- $S=r\theta$ [Arc = Radius $\times$ Angle in radians]
- Area of a sector $A=\frac{1}{2}r^2\theta$
- Fundamental identities of trigonometry
- $\cos^2\theta+\sin^2\theta=1$
- $1+\tan^2\theta=\sec^2\theta$
- $\cot^2\theta+1=cosec^2\theta$
-
$\theta$$0^0$$30^0$$45^0$$60^0$$90^0$$180^0$$\sin \theta$$0$$\frac{1}{2}$$\frac{1}{\sqrt{2}}$$\frac{\sqrt{3}}{2}$$1$$0$$\cos \theta$$1$$\frac{\sqrt{3}}{2}$$\frac{1}{\sqrt{2}}$$\frac{1}{2}$$0$$-1$$\tan \theta$$0$$\frac{1}{\sqrt{3}}$$1$$\sqrt{3}$$\infty$$0$
- $\sin(90^0-\theta)=\cos \theta$
$\cos(90^0-\theta)=\sin \theta$
- $\sin(-\theta)=-\sin \theta$
$\cos(-\theta)=\cos \theta$
- $\sin(180^0-\theta)=\sin \theta$
$\cos(180^0-\theta)=-\cos\theta$
- $\sin(90^0+\theta)=\cos \theta$
$\cos(90^0+\theta)=-\sin \theta$
- $\sin(180^0+\theta)=-\sin \theta$
$\cos(180^0+\theta)=-\cos\theta$
- $\sin(A\pm B)=\sin(A)\cos(B)\pm \cos(A)\sin(B)$
$\cos(A\pm B)=\cos(A)\cos(B)\mp \sin(A)\sin(B)$
- $\color{green}{\sin(2A)=2\sin(A)\cos(A)}$
- $\color{green}{\cos(2A)=\cos^2(A)-\sin^2(A)=2\cos^2(A)-1=1-2\sin^2(A)}$
- $\color{green}{\tan(2A)=\cfrac{2\tan(A)}{1-\tan^2(A)}}$
- $\color{blue}{\sin(3A)=3\sin(A)-4\sin^3(A)}$
- $\color{blue}{\cos(3A)=4\cos^3(A)-3\cos(A)}$
- $\color{blue}{\tan(3A)=\cfrac{3\tan(A)-\tan^3(A)}{1-3\tan^2(A)}}$
- $\sin C+\sin D=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2} \right)$
- $\sin C-\sin D=2\cos\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)$
- $\cos C+\cos D=2\cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$
- $\cos C-\cos D=2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{D-C}{2}\right)$
- $2\sin{ }A \cos{ }B=\sin(A+B)+\sin(A-B)$
- $2\cos A\sin B=\sin(A+B)-\sin(A-B)$
- $2\cos A\cos B=\cos(A+B)+\cos(A-B)$
- $2\sin A\sin B=\cos(A-B)-\cos(A+B)$
- If $\sin\theta=\sin\alpha$, then $\theta=n\pi+(-1)^n\alpha$
If $\cos\theta=\cos\alpha$, then $\theta=2n\pi\pm\alpha$
If $\tan\theta=\tan\alpha$, then $\theta=n\pi+\alpha$
Here $n\in\mathbb{Z}$
- For an $ABC$ triangle, in the standard notation,
- $\cfrac{\sin A}{a}=\cfrac{\sin B}{b}=\cfrac{\sin C}{c}$
- $\cos A=\cfrac{b^2+c^2-a^2}{2bc}$
- $a=b$ $\cos C+c$ $\cos B$
- $\tan\left(\cfrac{B-C}{2}\right)=\left(\cfrac{b-c}{b+c}\right)\cot\cfrac{A}{2}$
- $\cos\cfrac{A}{2}=\sqrt{\cfrac{s(s-a)}{bc}}$
Here $s$ is the double of perimeter, of the $ABC$ triangle $(2s=a+b+c)$
Mathematics has become a hard subject for most students. This blog tries to provide some theories of mathematics simply, which might easy to understand. There are lot of ares in mathematics and this blog will NOT cover all the areas as this is only to provide some basic knowledge of certain mathematical concepts. Please enable JAVA Scripts to see the equations properly.
Sunday, September 6, 2015
Trigonometry 01 (Useful Formulas in Trigonometry)
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