# 1-2sina*cosa÷2=sin^2(45^@-a)?

Feb 15, 2018

See below for a possible answer.

#### Explanation:

${\sin}^{2} \left(45 - a\right) = \sin \left(45 - a\right) \sin \left(45 - a\right)$

Using the angle subtraction formula for sine:

$\sin \left(45 - a\right) \sin \left(45 - a\right) = \left(\sin 45 \cos a - \sin a \cos 45\right) \left(\sin 45 \cos a - \sin a \cos 45\right)$

$\sin 45 = \cos 45 = \frac{\sqrt{2}}{2}$ So:

$\left(\frac{\sqrt{2}}{2} \cos a - \frac{\sqrt{2}}{2} \sin a\right) \left(\frac{\sqrt{2}}{2} \cos a - \frac{\sqrt{2}}{2} \sin a\right)$

Factor out the $\frac{\sqrt{2}}{2}$ in both expressions:

$\frac{\sqrt{2}}{2} \left(\cos a - \sin a\right) \frac{\sqrt{2}}{2} \left(\cos a - \sin a\right)$

Multiply the square roots and the trig expressions:

$\frac{2}{4} \left({\cos}^{2} a - 2 \sin a \cos a + {\sin}^{2} a\right)$

Simplify fraction and rearrange in parentheses:

$\frac{1}{2} \left({\sin}^{2} a + {\cos}^{2} a - 2 \sin a \cos a\right)$

${\sin}^{2} a + {\cos}^{2} a = 1$ therefore:

${\sin}^{2} \left(45 - a\right) = \frac{1 - 2 \sin a \cos a}{2}$

QED

(Don't know how to get the degrees mark. I'll edit it if I figure it out, but all of the 45s above are 45degrees)