Whats the value ?

We find that $cos(A) + cos^{2}(A) = cos(A) \cdot (1 + cos(A)) = 0 \Rightarrow cos(A) = 0, -1$ . Now, we find that:

$sin^{2}(A) - sin^{4}(A) = sin^{2}(A) \cdot (1 - sin^{2}(A)) = (1 - cos^{2}(A)) \cdot (cos^{2}(A)) = \boxed{0}.$

• Note that where $n$ is a positive integer greater than 1, $(\sin\theta)^{n} = \sin^{n}\theta$ and $\sin^{n}\theta = (\sin\theta)^{n}$ .

(By addition/subtraction property of equality) Subtract $\cos^{}A$ from both sides of the given equation $\cos^{}A + \cos^{2}A = 0$ .

$\Rightarrow \cos^{}A + \cos^{2}A - \cos^{}A = 0 - \cos^{}A$

$\Rightarrow \cos^{2}A = -\cos^{}A$ …………[1]

Use the Pythagorean identity $\sin^{2}x + \cos^{2}x = 1$ .

$\Rightarrow \sin^{2}A + \cos^{2}A = 1$

$\Rightarrow \sin^{2}A = 1 - \cos^{2}A$ ……[2]

(By law of substitution) Substitute [1] into [2].

$\Rightarrow \sin^{2}A = 1 - (-\cos^{}A)$

$\Rightarrow \sin^{2}A = 1 + \cos^{}A$ ………[3]

(By properties of powers and exponents) $\sin^{4}A = (\sin^{2}A)^{2}$ …[4]

(By substitution) Substitute [3] into the RHS (right-hand side) of equation [4].

$\Rightarrow \sin^{4}A = (1 + \cos^{}A)^{2}$ …[5]

Substitute [3] and [5] into the expression $\sin^{2}A - \sin^{4}A$ , whose value the problem is asking for, and then evaluate.

$\sin^{2}A - \sin^{4}A$

$= (1 + \cos^{}A) - (1 + \cos^{}A)^{2}$

$= 1 + \cos^{}A - (1 + 2\cos^{}A + \cos^{2}A)$

$= 1 + \cos^{}A - 1 - 2\cos^{}A - \cos^{2}A$

$= 1 - 1 + \cos^{}A - 2\cos^{}A - \cos^{2}A$

$= -\cos^{}A - \cos^{2}A$

$= -(\cos^{}A + \cos^{2}A)$ ………[7]

Substitute the given equation of the problem into [7].

$\sin^{2}A - \sin^{4}A$

$= -(0)$

$= \boxed{0}$

Since $\cos^{}A + \cos^{2}A = 0$ , $\sin^{2}A - \sin^{4}A$ is also equal to $\boxed{zero}$ .