## anonymous one year ago Simplify the expression: Sin^2x-1/cos(-x)

1. SolomonZelman

Have you seen the following identity? $$\large\color{black}{ \displaystyle \sin^2\theta+\cos^2\theta=1 }$$

2. anonymous

Yes

3. SolomonZelman

Subtract $$\sin^2\theta$$ from both sides and tell me what you get.

4. SolomonZelman

$$\large\color{black}{ \displaystyle \sin^2\theta+\cos^2\theta =1 \ }$$ $$\large\color{black}{ \displaystyle \sin^2\theta+\cos^2\theta \color{blue}{-\sin^2\theta }=1 \color{blue}{-\sin^2\theta } }$$ $$\large\color{black}{ \displaystyle \cos^2\theta =1-\sin^2\theta }$$

5. SolomonZelman

So, based on the very last statement, you can now simplify your expression: $$\large\color{black}{ \displaystyle \frac{1-\sin^2x}{\cos x} =\frac{?}{\cos x} }$$

6. SolomonZelman

hello?

7. anonymous

Sinx?

8. SolomonZelman

We have concluded that: $$1-\sin^2\theta =\cos^2\theta$$ correct?

9. anonymous

Yes so the ? Represents cos^2 theta?

10. SolomonZelman

Yes, therefore it comes out that: $$\large\color{black}{ \displaystyle \frac{1-\sin^2x}{\cos x} =\frac{\cos^2x}{\cos x} }$$

11. SolomonZelman

and THAT $$\Uparrow$$ you can probably tell me how to simplify:D

12. anonymous

So the answer is cos x? :)

13. SolomonZelman

yes, just $$\cos x$$.

14. SolomonZelman

Any questions?

15. anonymous

Thank you for your help! i understand it now :D

16. SolomonZelman

Yes, one more note: $$\color{red}{\cos(-\theta)=\cos(\theta)}$$

17. SolomonZelman

So, lets quicly recap everything once more, okay?

18. SolomonZelman

$$\large\color{black}{ \displaystyle \frac{1-\sin^2(x)}{\cos (-x)} = }$$ $$\large\color{black}{ \displaystyle \frac{1-\sin^2(x)}{\cos (x)} = }$$ $$\large\color{black}{ \displaystyle \frac{\cos^2(x)}{\cos (x)} = \cos(x) }$$