if cosθ = tanθ then
sinθ = cos^2θ
sinθ = 1 - sin^2θ
sin^2θ = 1 - sinθ
so,
2sinθ +sin^2θ +sin^3θ +sin^4θ
= 2 - 2sin^2θ + 1 - sinθ + sinθ(1 - sinθ) + (1-sinθ)^2
= 3 - 2sinθ + 1 - 2sin^2θ
= 3 - 2sinθ + sinθ - 1 + sinθ
= 2
sinθ = cos^2θ
sinθ = 1 - sin^2θ
sin^2θ = 1 - sinθ
so,
2sinθ +sin^2θ +sin^3θ +sin^4θ
= 2 - 2sin^2θ + 1 - sinθ + sinθ(1 - sinθ) + (1-sinθ)^2
= 3 - 2sinθ + 1 - 2sin^2θ
= 3 - 2sinθ + sinθ - 1 + sinθ
= 2
so
sin θ = cos^2 θ
so
first two terms
2 cos^2 + sin^2 = cos^2 + 1
last two terms
sin^2 (sin + sin^2) = sin^2 (sin + 1 - cos^2) but cos^2 = sin
so sin^2 (sin + 1 - sin) = sin^2
so the sum is
cos^2 + 1 + sin^2 = 1+1 = 2
Using the definitions of cosine and tangent, we can rewrite the equation as:
cosθ = sinθ / cosθ
Cross-multiplying gives us:
cos^2θ = sinθ
Since cosine squared is equal to 1 - sin squared (from the Pythagorean identity), we can substitute that in:
1 - sin^2θ = sinθ
Rearranging the equation, we get:
sin^2θ + sinθ - 1 = 0
Now, we have a quadratic equation in terms of sinθ. We can solve this equation using factoring or the quadratic formula.
For simplicity, let's substitute a = sinθ:
a^2 + a - 1 = 0
Factoring won't work in this case, so we will use the quadratic formula:
a = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values:
a = (-(1) ± √(1^2 - 4(1)(-1))) / 2(1)
Simplifying further:
a = (-1 ± √(1 + 4)) / 2
a = (-1 ± √5) / 2
We have two possible solutions for a (sinθ):
a1 = (-1 + √5) / 2
a2 = (-1 - √5) / 2
Now, we can find the value of 2sinθ + sin^2θ + sin^3θ + sin^4θ by substituting each value of a back into the expression.
Let's start with a1:
2sinθ + sin^2θ + sin^3θ + sin^4θ
= 2(a1) + (a1)^2 + (a1)^3 + (a1)^4
= 2((-1 + √5) / 2) + ((-1 + √5) / 2)^2 + ((-1 + √5) / 2)^3 + ((-1 + √5) / 2)^4
Similarly, we can calculate the value for a2.
After calculating the values for both a1 and a2, we can check which option (A, B, C, D) matches the final result.