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Thread: Trigonometric Functions and Inverse Trigonometric Functions

  1. #1
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    Trigonometric Functions and Inverse Trigonometric Functions

    How would you derive the relationships between the trigonometric functions and the inverse trigonometric functions, such as:

    \sin(\arccos(x)) = \sqrt{1 - x^2}

    \sin(\arctan(x)) = \frac{x}{\sqrt{1 + x^2}}
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    Quote Originally Posted by NOX Andrew View Post
    How would you derive the relationships between the trigonometric functions and the inverse trigonometric functions, such as:

    \sin(\arccos(x)) = \sqrt{1 - x^2}

    \sin(\arctan(x)) = \frac{x}{\sqrt{1 + x^2}}
    Hi

    Using \cos^2(\arccos(x))+\sin^2(\arccos(x)) = 1 and \cos(\arccos(x)) = x

    it comes x^2+\sin^2(\arccos(x)) = 1
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  3. #3
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    Quote Originally Posted by NOX Andrew View Post
    How would you derive the relationships between the trigonometric functions and the inverse trigonometric functions, such as:

    \sin(\arccos(x)) = \sqrt{1 - x^2}

    \sin(\arctan(x)) = \frac{x}{\sqrt{1 + x^2}}
    Use \sin^2 \alpha = \frac{\tan^2 \alpha}{1+\tan^2 \alpha} applied to \alpha = \arctan(x)
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