Proving Trigonometric Identities

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Proving Trigonometric Identities

Mathematics
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|dw:1436392327801:dw|
Start working with the left side, \[\cos(2 \theta) = \cos^2 \theta - \sin^2 \theta\]
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]}

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Other answers:

Ah, are you trying to use latex? Because I don't see it haha
You can draw it out
\[1 - {\sin^2(\theta) \over \cos ^2 (\theta)} + {\sin^2(\theta) \over \cos ^2 (\theta)} \] is this it?
For the right side you should have \[\large \frac{ -\left( \frac{ \sin(\theta) }{ \cos(\theta) } \right)^2+1 }{ \left( \frac{ \sin(\theta) }{ \cos(\theta) } \right)^2+1 }\]
If you continue to simplify it you should get \[\frac{ \cos^2(\theta)-\sin^2(\theta) }{ \cos^2(\theta)+\sin^2(\theta) }\]
You can then work to get that on the left side using what I told you earlier
how did it become cos - sin?
is it cuz 1 - sin is cos
and 1 - cos=sin
Well I'm not going to use latex for this, I'll just draw it out haha |dw:1436393154153:dw|
|dw:1436393252856:dw| similarly you can do the same to the numerator and simplify to what we have above
where'd the extra cos come from?
|dw:1436393543228:dw|
What extra cos?
I just used \[\frac{ a }{ b } \pm \frac{ c }{ d } \implies \frac{ ad \pm bc }{ bd }\]
|dw:1436393625814:dw|
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
\(\large \cos 2\theta ~~~~~~~~~~~~~~~~= \dfrac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \) \(\large \cos^2 \theta - \sin^2 \theta ~~= \dfrac{1 - \frac{\sin^2 \theta}{\cos^2 \theta}}{1 +\frac{\sin^2 \theta}{\cos^2 \theta} }\) Now multiply the rgight side by \(\dfrac{\cos^2 \theta}{\cos ^2 \theta} \)
^right side
Same thing
|dw:1436394083007:dw|
|dw:1436394195835:dw|
\(\large \cos^2 \theta - \sin^2 \theta ~~= \dfrac{1 - \frac{\sin^2 \theta}{\cos^2 \theta}}{1 +\frac{\sin^2 \theta}{\cos^2 \theta} } \times \dfrac{ \cos^2 \theta} {\cos^2 \theta}\) \(\large \cos^2 \theta - \sin^2 \theta ~~= \dfrac{\cos^2 \theta - \sin^2 \theta}{\color{red}{\cos^2 \theta + \sin^2 \theta}} \) We use the identity: \(\cos^2 \theta + \sin^2 \theta = 1\) \(\large \cos^2 \theta - \sin^2 \theta ~~= \dfrac{\cos^2 \theta - \sin^2 \theta}{\color{red}{1}} \) \(\large \cos^2 \theta - \sin^2 \theta ~~= \cos^2 \theta - \sin^2 \theta \)
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 \[\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)\]
You're welcome.

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