Math Formulas for Class 11 Chapter 3 Trigonometric Functions

Here we are providing Math Formulas for Class 11 Chapter 3 Trigonometric Functions. These formulas are important for all students studying maths as well as physics. Students studying physics also requires these formulas while deriving some physics formula and it is very useful for solving physics problems also. Students are suggested to remember these formulas.

RELATION BETWEEN SYSTEM OF MEASUREMENT OF ANGLES:

$$ \begin{aligned} & \frac{\mathrm{D}}{90}=\frac{\mathrm{G}}{100}=\frac{2 C}{\pi} \\ & 1 \text { Radian }=\frac{180}{\pi} \text { degree } \approx 57^{\circ} 17^{\prime} 15^{\prime \prime} \text { (approximately) } \\ & 1 \text { degree }=\frac{\pi}{180} \text { radian } \approx 0.0175 \text { radian } \end{aligned} $$

BASIC TRIGONOMETRIC IDENTITIES:

(a) $$\sin ^2 \theta+\cos ^2 \theta=1$$ or $$\sin ^2 \theta=1-\cos ^2 \theta$$ or $$\cos ^2 \theta=1-\sin ^2 \theta$$ (b) $$\sec ^2 \theta-\tan ^2 \theta=1$$ or $$\sec ^2 \theta=1+\tan ^2 \theta$$ or $$\tan ^2 \theta=\sec ^2 \theta-1$$ (c) If $$\sec \theta+\tan \theta=\mathrm{k} \\ \Rightarrow \sec \theta-\tan \theta=\frac{1}{\mathrm{k}} \\ \Rightarrow 2 \sec \theta=\mathrm{k}+\frac{1}{\mathrm{k}}$$ (d) $$\operatorname{cosec}^2 \theta-\cot ^2 \theta=1$$ or $$\operatorname{cosec}^2 \theta=1+\cot ^2 \theta$$ or $$\cot ^2 \theta=\operatorname{cosec}^2 \theta-1$$ (e) If $$\operatorname{cosec} \theta+\cot \theta=k \\ \Rightarrow \operatorname{cosec} \theta-\cot \theta=\frac{1}{k} \\ \Rightarrow 2 \operatorname{cosec} \theta=k+\frac{1}{k}$$

SIGNS OF TRIGONOMETRIC FUNCTIONS IN DIFFERENT QUADRANTS:

TRIGONOMETRIC FUNCTIONS OF ALLIED ANGLES:

$$ \begin{aligned} & \sin (2 \mathrm{n} \pi+\theta)=\sin \theta, \cos (2 \mathrm{n} \pi+\theta)=\cos \theta \text {, where } \mathrm{n} \in \mathrm{I}\\ & \begin{array}{ll} \sin (-\theta)=-\sin \theta & \\ \cos (-\theta)=\cos \theta \\ \sin \left(90^{\circ}-\theta\right)=\cos \theta & \\ \cos \left(90^{\circ}-\theta\right)=\sin \theta \\ \sin \left(90^{\circ}+\theta\right)=\cos \theta & \\ \cos \left(90^{\circ}+\theta\right)=-\sin \theta \\ \sin \left(180^{\circ}-\theta\right)=\sin \theta & \\ \cos \left(180^{\circ}-\theta\right)=-\cos \theta \\ \sin \left(180^{\circ}+\theta\right)=-\sin \theta & \\ \cos \left(180^{\circ}+\theta\right)=-\cos \theta \\ \sin \left(270^{\circ}-\theta\right)=-\cos \theta & \\ \cos \left(270^{\circ}-\theta\right)=-\sin \theta \\ \sin \left(270^{\circ}+\theta\right)=-\cos \theta & \\ \cos \left(270^{\circ}+\theta\right)=\sin \theta \end{array} \end{aligned} $$

IMPORTANT TRIGONOMETRIC FORMULAE:

(i) $$\sin (\mathrm{A}+\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B}$$ (ii) $$\quad \sin (A-B)=\sin A \cos B-\cos A \sin B$$ (iii) $$\cos (A+B)=\cos A \cos B-\sin A \sin B$$ (iv) $$\quad \cos (A-B)=\cos A \cos B+\sin A \sin B$$ (v) $$\tan (\mathrm{A}+\mathrm{B})=\frac{\tan \mathrm{A}+\tan \mathrm{B}}{1-\tan \mathrm{A} \tan \mathrm{B}}$$ (vi) $$\quad \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$$ (vii) $$\quad \cot (\mathrm{A}+\mathrm{B})=\frac{\cot \mathrm{B} \cot \mathrm{A}-1}{\cot \mathrm{B}+\cot \mathrm{A}}$$ (viii) $$\cot (\mathrm{A}-\mathrm{B})=\frac{\cot \mathrm{B} \cot \mathrm{A}+1}{\cot \mathrm{B}-\cot \mathrm{A}}$$ (ix) $$2 \sin A \cos B=\sin (A+B)+\sin (A-B)$$ (x) $$2 \cos A \sin B=\sin (A+B)-\sin (A-B)$$ (xi) $$2 \cos A \cos B=\cos (A+B)+\cos (A-B)$$ (xii) $$2 \sin A \sin B=\cos (A-B)-\cos (A+B)$$ (xiii) $$\sin \mathrm{C}+\sin \mathrm{D}=2 \sin \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \cos \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)$$ (xiv) $$\sin \mathrm{C}-\sin \mathrm{D}=2 \cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)$$ (xv) $$\quad \cos \mathrm{C}+\cos \mathrm{D}=2 \cos \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right) \cos \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)$$ (xvi) $$\quad \cos C-\cos D=2 \sin \left(\frac{C+D}{2}\right) \sin \left(\frac{D-C}{2}\right)$$ (xvii) $$\sin 2 \theta=2 \sin \theta \cos \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta}$$ (xviii) $$\cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta=2 \cos ^2 \theta-1=1-2 \sin ^2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}$$ (xix) $$1+\cos 2 \theta=2 \cos ^2 \theta$$ or $$|\cos \theta|=\sqrt{\frac{1+\cos 2 \theta}{2}}$$ (xx) $$1-\cos 2 \theta=2 \sin ^2 \theta$$ or $$|\sin \theta|=\sqrt{\frac{1-\cos 2 \theta}{2}}$$ (xxi) $$\tan \theta=\frac{1-\cos 2 \theta}{\sin 2 \theta}=\frac{\sin 2 \theta}{1+\cos 2 \theta}$$ (xxii) $$\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta}$$ (xxiii) $$\sin 3 \theta=3 \sin \theta-4 \sin ^3 \theta$$ (xxiv) $$\cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta$$ (xxv) $$\tan 3 \theta=\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}$$ (xxvi) $$\sin ^2 A-\sin ^2 B=\sin (A+B) \cdot \sin (A-B)=\cos ^2 B-\cos ^2 A$$ (xxvii) $$\cos ^2 A-\sin ^2 B=\cos (A+B) \cdot \cos (A-B)$$

Leave a Reply

Download Updated White Label Product Brochures (2023-24) 

%d bloggers like this:
search previous next tag category expand menu location phone mail time cart zoom edit close