Revision: 12th Std >> Trigonometric Functions MAH-MHT CET (PCM/PCB) Trigonometric Functions

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.

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

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

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

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

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

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

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

i. \[\sin^{-1}\frac{1}{x}=\mathrm{cosec}^{-1}\] if x ≥ 1 or x ≤ −1
\[\cos^{-1}\frac{1}{x}=\sec^{-1}x\] if x ≥ 1 or x ≤ −1
\[\tan^{-1}\frac{1}{x}=\cot^{-1}x\] if x > 0

ii. sin⁻¹(−x) = −sin⁻¹x, for x ∈ [−1, 1]
tan⁻¹(−x) = −tan⁻¹x, for x ∈ R
cosec⁻¹(−x) = −cosec⁻¹x, for x ≥ 1
cos⁻¹(−x) = π − cos⁻¹x, for x ∈ [−1, 1]
sec⁻¹(−x) = π − sec⁻¹x, for x ≥ 1
cot⁻¹(−x) = π − cot⁻¹x, for x ∈ R

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

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

\[\sec^{-1}x+\cos\sec^{-1}x=\frac{\pi}{2},\] for |x| ≥ 1

\[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\frac{x+y}{1-xy}\right),\] for x > 0, y > 0 and xy < 1

\[\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right),\] for x, y > 0 and xy > 1

\[\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right),\] for x, y > 0

\[2\tan^{-1}x=\sin^{-1}\left(\frac{2x}{1+x^{2}}\right),\] if −1 ≤ x ≤ 1

\[2\tan^{-1}x=\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right),\] if x > 0

\[2\tan^{-1}x=\tan^{-1}\left(\frac{2x}{1-x^{2}}\right),\] if −1 < x < 1