Fun side of trigonometry

Geometry Level 2

If the measures of angles of a triangle is x,y and 90 then find , $\sin^2x+\sin^2y-(\cos^2x+\cos^2y)$ =

1 0 -1 can't be determined

3 solutions

Gabriel Gomes
Jun 26, 2016

Since the angles are 90, x and y, and knowing that the sum of interior angles on a triangle is 180, we can state that $y = 90 - x$ .

On the given equation, we have:

$\sin^{2}x + \sin^{2} (90-x) - \cos^{2} x - \cos^{2} (90-x)$

$\sin^{2} x + \cos^{2} x - \cos^{2} x - \sin^{2} x$

$1 - 1$

$\boxed{0}$

Thanks for sharing your approach, Gabriel. You have explained it nicely. I upvoted your solution (+1) :)

Pranshu Gaba - 4 years, 11 months ago

Robin Kashyap
Oct 19, 2016

We have identity cos2x=(cosx)^2-(sinx)^2

Let value of above expression be Z

Z= -cos2x-cos2y

Z= -cos2x + cos2y (since x+y=90) Z=0

Kiran Patil
Oct 24, 2014

i considered angles x and y as a 45. then substituted values of the cos 45= sin 45= 1/2. after that i got solution as a zero.