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Math Help - Derivative of sine in degrees?

  1. #1
    Member integral's Avatar
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    Derivative of sine in degrees?

    Derivative of sine in degrees?-matrix_result.jpg I was reading this and I could not figure out why \frac{d}{dx}sin(x)=\frac{\pi}{180}cos(x) in degrees.

    Could anyone explain this to me?
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  2. #2
    Master Of Puppets
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    Re: Derivative of sine in degrees?

    Did you consider

    \displaystyle \pi^c = 180^o \implies \frac{\pi^c}{180} = 1^o
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  3. #3
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    Re: Derivative of sine in degrees?

    Let \sin x and \cos x be the regular trigonometric functions, and let \mathop{\mathrm{sind}} x and \mathop{\mathrm{cosd}} x be functions that return the sine and cosine, respectively, of the angle of x degrees. That is, \mathop{\mathrm{sind}} x=\sin \frac{\pi x}{180} and \mathop{\mathrm{cosd}} x=\cos \frac{\pi x}{180}. For example, \mathop{\mathrm{sind}} 45 = \sin\frac{45\pi}{180}=\sin\frac{\pi}{4}=\sqrt{2}/2, as expected.

    Then, according to the chain rule, (\mathop{\mathrm{sind}} x)'=\left(\sin \frac{\pi x}{180}\right)'=\cos\frac{\pi x}{180}\cdot\frac{\pi}{180}=\mathop{\mathrm{cosd}} x\cdot\frac{\pi}{180}.
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  4. #4
    Member integral's Avatar
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    Re: Derivative of sine in degrees?

    Oh my goodness that makes so much sense. Thank you emakarov.
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