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anonymous
 one year ago
Proving Trigonometric Identities
anonymous
 one year ago
Proving Trigonometric Identities

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436392327801:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Start working with the left side, \[\cos(2 \theta) = \cos^2 \theta  \sin^2 \theta\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I broke down the right side to 1  {(sin^2[theta] over cos ^2 [theta]} divided by 1 + {(sin^2[theta] over cos ^2 [theta]}

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Ah, are you trying to use latex? Because I don't see it haha

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0You can draw it out

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0\[1  {\sin^2(\theta) \over \cos ^2 (\theta)} + {\sin^2(\theta) \over \cos ^2 (\theta)} \] is this it?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0For the right side you should have \[\large \frac{ \left( \frac{ \sin(\theta) }{ \cos(\theta) } \right)^2+1 }{ \left( \frac{ \sin(\theta) }{ \cos(\theta) } \right)^2+1 }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0If you continue to simplify it you should get \[\frac{ \cos^2(\theta)\sin^2(\theta) }{ \cos^2(\theta)+\sin^2(\theta) }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0You can then work to get that on the left side using what I told you earlier

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did it become cos  sin?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it cuz 1  sin is cos

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Well I'm not going to use latex for this, I'll just draw it out haha dw:1436393154153:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436393252856:dw similarly you can do the same to the numerator and simplify to what we have above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where'd the extra cos come from?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436393543228:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I just used \[\frac{ a }{ b } \pm \frac{ c }{ d } \implies \frac{ ad \pm bc }{ bd }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436393625814:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I 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

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \cos 2\theta ~~~~~~~~~~~~~~~~= \dfrac{1  \tan^2 \theta}{1 + \tan^2 \theta} \) \(\large \cos^2 \theta  \sin^2 \theta ~~= \dfrac{1  \frac{\sin^2 \theta}{\cos^2 \theta}}{1 +\frac{\sin^2 \theta}{\cos^2 \theta} }\) Now multiply the rgight side by \(\dfrac{\cos^2 \theta}{\cos ^2 \theta} \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436394083007:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436394195835:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \cos^2 \theta  \sin^2 \theta ~~= \dfrac{1  \frac{\sin^2 \theta}{\cos^2 \theta}}{1 +\frac{\sin^2 \theta}{\cos^2 \theta} } \times \dfrac{ \cos^2 \theta} {\cos^2 \theta}\) \(\large \cos^2 \theta  \sin^2 \theta ~~= \dfrac{\cos^2 \theta  \sin^2 \theta}{\color{red}{\cos^2 \theta + \sin^2 \theta}} \) We use the identity: \(\cos^2 \theta + \sin^2 \theta = 1\) \(\large \cos^2 \theta  \sin^2 \theta ~~= \dfrac{\cos^2 \theta  \sin^2 \theta}{\color{red}{1}} \) \(\large \cos^2 \theta  \sin^2 \theta ~~= \cos^2 \theta  \sin^2 \theta \)

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0wow... 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 \[\cos(2\theta) = \cos^2(\theta)  \sin^2(\theta)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks guys @campbell_st @Astrophysics @mathstudent55
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