Revision: Trigonometric Functions Maths and Stats HSC Science (General) 12th Standard Board Exam Maharashtra State Board

An equation involving trigonometric functions of a variable is called a trigonometric equation.

e.g. cos²θ − sinθ = 1/2, tan mθ = cot nθ, etc. are 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.

\[\mathrm{In~\Delta ABC,~\frac{a}{\sin A}=\frac{b}{sinB}=\frac{c}{sinC}}\]

In ΔABC,

i.  \(\mathrm{cos}A=\frac{\mathrm{b}^2+\mathrm{c}^2-\mathrm{a}^2}{2\mathrm{b}\mathrm{c}}\)

ii. \[\mathrm{cos}\mathrm{B}=\frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2\mathrm{c}\mathrm{a}}\]

iii. \[\mathrm{cos}\mathrm{C}=\frac{\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{c}^{2}}{2\mathrm{ab}}\]

• x = r cosθ
• y = r sinθ
• \[\tan\theta=\frac{y}{x}\]
• \[\mathbf{r}=\sqrt{x^2+y^2}\]

In ΔABC,

In ΔABC, if a + b + c = 2s, then

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

\[\sin\frac{\mathrm{B}}{2}=\sqrt{\frac{(\mathrm{s-c})(\mathrm{s-a})}{\mathrm{ca}}}\]

\[\sin\frac{\mathrm{C}}{2}=\sqrt{\frac{(\mathrm{s-a})(\mathrm{s-b})}{\mathrm{ab}}}\]

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

\[\cos\frac{\mathrm{B}}{2}=\sqrt{\frac{\mathrm{s(s-b)}}{\mathrm{ca}}}\]

\[\cos\frac{\mathrm{C}}{2}=\sqrt{\frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}}\]

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

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

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

In ΔABC,

i. \[\tan\left(\frac{\mathrm{A-B}}{2}\right)=\left(\frac{\mathrm{a-b}}{\mathrm{a+b}}\right)\cot\frac{\mathrm{C}}{2}\]

ii. \[\tan\left(\frac{\mathrm{B-C}}{2}\right)=\left(\frac{\mathrm{b-c}}{\mathrm{b+c}}\right)\cot\frac{\mathrm{A}}{2}\]

iii. \[\tan\left(\frac{\mathrm{C-A}}{2}\right)=\left(\frac{\mathrm{c-a}}{\mathrm{c+a}}\right)\cot\frac{\mathrm{B}}{2}\]

Area of ΔABC = \[\frac{1}{2}\mathrm{ab~sinC}\]

                         = \[=\frac{1}{2}\mathrm{bc~sinA}=\frac{1}{2}\mathrm{ac~sinB}\]

Heron’s Formula:

The area of ΔABC = \[\sqrt{\mathrm{s(s-a)(s-b)(s-c)}}\] 

where, 2s = a + b + c

Direct Identities

• sin⁻¹(sin θ) = θ, if −π/2 ≤ θ ≤ π/2
• cos⁻¹(cos θ) = θ, if 0 ≤ θ ≤ π
• tan⁻¹(tan θ) = θ, if −π/2 < θ < π/2

Inverse Identities

• sin(sin⁻¹x) = x, if −1 ≤ x ≤ 1
• cos(cos⁻¹x) = x, if −1 ≤ x ≤ 1
• tan(tan⁻¹x) = x, for all real x

Other Important Ones

• sec⁻¹(sec θ) = θ, if 0 ≤ θ ≤ π, θ ≠ π/2
• cosec⁻¹(cosec θ) = θ, if −π/2 ≤ θ ≤ π/2, θ ≠ 0
• cot⁻¹(cot θ) = θ, if 0 < θ < π

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

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

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

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

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

 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

b = ccosA + acosC

c = acos⁡B + bcos⁡A

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

By sine rule, `a/(sin A) = b/(sin B) = c/(sin C) = k`

∴ a = k sin A, b = k sin B, c = k sin C

RHS = `((a - b)/(a + b)) cot (C/2)`

= `((k sin A - k sin B)/(k sin A + k sin B)) cot(C/2)`

= `((sin A - sin B)/(sin A + sin B)) cot (C/2)`

= `(2 cos ((A + B)/2)*sin((A - B)/2))/(2 sin ((A + B)/2)*cos((A - B)/2)) xx (cos(C/2))/(sin(C/2))`

= `(cos(pi/2 - C/2)*sin((A - B)/2))/(sin(pi/2 - C/2)*cos((A - B)/2)) xx (cos (C/2))/(sin(C/2))`     ...[∵A + B + C = π]

= `(sin(C/2))/(cos(C/2)) xx tan ((A - B)/2) xx (cos (C/2))/(sin(C/2))`

= `tan ((A - B)/2)` = LHS

• x = r cos θ

• y = r sin θ

• r² = x² + y²

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