# How does cosx^2 -1 = -sinx^2? Q concerning limits.

• May 2nd 2013, 03:34 PM
lukasaurus
How does cosx^2 -1 = -sinx^2? Q concerning limits.
http://i.imgur.com/pvhhgrA.jpg

How does that work? I've tried it with various values on the calculator and it's not equal. eg cos4 - 1 is not equal to -sin4 (if theta = 2). If neccesary, I can post the entire question. It concerns limits.
• May 2nd 2013, 03:37 PM
MarkFL
Re: How does cosx^2 -1 = -sinx^2? Q concerning limits.
Are you squaring the trig. functions when you test with your calculator?
• May 2nd 2013, 03:54 PM
lukasaurus
Re: How does cosx^2 -1 = -sinx^2? Q concerning limits.
I've got an old fx-82. I don't have any option to square the actual trig function, but I should be able to square the x value right?
let theta = x (because it's quicker to write)
so let x = 2

cos^2(x) -1 = cos4 -1 = -0.002
-sin^2(x) = -sin4 = -0.07

No idea what I am doing wrong..

edit: Actually, it should be (cosx)^2 right? not what I am doing. I think I got it.
• May 2nd 2013, 05:32 PM
HallsofIvy
Re: How does cosx^2 -1 = -sinx^2? Q concerning limits.
You may be confusing \$\displaystyle cos^2(x)\$ with \$\displaystyle cos(x^2)\$.

If x= 2 radians, then cos(x)= cos(2)= -0.416147 so that \$\displaystyle cos^2(2)= 0.17318\$.

But \$\displaystyle cos(x^2)= cos(4)= -0.6536436\$.