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anonymous

  • one year ago

Proving Trigonometric Identities

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  1. anonymous
    • one year ago
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    |dw:1436392327801:dw|

  2. Astrophysics
    • one year ago
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    Start working with the left side, \[\cos(2 \theta) = \cos^2 \theta - \sin^2 \theta\]

  3. anonymous
    • one year ago
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    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
    • one year ago
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    Ah, are you trying to use latex? Because I don't see it haha

  5. Astrophysics
    • one year ago
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    You can draw it out

  6. Astrophysics
    • one year ago
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    \[1 - {\sin^2(\theta) \over \cos ^2 (\theta)} + {\sin^2(\theta) \over \cos ^2 (\theta)} \] is this it?

  7. Astrophysics
    • one year ago
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    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
    • one year ago
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    If you continue to simplify it you should get \[\frac{ \cos^2(\theta)-\sin^2(\theta) }{ \cos^2(\theta)+\sin^2(\theta) }\]

  9. Astrophysics
    • one year ago
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    You can then work to get that on the left side using what I told you earlier

  10. anonymous
    • one year ago
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    how did it become cos - sin?

  11. anonymous
    • one year ago
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    is it cuz 1 - sin is cos

  12. anonymous
    • one year ago
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    and 1 - cos=sin

  13. Astrophysics
    • one year ago
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    Well I'm not going to use latex for this, I'll just draw it out haha |dw:1436393154153:dw|

  14. Astrophysics
    • one year ago
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    |dw:1436393252856:dw| similarly you can do the same to the numerator and simplify to what we have above

  15. anonymous
    • one year ago
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    where'd the extra cos come from?

  16. Astrophysics
    • one year ago
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    |dw:1436393543228:dw|

  17. Astrophysics
    • one year ago
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    What extra cos?

  18. Astrophysics
    • one year ago
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    I just used \[\frac{ a }{ b } \pm \frac{ c }{ d } \implies \frac{ ad \pm bc }{ bd }\]

  19. anonymous
    • one year ago
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    |dw:1436393625814:dw|

  20. Astrophysics
    • one year ago
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    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
    • one year ago
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    @mathstudent55

  22. mathstudent55
    • one year ago
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    \(\large \cos 2\theta ~~~~~~~~~~~~~~~~= \dfrac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \) \(\large \cos^2 \theta - \sin^2 \theta ~~= \dfrac{1 - \frac{\sin^2 \theta}{\cos^2 \theta}}{1 +\frac{\sin^2 \theta}{\cos^2 \theta} }\) Now multiply the rgight side by \(\dfrac{\cos^2 \theta}{\cos ^2 \theta} \)

  23. mathstudent55
    • one year ago
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    ^right side

  24. Astrophysics
    • one year ago
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    Same thing

  25. anonymous
    • one year ago
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    |dw:1436394083007:dw|

  26. anonymous
    • one year ago
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    |dw:1436394195835:dw|

  27. mathstudent55
    • one year ago
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    \(\large \cos^2 \theta - \sin^2 \theta ~~= \dfrac{1 - \frac{\sin^2 \theta}{\cos^2 \theta}}{1 +\frac{\sin^2 \theta}{\cos^2 \theta} } \times \dfrac{ \cos^2 \theta} {\cos^2 \theta}\) \(\large \cos^2 \theta - \sin^2 \theta ~~= \dfrac{\cos^2 \theta - \sin^2 \theta}{\color{red}{\cos^2 \theta + \sin^2 \theta}} \) 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
    • one year ago
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    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
    • one year ago
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    Thanks guys @campbell_st @Astrophysics @mathstudent55

  30. mathstudent55
    • one year ago
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    You're welcome.

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