मराठी

Overview of Trigonometric Functions

Advertisements

Topics

Estimated time: 22 minutes
Maharashtra State Board: Class 12

Definition: Trigonometric Equations

Trigonometric Equations:

An equation involving trigonometric functions (or functions) is called a trigonometric equation.

Solution of the Trigonometric Equation:

A value of a variable in a trigonometric equation which satisfies the equation is called a solution of the trigonometric equation.

Maharashtra State Board: Class 12

Formula: Trigonometric Equations

General Solutions

  • sin θ = sin α ⇒ θ = nπ + (−1)ⁿα

  • cos θ = cos α ⇒ θ = 2nπ ± α

  • tan θ = tan α ⇒ θ = nπ + α

Special Results

  • sin θ = 0 ⇒ θ = nπ

  • cos θ = 0 ⇒ θ = (2n + 1)π/2

  • tan θ = 0 ⇒ θ = nπ

Squared Forms

  • sin²θ = sin²α ⇒ θ = nπ + α

  • cos²θ = cos²α ⇒ θ = nπ + α

  • tan²θ = tan²α ⇒ θ = nπ + α

Maharashtra State Board: Class 12

Key Points: Polar Co-ordinates

  • x = r cos θ

  • y = r sin θ

  • r² = x² + y²

\[r=\sqrt{x^2+y^2}\]

Maharashtra State Board: Class 12

Formula: Solution of Triangle

 The Sine Rule:

 \[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\]

The Cosine Rule:

\[a^2=b^2+c^2-2bc\cos A\]

\[b^2=c^2+a^2-2ca\cos B\]

\[c^2=a^2+b^2-2ab\cos C\]

Also:

\[\cos A=\frac{b^2+c^2-a^2}{2bc}\]

The projection Rule:

a = bcosC + ccosB

c = acos⁡B + bcos⁡A

Maharashtra State Board: Class 12

Formula: Half Angle Formulas

\[\sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}}\]

\[\cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}}\]

\[\tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\]

Maharashtra State Board: Class 12

Formula: Heron’s Formula

\[\mathrm{Area}=\sqrt{s(s-a)(s-b)(s-c)}\]

Maharashtra State Board: Class 12

Formula: Napier’s Analogy

\[\begin{aligned}
\tan\frac{B-C}{2}=\frac{b-c}{b+c}\cot\frac{A}{2}
\end{aligned}\]

Maharashtra State Board: Class 12

Key Point: Principal Value Ranges

Function Principal Range
sin⁻¹x \[[-\frac{\pi}{2},\frac{\pi}{2}]\]
cos⁻¹x \[[0,\pi]\]
tan⁻¹x \[(-\frac{\pi}{2},\frac{\pi}{2})\]
cot⁻¹x \[(0,\pi)\]
sec⁻¹x \[[0,\pi]-\{\frac{\pi}{2}\}\]
cosec⁻¹x \[[-\frac{\pi}{2},\frac{\pi}{2}]-\{0\}\]
Maharashtra State Board: Class 12

Formula: Identities

\[\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\]

\[\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\]

\[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\frac{x+y}{1-xy}\right)\quad(xy<1)\]

\[\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)\] (xy>1)

\[\tan^{-1}x+\tan^{-1}y=\frac{\pi}{2}\]

\[\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right)\]

Maharashtra State Board: Class 12

Formula: Symmetry Properties

\[\sin^{-1}(-x)=-\sin^{-1}x\]

\[\cos^{-1}(-x)=\pi-\cos^{-1}x\]

\[\tan^{-1}(-x)=-\tan^{-1}x\]

\[\mathrm{cosec}^{-1}(-x)=-\mathrm{cosec}^{-1}(x)\]

\[\sec^{-1}(-x)=\pi-\sec^{-1}(x)\]

\[\cot^{-1}(-x)=\pi-\cot^{-1}(x)\]

Maharashtra State Board: Class 12

Formula: Reciprocal Relations

\[\sin^{-1}x=\mathrm{cosec}^{-1}\left(\frac{1}{x}\right)\]

\[\cos^{-1}x=\sec^{-1}\left(\frac{1}{x}\right)\]

\[\tan^{-1}x=\cot^{-1}\left(\frac{1}{x}\right)\]

Advertisements
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×