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rahrah

  • 2 years ago

(cosx)(tanx+sinx cotx)=sinx+cos^2x Prove the identity

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  1. RD⁴²
    • 2 years ago
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    Are you sure it's not (cosx)(tanx+sinx cosx) = sinx + sinx cos²x ?

  2. rahrah
    • 2 years ago
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    oh crap, its actually (cosx)(tanx+sinx cotx)=sinx+cos^2x

  3. RD⁴²
    • 2 years ago
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    We know that \(\tan x = \dfrac{\sin x}{\cos x}\) and \(\cot x = \dfrac{\cos x}{\sin x}\) So \((\cos x)(\tan x+\sin x \cot x) = (\cos x)\left(\dfrac{\sin x}{\cos x} + \sin x\cdot\dfrac{\cos x}{\sin x}\right)\) \(=(\cos x)\left(\dfrac{\sin x}{\cos x} + \cos x\right)\) \( = \boxed{\sin x + \cos^2 x}\)

  4. RD⁴²
    • 2 years ago
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    \[\Huge\text{Q.E.D.}\]

  5. rahrah
    • 2 years ago
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    Q.E.D. ? Thanks a lot for the help otherwise

  6. RD⁴²
    • 2 years ago
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    http://en.wikipedia.org/wiki/Q.E.D.

  7. rahrah
    • 2 years ago
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    wait, how'd you get sinx + cos^2 at the end? shouldn't sinx/cosx cross out the + cosx?

  8. RD⁴²
    • 2 years ago
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    No, because it's just addition. you can't cancel cos x out like that.|dw:1362452530886:dw|

  9. rahrah
    • 2 years ago
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    shouldnt that make the end result be tanx cosx + (cosx)^2? how did we get sin from the tan?

  10. RD⁴²
    • 2 years ago
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    \(\tan x = \dfrac{\sin x}{\cos x}\) right? So \(\tan x\cos x\ = \dfrac{\sin x}{\cos x}\cdot \cos x = \sin x\) And cos²x is just another way to write (cos x)²

  11. rahrah
    • 2 years ago
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    oh, i get it now. Thanks a lot for all your help!

  12. RD⁴²
    • 2 years ago
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    Glad I helped.

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