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

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) (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 (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=5 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.618033988749895, -1.61803398874989. Here's your graph:

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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
======================
Do the same for the other root, cos(x) = -1/2 - sqrt(5)/2
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