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Mathematics LibreTexts

Fundamental Trigonometric Identities

  • Page ID
    619
  • [ "article:topic" ]

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    "arc" Identities

    \[\arctan\theta=\tan^{-1}\theta\]

    \[\arcsin\theta=\sin^{-1}\theta\]

    \[\arccos\theta=\cos^{-1}\theta\]

    Quotient and reciprocal identities

    \[\tan\theta=\dfrac{\sin\theta}{\cos\theta}\]

    \[\cot\theta=\dfrac{\cos\theta}{\sin\theta}= \dfrac{\csc\theta}{\sec\theta}= \dfrac{1}{\tan\theta}\]

    \[\sec\theta=\dfrac{1}{\cos\theta}\]

    \[\csc\theta=\dfrac{1}{\sin\theta}\]

    Cofunction Function identities

    \[\sin\theta = \cos(\dfrac{\pi}{2} - \theta)\]

    \[\cos\theta = \sin(\dfrac{\pi}{2} - \theta)\]

    Even/Odd Functions

    \[\cos(-\theta) = \cos(\theta)\]

    \[\sin(-\theta) = -\sin(\theta)\]

    Pythagorean identities

    \[\sin^2\theta+\cos^2\theta=1\]

    \[\tan^2\theta+1=\sec^2\theta\]

    \[1+\cot^2\theta=\csc^2\theta\]

    Angle sum and difference identities

    \[\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha\]

    \[\sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha\]

    \[\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta\]

    \[\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta\]

    \[\tan(\alpha+\beta) = \dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\]

    \[\tan(\alpha-\beta) = \dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}\]

    Double-angle identities

    \[\sin2\theta=2\sin\theta\cos\theta\]

    \[\cos2\theta=\cos^2\theta-\sin^2\theta = 2\cos^2\theta-1 = 1-2\sin^2\theta\]

    \[\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}\]

    Half-angle identities

    \[\sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{2}}\]

    \[\cos\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1+\cos\theta}{2}}\]

    \[\tan\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}} = \dfrac{\sin\theta}{1+\cos\theta} = \dfrac{1-\cos\theta}{\sin\theta}\]

    Reduction formulas

    \[\sin^2\theta=\dfrac{1-\cos2\theta}{2}\]

    \[\cos^2\theta=\dfrac{1+\cos2\theta}{2}\]

    \[\tan^2\theta=\dfrac{1-\cos2\theta}{1+\cos2\theta} = \dfrac{\sin2\theta}{1+\cos2\theta} = \dfrac{1-\cos2\theta}{\sin2\theta}\]