If cosθ + cos^2 θ = 1,
then
sin^12 θ + 3 sin^10 θ + 3 sin^8 θ+ sin^6 θ + 2 sin^4 θ + 2 sin^2 θ – 2 =?
since sin^2 + cos^2 = 1,
sin^2 θ = cos θ
since cosθ + cos^2 θ = 1,
cosθ = ±(√5-1)/2
Just plug that in for sin^2 θ, and you have
((√5-1)/2)^6 + 3((√5-1)/2)^5 + 3((√5-1)/2)^4 + ((√5-1)/2)^3 + 2((√5-1)/2)^2 + 2((√5-1)/2) - 2
= 1
Thank you wolframalpha!
Now, how can we get that for ourselves?
If you let x = sin^2 θ, the above is
x^3 (x+1)^3 + 2(x^2+x-1)
= ((√5-1)/2)^3 ((√5+1)/2)^3 + 2(x^2+x-1)
But, since (√5-1)/2 is a root of x^2+x-1 = 0 (from our original condition),
= (5-1)^3/64 + 2(0)
= 1
To find the value of sin^12 θ + 3 sin^10 θ + 3 sin^8 θ + sin^6 θ + 2 sin^4 θ + 2 sin^2 θ - 2, we can use the Pythagorean identity and the given equation cosθ + cos^2 θ = 1.
First, let's rewrite sin^12 θ + 3 sin^10 θ + 3 sin^8 θ + sin^6 θ + 2 sin^4 θ + 2 sin^2 θ - 2 in terms of cosθ.
Using the Pythagorean identity, sin^2 θ = 1 - cos^2 θ, we can express each power of sin as a function of cos:
sin^12 θ = (sin^2 θ)^6 = (1 - cos^2 θ)^6,
sin^10 θ = (sin^2 θ)^5 = (1 - cos^2 θ)^5,
sin^8 θ = (sin^2 θ)^4 = (1 - cos^2 θ)^4,
sin^6 θ = (sin^2 θ)^3 = (1 - cos^2 θ)^3,
sin^4 θ = (sin^2 θ)^2 = (1 - cos^2 θ)^2,
sin^2 θ = 1 - cos^2 θ.
Now we can substitute these expressions into the given equation:
sin^12 θ + 3 sin^10 θ + 3 sin^8 θ + sin^6 θ + 2 sin^4 θ + 2 sin^2 θ - 2
= (1 - cos^2 θ)^6 + 3(1 - cos^2 θ)^5 + 3(1 - cos^2 θ)^4 + (1 - cos^2 θ)^3 + 2(1 - cos^2 θ)^2 + 2(1 - cos^2 θ) - 2.
Now let's simplify this expression:
Using the binomial theorem, we can expand each term:
= (1 - 6cos^2 θ + 15cos^4 θ - 20cos^6 θ + 15cos^8 θ - 6cos^10 θ + cos^12 θ)
+ 3(1 - 5cos^2 θ + 10cos^4 θ - 10cos^6 θ + 5cos^8 θ - cos^10 θ)
+ 3(1 - 4cos^2 θ + 6cos^4 θ - 4cos^6 θ + cos^8 θ)
+ (1 - 3cos^2 θ + 3cos^4 θ - cos^6 θ)
+ 2(1 - 2cos^2 θ + cos^4 θ)
+ 2(1 - cos^2 θ)
- 2.
Now combine like terms:
= 1 - 6cos^2 θ + 15cos^4 θ - 20cos^6 θ + 15cos^8 θ - 6cos^10 θ + cos^12 θ
+ 3 - 15cos^2 θ + 30cos^4 θ - 30cos^6 θ + 15cos^8 θ - 3cos^10 θ
+ 3 - 12cos^2 θ + 18cos^4 θ - 12cos^6 θ + 3cos^8 θ
+ 1 - 3cos^2 θ + 3cos^4 θ - cos^6 θ
+ 2 - 4cos^2 θ + 2cos^4 θ
+ 2 - 2cos^2 θ
- 2.
Now simplify further:
= cos^12 θ - 6cos^2 θ - 3cos^10 θ + 12cos^4 θ - 8cos^6 θ + 33cos^8 θ - 2cos^6 θ - 2cos^2 θ.
Next, we can use the given equation cosθ + cos^2 θ = 1 to substitute for cos^2 θ:
cosθ + cos^2 θ = 1
cosθ = 1 - cos^2 θ.
Now substitute this equation into the expression:
= cos^12 θ - 6(1 - cos^2 θ) - 3cos^10 θ + 12cos^4 θ - 8cos^6 θ + 33cos^8 θ - 2cos^6 θ - 2cos^2 θ
= cos^12 θ - 6 + 6cos^2 θ - 3cos^10 θ + 12cos^4 θ - 8cos^6 θ + 33cos^8 θ - 2cos^6 θ - 2cos^2 θ
= cos^12 θ + 6cos^2 θ - 3cos^10 θ + 10cos^4 θ - 10cos^6 θ + 33cos^8 θ.
Therefore, sin^12 θ + 3 sin^10 θ + 3 sin^8 θ + sin^6 θ + 2 sin^4 θ + 2 sin^2 θ - 2 is equivalent to cos^12 θ + 6cos^2 θ - 3cos^10 θ + 10cos^4 θ - 10cos^6 θ + 33cos^8 θ.