## anonymous one year ago True or False? tan^2x= 1+ cos2x / 1-cos2x

1. SolomonZelman

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

2. SolomonZelman

verify or prove ?

3. P0sitr0n

Multiply both sides by 1-cos^2 x and simplify , keeping in mind that $tan(x)=\frac{sin(x)}{cos(x)}$

4. anonymous

It does not specify on the question itself but I would assume proving.

5. SolomonZelman

(verify = manipulate 1 side at a time only prove= both sides can be manipulate) this is a traditional interpretation/definition of these words.

6. SolomonZelman

ok, lets just in case verify it. will play 1 side at a time

7. anonymous

Ok

8. SolomonZelman

$$\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }$$ can you re-write the bottom ?

9. SolomonZelman

use the fact that $$\large\color{black}{ \displaystyle \sin^2(x)+\cos^2(x)=1 }$$

10. anonymous

So you would be able to re-write the bottem

11. SolomonZelman

yes, in the fact I just offered, subtract cos^2(x) from both sides, and tell me what you get

12. anonymous

From the sin^2(x) one?

13. SolomonZelman

yes, from the sin^2(x)+cos^2(x)=1, there subtract cos^2(x) from both sides

14. anonymous

$\sin ^{2}(x)=-\cos ^{2}(x)+1$

15. SolomonZelman

yes sin^2(x)=1-cos^2(x)

16. SolomonZelman

now, we have $$\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }$$ can you re-write the bottom for me?

That is false As 1+cos2x=1+2cos^2x-1 =2cos^2x And 1-cos2x=1-(1-2sin^2x) =2sin^2x Put these values you will get the answer.

18. Loser66

just put x = pi/4 , you can see how it is false

19. anonymous

When I rearrange this @SolomonZelman, will I be taking the -cos^2(x) away from the denominator and onto the numerator? I am not very good at trig

20. SolomonZelman

$$\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }$$ $$\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{\sin^2(x)} }$$ that is what I meant

21. SolomonZelman

anyway, i think plugging in value to disprove is valid enough.

22. SolomonZelman

unless u want to go and do it without plug ins

23. anonymous

Would I just have to plug in pi/4 to the equation?

But that's not a proper method

25. anonymous

Oh ok

26. anonymous

Well thank you guys for the help and time

27. Loser66

Who said : proving the invalid of an expression by a counterexample is not a proper method?

28. Loser66

That is the MOST proper method to prove something wrong.

29. SolomonZelman

$$\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }$$ $$\large\color{black}{ \displaystyle \frac{\sin^2x}{\cos^2x}=\frac{ (1+\cos^2x) }{(1-\cos^2x)} }$$ $$\large\color{black}{ \displaystyle \frac{\sin^2x}{\cos^2x}=\frac{ (1+\cos^2x) }{(1-\cos^2x)} }$$ $$\large\color{black}{ \displaystyle \sin^2x(1-\cos^2x)=\cos^2x(1+\cos^2x) }$$ $$\large\color{black}{ \displaystyle \sin^2x-\sin^2x\cos^2x=\cos^2x+\cos^2x\cos^2x}$$ $$\large\color{black}{ \displaystyle \sin^2x=\sin^2x\cos^2x+\cos^2x+\cos^2x\cos^2x}$$ $$\large\color{black}{ \displaystyle \sin^2x=\cos^2x(\sin^2x+1+\cos^2x)}$$ $$\large\color{black}{ \displaystyle \sin^2x=\cos^2x(2)}$$ $$\large\color{black}{ \displaystyle \sin^2x=2(1- \sin^2x)}$$ $$\large\color{black}{ \displaystyle \sin^2x=2- 2\sin^2x}$$ $$\large\color{black}{ \displaystyle 3\sin^2x=2}$$ $$\large\color{black}{ \displaystyle \sin^2x=2/3}$$ $$\large\color{black}{ \displaystyle \sin x=\pm\sqrt{2/3}}$$ $$\large\color{black}{ \displaystyle x=\sin^{-1}\left(\pm\sqrt{2/3}\right)}$$

30. SolomonZelman

if your definition were to be true for all numbers, then this solution for x would be such that includes all numbers. $$\large\color{black}{ \displaystyle \sin^{-1}\left(\pm\sqrt{2/3}\right)}$$ is no way a representation of all [real] numbers!

31. anonymous

*Takes a picture for future reference* Thank you for the explanation! This will help me 200x better than before!

32. SolomonZelman

i took everything down piece by piece. it isn't that bad usually though