Practice: Basic Trigonometric Identity

$\sin 0 = 0\ \Rightarrow\ \sin^2 0 = 0^2 = 0$

$\cos 90^\circ = 0\ \Rightarrow \ \cos^2 90^\circ = 0^2 = 0$

$0 + 0 = \boxed{0}$

00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 Yeah, the solution is zero. -_- ...... ;-)

We know, $sin(x)=cos(90-x)$

Therefore, $sin^2(0)+cos^2(90)=2\cdot cos(90)=0$

This is mosquito-nuking,really.Just figure out the values of $sin(90)$ and $cos(90)$ individually and plug them in.

sin 0=0; sin^2 0=0

cos 90=0 cos^2=0

0+0=0 answer

Sin (90-0)= cos0 according to that solve..

Silly question

sin0=0 cos90=0

sin 0=0 and cos 90=0 therefore answer is 0

sin^2 0 Deg + cos^2 90 Deg ->0^2+0^2 -> 0+0-> 0

it's question for little baby :p

$sin^{2}$ 0 = 0

Also, $cos^{2}$ 90 = 0

Hence the sum is 0

SIN 0 AND COS 90 BOTH ARE ZERO THEN THE ANSWER BECOME ZERO

sin 0 = 0 cos 90 = 0 Putting these values, (0) + (0) = 0

sin0=0 cos0=0

0 + 0 = 0

since value of sin 0 and cos 90 are zero the term becomes zero

As $\sin 0^{\circ} = 0$ and $\cos 90^{\circ} = 0$ , the sum of their squares will be $\boxed{0.}$ .