A geometry problem by Charlton Teo

$-\dfrac {\sin^4{\theta} +\cos^4 {\theta} -1 } {2}$

$= -\dfrac {\sin^2{\theta} (1 - \cos^2 {\theta}) +\cos^2 {\theta} (1 - \sin^2 {\theta}) -1 } {2}$

$= -\dfrac {\sin^2{\theta} - \sin^2 {\theta} \cos^2 {\theta} +\cos^2 {\theta} - \sin^2 {\theta} \cos^2 {\theta} -1 } {2}$

$= -\dfrac {1 - 2 \sin^2 {\theta} \cos^2 {\theta} -1 } {2}$

$= \boxed {\sin^2 {\theta} \cos^2 {\theta}}$

The expression can be written as $\large -\frac{(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta• \cos^2\theta-1}{2}$ $= -\frac{-2 \sin^2\theta• \cos^2\theta}{2}=\huge\boxed{\sin^2 \theta •\cos^2 \theta}$ Simpil!!

Plug in $\theta = 0$ . The first answer choice is undefined and the rest are nonzero.

Distribute the -1 in and we get:

$\frac { 1-{ sin }^{ 4 }\theta -{ cos }^{ 4 }\theta }{ 2 } =\frac { (1-sin^{ 2 }\theta )(1+sin^{ 2 }\theta )-cos^{ 4 }\theta }{ 2 }$

Since $1-sin^{ 2 }\theta = cos^{2}\theta$ , we can factor that out:

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

Finally, since $1-cos^{2}\theta = sin^{2}\theta$ , we get:

$\frac { cos^{ 2 }\theta (sin^{ 2 }\theta +sin^{ 2 }\theta ) }{ 2 } =\boxed{sin^{ 2 }\theta cos^{ 2 }\theta}$