin^8 x + sin^6 x = ?

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Question 980725: If cos^2 x + cos x = 1. so, sin^12 x + 3 sin^10 x + 3 sin^8 x + sin^6 x = ?
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
If cos^2 x + cos x = 1. so, sin^12 x + 3 sin^10 x + 3 sin^8 x + sin^6 x = ?
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Solve for cos(x)
cos^2 x + cos x = 1
cos^2 x + cos x - 1 = 0
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Sub u for cos(x)
u^2 + u - 1 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B1x%2B-1+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A1%2A-1=5.

Discriminant d=5 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+5+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+5+%29%29%2F2%5C1+=+0.618033988749895
x%5B2%5D+=+%28-%281%29-sqrt%28+5+%29%29%2F2%5C1+=+-1.61803398874989

Quadratic expression 1x%5E2%2B1x%2B-1 can be factored:
1x%5E2%2B1x%2B-1+=+%28x-0.618033988749895%29%2A%28x--1.61803398874989%29
Again, the answer is: 0.618033988749895, -1.61803398874989. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-1+%29

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cos(x) = -1/2 + sqrt(5)/2
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Find sin^2(x)
sin^2 + cos^2 = 1
sin^2(x) = 1 - (1/4 - sqrt(5)/2 + 5/4) = -1/2 - sqrt(5)/2
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sin^12 x + 3 sin^10 x + 3 sin^8 x + sin^6 x = sin^6(x)*(sin^6 + 3sin^4 + 3sin^2 + 1)
= sin^6(x)*(sin^2 + 1)^3
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sin^6 = (sin^2)^3 = (1 + sqrt(5))/2
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(1/2)*(1 + sqrt(5))*(-1/2)*(1 - sqrt(5))
= -(1/4)*(1 - 5)
= 1
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Do the same for the other root, cos(x) = -1/2 - sqrt(5)/2