Short proof (trig function)

Question

Prove that $\sin^2x=1/2(1-\cos(2x))$ is true for every $x \in \Bbb R$.

Useful information: $\cos(2x)=\cos^2x -\sin^2x$

Answer 1

Consider that $sin^{2}(x) + cos^{2}(x) = 1$ and $cos(2x) = cos^{2}(x) - sin^{2}(x)$, we have

$$1 = cos^{2}(x) + sin^{2}(x) \Rightarrow$$ $$1 - cos^{2}(x) = sin^{2}(x) \Rightarrow$$ $$1 - (cos(2x) + sin^{2}(x)) = sin^{2}(x) \Rightarrow$$ $$1 - cos(2x) - sin^{2}(x) = sin^{2}(x)\Rightarrow$$ $$1 - cos(2x) = 2sin^{2}(x) \Rightarrow$$ $$sin^{2}(x) = \frac{1-cos(2x)}{2}$$

Answer 2

hint

$$\sin^2(x)=\frac{\sin^2(x)+\sin^2(x)}{2}$$

$$=\frac{\sin^2(x)+1-\cos^2(x)}{2}$$ $$=\frac {1-(\cos(x)\cos(x)-\sin(x)\sin(x))}{2}$$

Answer 3

Hint: From the given formula, you deduce, with Pythagoras, the other two standard formulæ: $$\cos 2x=2\cos^2x-1=1-2\sin^2x,$$ whence the linearisation formulæ.