## anonymous one year ago Proving Trigonometric Identities

1. anonymous

|dw:1436392327801:dw|

2. Astrophysics

Start working with the left side, $\cos(2 \theta) = \cos^2 \theta - \sin^2 \theta$

3. anonymous

I broke down the right side to 1 - {(sin^2[theta] over cos ^2 [theta]} divided by 1 + {(sin^2[theta] over cos ^2 [theta]}

4. Astrophysics

Ah, are you trying to use latex? Because I don't see it haha

5. Astrophysics

You can draw it out

6. Astrophysics

$1 - {\sin^2(\theta) \over \cos ^2 (\theta)} + {\sin^2(\theta) \over \cos ^2 (\theta)}$ is this it?

7. Astrophysics

For the right side you should have $\large \frac{ -\left( \frac{ \sin(\theta) }{ \cos(\theta) } \right)^2+1 }{ \left( \frac{ \sin(\theta) }{ \cos(\theta) } \right)^2+1 }$

8. Astrophysics

If you continue to simplify it you should get $\frac{ \cos^2(\theta)-\sin^2(\theta) }{ \cos^2(\theta)+\sin^2(\theta) }$

9. Astrophysics

You can then work to get that on the left side using what I told you earlier

10. anonymous

how did it become cos - sin?

11. anonymous

is it cuz 1 - sin is cos

12. anonymous

and 1 - cos=sin

13. Astrophysics

Well I'm not going to use latex for this, I'll just draw it out haha |dw:1436393154153:dw|

14. Astrophysics

|dw:1436393252856:dw| similarly you can do the same to the numerator and simplify to what we have above

15. anonymous

where'd the extra cos come from?

16. Astrophysics

|dw:1436393543228:dw|

17. Astrophysics

What extra cos?

18. Astrophysics

I just used $\frac{ a }{ b } \pm \frac{ c }{ d } \implies \frac{ ad \pm bc }{ bd }$

19. anonymous

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20. Astrophysics

I think you should just play around and see what you get, this one is a bit messy, so the only way you'll learn is by messing around with the algebra and trig functions

21. anonymous

@mathstudent55

22. mathstudent55

$$\large \cos 2\theta ~~~~~~~~~~~~~~~~= \dfrac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$$ $$\large \cos^2 \theta - \sin^2 \theta ~~= \dfrac{1 - \frac{\sin^2 \theta}{\cos^2 \theta}}{1 +\frac{\sin^2 \theta}{\cos^2 \theta} }$$ Now multiply the rgight side by $$\dfrac{\cos^2 \theta}{\cos ^2 \theta}$$

23. mathstudent55

^right side

24. Astrophysics

Same thing

25. anonymous

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26. anonymous

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27. mathstudent55

$$\large \cos^2 \theta - \sin^2 \theta ~~= \dfrac{1 - \frac{\sin^2 \theta}{\cos^2 \theta}}{1 +\frac{\sin^2 \theta}{\cos^2 \theta} } \times \dfrac{ \cos^2 \theta} {\cos^2 \theta}$$ $$\large \cos^2 \theta - \sin^2 \theta ~~= \dfrac{\cos^2 \theta - \sin^2 \theta}{\color{red}{\cos^2 \theta + \sin^2 \theta}}$$ We use the identity: $$\cos^2 \theta + \sin^2 \theta = 1$$ $$\large \cos^2 \theta - \sin^2 \theta ~~= \dfrac{\cos^2 \theta - \sin^2 \theta}{\color{red}{1}}$$ $$\large \cos^2 \theta - \sin^2 \theta ~~= \cos^2 \theta - \sin^2 \theta$$

28. campbell_st

wow... what a long and troublesome problem start with $\sin^2 + \cos^2 = 1$ divide every term by cos^2 $\tan^2 + 1 = \frac{1}{cos^2}$ so your problem becomes $\cos(2\theta) = \frac{1 - \tan^2(\theta)}{\frac{1}{\cos^2(\theta)}}$ remember, dividing by a fraction, flip and multiply $\cos(2\theta) = (1 - \tan^2(\theta)) \times \frac{\cos^2(\theta)}{1}$ and you get $\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$

29. anonymous

Thanks guys @campbell_st @Astrophysics @mathstudent55

30. mathstudent55

You're welcome.