# Please Bare with Me.. Can you prove this identity?

• Apr 18th 2010, 09:20 PM
Please Bare with Me.. Can you prove this identity?
1 - tanx sinx = cos2x + sin²x

I think i finally created a proper iidentity.. I just need somebody to try and prove it!

Thanks
• Apr 18th 2010, 09:30 PM
Quote:

1 - tanx sinx = cos2x + sin²x

RHS
cos2x = cos²x - sin²x

Therefore, the RHS = cos²x

LHS
1 - tanx sinx = cos²x + sin²x - tanx sinx

LHS =! RHS unless sin²x = tanx sinx
• Apr 18th 2010, 09:36 PM
Im not sure what you mean?
• Apr 18th 2010, 09:41 PM
harish21
Quote:

1 - tanx sinx = cos2x + sin²x

I think i finally created a proper iidentity.. I just need somebody to try and prove it!

Thanks

$\displaystyle LHS = 1 - (tanx \times sinx)$

$\displaystyle = 1 - \frac{sinx}{cosx} \times sinx$

$\displaystyle = 1-\frac{sin^2x}{cosx}$

$\displaystyle RHS = cos2x + sin^2x = cos^2x-sin^2x+sin^2x =cos^2x$

So the identity you have created cannot be equal.

however, if you change your LHS to :

$\displaystyle 1 - (tanx \times sinx \times cosx)$

$\displaystyle = 1 - (\frac{sinx}{cosx} \times sinx \times cosx)$

$\displaystyle = 1-(sinx \times sinx) = 1-(sin^2x) = cos^2x$

In this case, your RHS will be equal to LHS giving you a proper identity.