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

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

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

Let `cos^(-1)  4/5` = x

Then, cos x = `4/5`

⇒ sin x = `sqrt (1 - (4/5)^2)`

⇒ sin x = `sqrt(1 - 16/25)`

⇒ sin x = `sqrt(9/25)`

⇒ sin x = `3/5`

∴ tan x = `3/4` ⇒ x = `tan^(-1)  3/4`

∴ `cos^(-1)  4/5 =  tan^(-1)  3/4`   ...(1)

Now let `cos^(-1)  12/13` = y

Then cos y = `12/13`

⇒ sin y = `5/13`

∴ tan y = `5/12` ⇒ y = `tan^(-1)  5/12`

∴ `cos^(-1)  12/13 = tan^(-1)  5/12`  ....(2)

Let `cos^(-1)  33/65` = z

Then cos z = `33/65`

⇒ sin z = `56/65`

∴ tan z = `56/33` ⇒ z = `tan^(-1)  56/33`

∴ `cos^(-1)  33/65 = tan^(-1)  56/33`  ....(3)

Now, we will prove that:

L.H.S = `cos^(-1)  4/5 + cos^(-1)  12/13`

= `tan^(-1)  3/4 + tan^(-1)  5/12`  ....[Using (1) and (2)]

= `tan^(-1)  (3/4 + 5/12)/(1 - 3/4 * 5/12)    ....[tan^(-1) x + tan^(-1) y = tan^(-1)  (x + y)/(1 - xy)]`

= `tan^(-1)  (36+20)/(48-15)`

= `tan^(-1)  56/33`

= `cos^(-1)  33/65`   .....[by (3)]

= R.H.S.

Let x = cos θ

Then, cos⁻¹x =  θ

We have,

R.H.S = cos⁻¹(4x³ − 3x)

⇒ cos⁻¹(4 cos³θ − 3 cos θ)

⇒ cos⁻¹(cos 3θ) = cos⁻¹(4x³ − 3x)

⇒ 3θ =  cos⁻¹(4x³ − 3x)

⇒ 3 cos⁻¹x = cos⁻¹(4x³ − 3x)

R.H.S = L.H.S

`sin^-1  8/17 + sin^-1  3/5`

= `tan^-1  8/sqrt(17^2 - 8^2) + tan^-1  3/sqrt(5^2 - 3^2)  ...[sin^-1  p/h = tan^-1  p/sqrt(h^2 - p^2)]`

= `tan^-1  8/sqrt(289 - 64) + tan^-1  3/sqrt(25 - 9)`

= `tan^-1  8/sqrt225 + tan^-1  3/sqrt16`

= `tan^-1  8/15 + tan^-1  3/4`

= `tan^-1  ((8/15 + 3/4)/(1 - 8/15 xx 3/4))  ...[tan^-1x + tan^-1y = tan^-1((x + y)/(1 - x xx y))]`

= `tan^-1[((32 + 45)/60)/(1 - 24/60)]`

= `tan^-1  77/36`

Let x = `cos^(-1)  12/13` and y = `sin^(-1)  3/5`

or cos x = `12/13` and sin y = `3/5`

sin x = `sqrt (1 - cos^2 x)` and cos y = `sqrt(1 - sin^2 y)`

Now, sin x = `sqrt(1 - 144/169)` and cos y = `sqrt( 1 - 9/25)`

⇒ sin x = `5/13` and cos y = `4/5`

We know that,

sin (x + y) = sin x cos y + cos x sin y

= `5/13 xx 4/5 + 12/13 xx 3/5 `

= `20/65 + 36/65 `

= `56/65`

⇒ x + y = `sin ^-1(56/65)`

or, `cos^-1(12/13) + sin^-1 (3/5)`

= `sin^-1(56/65)`

Let `sin^(-1)  5/13` = x and `cos^(-1)  3/5` = y

⇒ sin x = `5/13 ` and cos y = `3/5`

or tan x = `5/12` and tan y = `4/3`

⇒ x = `tan^-1  5/12` and y = `tan^(-1)  4/3`

x + y = `tan^-1  5/12 + tan^-1  4/3`

= `tan^-1 ((5/12 + 4/3)/(1 - 5/12 xx 4/3))`

= `tan^(-1) ((15+48)/(36-20))`

= `tan^(-1)  63/16`

Let x = tan² θ

⇒ `sqrtx` = tan θ

⇒ θ = `tan^(-1) sqrtx`  ...(1)

∴ `(1-x)/(1+x)`

= `(1-tan^2 θ)/(1 + tan^2 θ)`

= cos 2θ

Now we have,

R.H.S = `1/2 cos^(-1)  (1-x)/(1+x)`

= `1/2 cos^(-1)(cos 2θ)`

= `1/2 xx 2θ`

= θ

= `tan^(-1) sqrtx`  ....[From (1)]

R.H.S. = L.H.S.

L.H.S. = `cot^(-1)  ((sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1- sinx)))`

= `cot^(-1)  (sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1 - sin x)) xx (sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1 - sin x))`

= `cot^(-1)  ((1+sinx) + (1-sinx) + 2sqrt(1 - sin^2 x))/((1+sinx) - (1 - sinx)`

= `cot^(-1)  (2(1 + cos x))/(2sin x)`

= `cot^(-1)  (1+ cosx)/sin x`

= `cot^(-1)  (2 cos^2  x/2)/(2sin  x/2 cos  x/2)`

= `cot^-1 (cot  x/2)`

= `x/2`

L.H.S. = R.H.S.

Let x = sin θ

Then, sin⁻¹ x = θ

We have

R.H.S = sin⁻¹ (3x − 4x³) = sin⁻¹ (3 sin θ − 4 sin³θ)

= sin⁻¹ (sin 3θ) = sin⁻¹ (3 sin θ − 4 sin³θ)

= 3θ = sin⁻¹ (3 sin θ − 4 sin³θ)

= 3 sin⁻¹ x = sin⁻¹ (3 sin θ − 4 sin³θ)

R.H.S = L.H.S

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
fcot⁻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