Trigonometry Compound Angle/ Multiple Angle Identities/Range of functions
Compound Angle:
$$ \sin(A+B)=\sin A\cos B+\cos A\sin B $$
$$ \sin(A-B)=\sin A\cos B-\cos A\sin B $$
$$ \cos(A+B)=\cos A\cos B-\sin A\sin B $$
$$ \cos(A-B)=\cos A\cos B+\sin A\sin B $$
$$ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} $$
$$ \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B} $$ Multiple Angle:
$$ \sin2\theta = 2\sin\theta\cos\theta $$
$$ \cos2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1 = \cos^2\theta - \sin^2\theta $$
$$ \sin3\theta = 3\sin\theta - 4\sin^3\theta $$
$$ \cos3\theta = 4\cos^3\theta - 3\cos\theta $$
$$ \tan2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} $$
Range:
$$ -1 \le \sin\theta \le 1 \;\;\implies\;\; \sin\theta \in [-1,\,1] $$
$$ -1 \le \cos\theta \le 1 \;\;\implies\;\; \cos\theta \in [-1,\,1] $$
$$ -\infty < \tan\theta < \infty \;\;\implies\;\; \tan\theta \in (-\infty,\,\infty) \;\;\implies\;\; \tan\theta \in \mathbb{R} $$
$$ -\infty < \cot\theta < \infty \;\;\implies\;\; \cot\theta \in \mathbb{R} $$
$$ |\sec\theta|\ge1 \;\;\implies\;\; \sec\theta \in (-\infty,\,-1] \,\cup\,[1,\,\infty) $$
$$ |\csc\theta|\ge1 \;\;\implies\;\; \csc\theta \in (-\infty,\,-1] \,\cup\,[1,\,\infty) $$
$$ \bigl|a\sin x + b\cos x\bigr| \le \sqrt{a^2 + b^2} $$
$$ -\,\sqrt{a^2 + b^2}\;\le\;a\sin x + b\cos x\;\le\;\sqrt{a^2 + b^2} $$
$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R $$
where \(R\) is the radius of the circumscribed circle of \(\triangle ABC\), given by
$$ R \;=\;\frac{a}{2\sin A}\;=\;\frac{b}{2\sin B}\;=\;\frac{c}{2\sin C}. $$
R is the radius of the circle which passes through all vertices of the triangle.
$$ \sin(A+B)=\sin A\cos B+\cos A\sin B $$
$$ \sin(A-B)=\sin A\cos B-\cos A\sin B $$
$$ \cos(A+B)=\cos A\cos B-\sin A\sin B $$
$$ \cos(A-B)=\cos A\cos B+\sin A\sin B $$
$$ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} $$
$$ \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B} $$ Multiple Angle:
$$ \sin2\theta = 2\sin\theta\cos\theta $$
$$ \cos2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1 = \cos^2\theta - \sin^2\theta $$
$$ \sin3\theta = 3\sin\theta - 4\sin^3\theta $$
$$ \cos3\theta = 4\cos^3\theta - 3\cos\theta $$
$$ \tan2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} $$
Range:
$$ -1 \le \sin\theta \le 1 \;\;\implies\;\; \sin\theta \in [-1,\,1] $$
$$ -1 \le \cos\theta \le 1 \;\;\implies\;\; \cos\theta \in [-1,\,1] $$
$$ -\infty < \tan\theta < \infty \;\;\implies\;\; \tan\theta \in (-\infty,\,\infty) \;\;\implies\;\; \tan\theta \in \mathbb{R} $$
$$ -\infty < \cot\theta < \infty \;\;\implies\;\; \cot\theta \in \mathbb{R} $$
$$ |\sec\theta|\ge1 \;\;\implies\;\; \sec\theta \in (-\infty,\,-1] \,\cup\,[1,\,\infty) $$
$$ |\csc\theta|\ge1 \;\;\implies\;\; \csc\theta \in (-\infty,\,-1] \,\cup\,[1,\,\infty) $$
$$ \bigl|a\sin x + b\cos x\bigr| \le \sqrt{a^2 + b^2} $$
$$ -\,\sqrt{a^2 + b^2}\;\le\;a\sin x + b\cos x\;\le\;\sqrt{a^2 + b^2} $$
$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R $$
where \(R\) is the radius of the circumscribed circle of \(\triangle ABC\), given by
$$ R \;=\;\frac{a}{2\sin A}\;=\;\frac{b}{2\sin B}\;=\;\frac{c}{2\sin C}. $$
R is the radius of the circle which passes through all vertices of the triangle.
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