Question

help .-.

Answer 1

\[\frac{ \sin^2x-1 }{ \cos(-x) }\]

Answer 2

simplify

Answer 3

cos(-x)=cosx
1=sin^2x+cos^2x
Therefore, it becomes sin^2x-sin^2x+cos^2x/cosx
=cos^2x/cosx=cosx

Answer 4

u can use
\[\large \sin^2x+\cos^2=1 \]
and the fact that \(\cos x\) is an even function.

Answer 5

i feel like your picture and username are connected because it looks like fruitcake and your picture looks like a circuit board so i just keep thinking to you give people circuitboard fruitcakes at Christmas time....

Answer 6

and your answers wrong ^.^

Answer 7

@helder_edwin how would i go about putting that formula into my equation?

Answer 8

first of all yeas my answer is wrong because there must be a minus before the cos^2x so the result becomes -cos(x)
And I like your perspective i didnt think like this before now :)

Answer 9

\[\large \frac{\sin^2x-1}{\cos(-x)}=\frac{-\cos^2x}{\cos x}=-\cos x \]

Answer 10

yay you both got the same answer c: okay kagitucak you give edwin a medal and ill give you a medal c: sound fair?

Answer 11

okay :)

Answer 12

you both helped equal so one of you give the other a medal thats only fair c: