How do you solve sin^2x-cos^2x?

Question

solomonzelman · Answer

It is a double angle identity:
$\large\color{slate}{  \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)        }$ 

what happens when $\large\color{slate}{  a   }$ and $\large\color{slate}{  b      }$ are same?

bossimbacon · Answer

can you help me solomon

solomonzelman · Answer

you could have messaged, but don't do this in this question, it is his/her question...

anonymous · Answer

When a and b are the same, it is 1.

anonymous · Answer

Is that correct?

solomonzelman · Answer

it is not 1,

tell me:
$\large\color{slate}{ \cos(x)\times\cos(x)=?  }$ 
$\large\color{slate}{ \sin(x)\times\sin(c)=?  }$

solomonzelman · Answer

For the second one I meant:
$\large\color{slate}{  \sin(x)\times\sin(x)        }$

anonymous · Answer

cos^2x 
sin^2x

solomonzelman · Answer

yes.

solomonzelman · Answer

$\large\color{slate}{  \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)        }$ 

when a and b are same,

$\large\color{slate}{  \cos(2x)=\cos(x+x)=\cos(x)\cos(x)-\sin(x)\sin(x) \=\cos^2(x)\sin^2(x)        }$

solomonzelman · Answer

oh, the last step another typo

solomonzelman · Answer

it should say cos^2(x)-sin^2(x)

solomonzelman · Answer

$\large\color{slate}{  \cos(2x)=\cos^2(x)-\sin^2(x)                   }$

solomonzelman · Answer

so your answer is?

anonymous · Answer

1

anonymous · Answer

1-cos^2x