# Question b8b69

Nov 2, 2016

yes,for $x = \frac{\pi}{4} , {\sin}^{2} x + {\cos}^{2} x = 2 \sin x \cos x$

#### Explanation:

here for$x = \frac{\pi}{4} , \sin x = \cos x = \left(\frac{1}{\sqrt{2}}\right)$
so ,$\sin x = \cos x$
$\sin x - \cos x = 0$
squaring both sides we get${\left(\sin x - \cos x\right)}^{2} = 0$
or ,${\sin}^{2} x + {\cos}^{2} x - 2 \sin x \cos x = 0$
or${\sin}^{2} x + {\cos}^{2} x = 2 \sin x \cos x$ proved.

Nov 5, 2016

While this relation is true for certain values of $x$ it is not valid in general.
I can not be proven as an identity because it is not, in general, true.

#### Explanation:

${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$ (this is an extension of the Pythagorean Theorem)

For x=0
color(white)("XXX")2sin(x)⋅cos(x)=2×0×1=0

sin^2(x)+cos^2(x)≠2sin(x)⋅cos(x)#