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續三角函數

Rad(弧度): π=180∘\pi=180^\circ

奇變偶不變 符號看象限

六個三角函數

sin⁡θ\sin\theta1sin⁡θ=cosec⁡θ\frac{1}{\sin\theta}=\cosec\theta
cos⁡θ\cos\theta1cos⁡θ=sec⁡θ\frac{1}{\cos\theta}=\sec\theta
tan⁡θ\tan\theta1tan⁡θ=cot⁡θ\frac{1}{\tan\theta}=\cot\theta

會提供

複角

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B

cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B

tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}

積化和

2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2 \sin A \cos B = \sin (A + B) + \sin (A - B)

2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2 \cos A \cos B = \cos (A + B) + \cos (A - B)

2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2 \sin A \sin B = \cos (A - B) - \cos (A + B)

和化積

sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A + \sin B = 2 \sin \frac{A + B}{2} \cos \frac{A - B}{2}

sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A - \sin B = 2 \cos \frac{A + B}{2} \sin \frac{A - B}{2}

cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A + \cos B = 2 \cos \frac{A + B}{2} \cos \frac{A - B}{2}

cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2\cos A - \cos B = -2 \sin \frac{A + B}{2} \sin \frac{A - B}{2}


不提供

三角恆等式

tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}

sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta

1+cot⁡2θ=cosec⁡2θ1 + \cot^2\theta = \cosec^2\theta

二倍角公式

sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2\sin\theta\cos\theta
cos⁡(2θ)=cos⁡2θ−sin⁡2θ\cos(2\theta) = \cos^2\theta - \sin^2\theta

cos⁡(2θ)=2cos⁡2θ−1\cos(2\theta) = 2\cos^2\theta - 1

cos⁡(2θ)=1−2sin⁡2θ\cos(2\theta) = 1 - 2\sin^2\theta
tan⁡(2θ)=2tan⁡θ1−tan⁡2θ\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}
sin⁡2θ=12(1−cos⁡2θ)\sin^2\theta = \frac12\left(1 - \cos 2\theta\right)

cos⁡2θ=12(1+cos⁡2θ)\cos^2\theta = \frac12\left(1 + \cos 2\theta\right)
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二項式定理