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anonymous

  • one year ago

sina-cosatanb/cosa+sinatanb = tan(a+b) Prove the identity

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  1. anonymous
    • one year ago
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    divide both the numerator and denominator by cosa

  2. welshfella
    • one year ago
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    are you sure the question is correct?

  3. anonymous
    • one year ago
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    Yeah, the question is correct

  4. anonymous
    • one year ago
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    Why divide cosa?

  5. anonymous
    • one year ago
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    What's the next step after that?

  6. welshfella
    • one year ago
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    I plugged in 2 values of a and b and didn't get the 2 sides to be equal

  7. anonymous
    • one year ago
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    Ohh

  8. anonymous
    • one year ago
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    What about this question? 2cos^23-1

  9. welshfella
    • one year ago
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    Yes there is a mistake in the first question.

  10. anonymous
    • one year ago
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    How do you solve 2cos^23-1

  11. welshfella
    • one year ago
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    is that supposed to be an equation? there is no variable

  12. anonymous
    • one year ago
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    I think you're meant to simplify it

  13. welshfella
    • one year ago
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    is the 3 degrees or radians?

  14. anonymous
    • one year ago
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    Woops I just realised I wrote the question wrong

  15. anonymous
    • one year ago
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    It's actually meant to be 2cos^23x-1

  16. anonymous
    • one year ago
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    Sorry

  17. anonymous
    • one year ago
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    X is the variable

  18. welshfella
    • one year ago
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    2 cos^2 3x = 1 so cos ^2 3x = 1/2 cos 3x = +/- 1 / sqrt2

  19. Hero
    • one year ago
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    Taking a look at your original question

  20. Hero
    • one year ago
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    The original question says to Prove The Identity, but the given trig equation is not an identity. Double check all the Variables to make sure they are correct.

  21. Hero
    • one year ago
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    Apparently the expression on the left side is equivalent to tan(a-b) not tan (a+b)

  22. Hero
    • one year ago
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    We can prove this identity: (sina-cosatanb )/(cosa+sinatanb) = tan(a - b)

  23. welshfella
    • one year ago
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    yes one way to do this would be to convert tan b to sin b / cos b and simplify

  24. welshfella
    • one year ago
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    and finally convert back to a tangent

  25. welshfella
    • one year ago
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    Hint: use compound angle formula for sin and cos

  26. Hero
    • one year ago
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    @welshfella give @Cococute a chance to respond first

  27. welshfella
    • one year ago
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    ok

  28. anonymous
    • one year ago
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    Thank you @welshfella and @Hero

  29. Hero
    • one year ago
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    YW

  30. anonymous
    • one year ago
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    I changed the tanb to sinb/cosb and got this: sina-cosaxsinb/cosb / cosa+sinaxsinb/cosb

  31. anonymous
    • one year ago
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    What's the next step?

  32. prizzyjade
    • one year ago
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    \[\frac{ \sin \alpha-\cos \alpha \tan \beta}{\cos \alpha+\sin \\[=\frac{ \frac{ \sin \alpha \cos \alpha }{ \cos^2 \beta }- \tan \beta }{ \cos^2 \alpha -\sin^2 \alpha \tan^2 \beta }\]alpha \tan \beta }= \tan(\alpha+\beta)\]\[\tan (\alpha+\beta)=\frac{ \sin \alpha -\cos \alpha \tan \beta}{ \cos \alpha+ \sin \alpha \tan \beta }\] \[=\frac{ \sin \alpha-\cos \alpha \tan \beta }{ \cos \alpha+ \sin \alpha \tan \beta } \times \frac{ \cos \alpha-\sin \alpha \tan \beta }{ \cos \alpha-\sin \alpha \tan \beta }\] \[=\frac{ \sin \alpha \cos \alpha-\sin ^{2}\alpha \tan \beta- \cos^2 \alpha \tan \beta +\cos \alpha \sin \alpha \tan^2 \beta }{ \cos^2 \alpha-\sin^2 \alpha \tan ^2 \beta}\] \[=\frac{ \sin \alpha \cos \alpha (1+\tan^2 \beta) -\tan \beta (\sin^2 \alpha + \cos^2 \alpha )}{ \cos^2 \alpha - \sin^2 \alpha \tan ^2 \beta}\]\[=\frac{ \sin \alpha \cos \alpha(\sec^2 \beta)- \tan \beta (1) }{ \cos^2 \alpha- \sin^2 \alpha \tan^2 \beta}\]

  33. prizzyjade
    • one year ago
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    \[=\frac{ \frac{ \sin \alpha \cos \alpha }{ \cos^2 \beta }- \tan \beta }{ \cos^2 \alpha - \sin^2 \alpha \tan^2 \beta }\] \[=\frac{\frac{ \sin \alpha \cos \alpha }{ \cos^2 \beta} -\frac{ \sin \beta}{ \cos \beta }}{\cos^2\alpha- \frac{ \sin^ 2 \alpha \sin^2 \beta }{\cos^2 \beta } }\]\[=\frac{ \frac{ \sin \alpha \cos \alpha- \sin \beta \cos \beta }{ \cos^2 \beta } }{ \frac{ \cos^2 \alpha \cos^2 \beta - \sin^2 \alpha \sin ^2 \beta }{\cos^2 \beta } }\]\[=\frac{ \sin \alpha \cos \alpha -\sin \beta \cos \beta }{ \cos^2 \alpha \cos^2 \beta - \sin^2 \alpha \sin^2 \beta }\]\[=\frac{ \frac{ \tan \alpha }{ 1+\tan^2 \alpha } - \frac{ \tan \beta }{ 1+ \tan^\beta } }{ 2-\sin^2 \alpha - \sin^ 2 \beta + \sin^2 \alpha \sin^2 \beta-\sin^2 \alpha \sin^2 \beta}\]

  34. anonymous
    • one year ago
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    work from left, sub in the addition formula for tan(A+B)

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