Solving trigonometry identities by simplifying terms

Question

Verify the identity by simplifying the left side.

$\sin^2x-\sin^2y=\cos^2y-\cos^2x$

p.s. trigonometric identity this is the way but I wouldn't call it "simplifying". — , Dec 07 '13 at 14:00
Sum $(\sin (y))^2+(\cos(x))^2$to both sides of the equality. — Git Gud, Dec 07 '13 at 14:15

score 2 · Answer 1 · answered Dec 07 '13 at 14:22

2

We have $$\displaystyle \sin^2x+\cos^2x=\sin^2y+\cos^2y$$ as both are equal to $1$

Now change the sides of $\displaystyle \sin^2y,\cos^2x$

answered Dec 07 '13 at 14:22

lab bhattacharjee

Related : http://math.stackexchange.com/questions/510011/how-to-prove-cot-2x-sec-2x-tan-2x-csc-2x/510028#510028 – lab bhattacharjee Dec 07 '13 at 15:16

score 1 · Answer 2 · answered Dec 07 '13 at 14:17

1

This can be verified by using $\cos^2(x)=1-\sin^2(x)$ and $\cos^2(y)=1-\sin^2(y)$.

answered Dec 07 '13 at 14:17

robjohn

Madrit Zhaku · Answer 3 · 2013-12-07T14:25:49.190

1

$$\sin^2 x-\sin^2 y=1-\sin^2 y-(1-\sin^2 x)=\sin^2 x-\sin^2 y,$$ because $$\cos^2 x=1-\sin^2 x,$$ and $$\cos^2 y=1-\sin^2 y$$

edited Dec 07 '13 at 14:25

answered Dec 07 '13 at 14:18

Madrit Zhaku

3 Answers3