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

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

Let x = cos θ.

Then, cos^–1x = θ.

We have,

R.H.S. = cos^–1(4x³ – 3x)

= cos^–1(4 cos³ θ – 3 cos θ)

= cos^–1(cos 3θ)

= 3θ

= 3cos^–1x

= L.H.S.

Let `sin^-1  8/17 = x`.

Then, `sin x = 8/17`

 ⇒ `cos x = sqrt(1 - (8/17)^2`

= `sqrt((225)/(289)`

= `15/17`

∴ `tan x= 8/15` ⇒ `x = tan^-1  8/15`

∴ `sin^-1  8/17 = tan^-1  8/15`   ...(1)

Now, let `sin^-1  3/5 = y`.

Then, `sin y = 3/5`

⇒ `cos y = sqrt(1 - (3/5)^2`

= `sqrt(16/25)`

= `4/5`

∴ `tan y = 3/5` ⇒ `y = tan^-1  3/4`

∴ `sin^-1  3/5 = tan^-1  3/4`   ...(2)

Now, we ahve:

L.H.S. = `sin^-1  8/17 + sin^-1  3/5`

= `tan^-1  8/15 + tan^-1  3/4`   ...[Using (1) and (2)]

= `tan^-1  (8/15 + 3/4)/(1 - 8/15 xx 3/4)`

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

= `tan^-1  77/36` = R.H.S.

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`

= `tan^(-1)  56/33`   ...[By (3)]

= R.H.S.

Let `sin^-1  3/5 = x`.

Then, `sin x = 3/5`

⇒ `cos x = sqrt(1 - (3/5)^2`

= `sqrt(16/25)`

= `4/5`

∴ `tan x = 3/4` ⇒ `x = tan^-1  3/4`

∴ `sin^-1  3/5 = 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 `sin^-1  56/65 = z`.

Then, `sin z = 56/65` ⇒ `cos z = 33/65`.

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

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

Now, we have:

L.H.S. = `cos^-1  12/13 + sin^-1  3/5`

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

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

= `tan^-1  (20 + 36)/(48 - 15)`

= `tan^-1  56/33`

= `sin^-1  56/65` = R.H.S.   ...[Using (3)]

Let `sin^(-1)  5/13 = x`.

Then, `sin x = 5/13` ⇒ `cos x = 12/13`.

∴ `tan x = 5/12` ⇒ `x = tan^-1  5/12`

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

Let `cos^-1  3/5 = y`.

Then, `cos y = 3/5` ⇒ `sin y = 4/5`.

∴ `tan y = 4/3` ⇒ `y = tan^-1  4/3`

∴ `cos^-1  3/5 = tan^-1  4/3`   ...(2)

Using (1) and (2), we have

R.H.S. = `sin^-1  5/13 + cos^-1  3/5`

= `tan^-1  5/12 + tan^-1  4/3`

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

= `tan^-1  ((15 + 48)/(36 - 20))`

= `tan^-1  63/16`

= L.H.S.

Let x = tan² θ.

Then, `sqrt(x) = tan θ`

⇒ `θ = tan^(-1) sqrtx`

∴ `(1 - x)/(1 + x) = (1 - tan^2θ)/(1 + tan^2θ)`

= 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) sqrt(x)`

= L.H.S.

Consider `(sqrt(1 + sinx) + sqrt(1 - sin x))/(sqrt(1 + sinx) - sqrt(1 - sinx))`

= `((sqrt(1 + sinx) + sqrt(1 - sinx))^2)/((sqrt(1 + sin x))^2 - (sqrt(1 - sin x))^2)`   ...(By rationalizing)

= `((1 + sinx) + (1 - sinx) + 2sqrt((1 + sinx)(1 - sinx)))/(1 + sinx - 1 + sinx)`

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

= `(1 + cosx)/(sin x)`

= `(2 cos^2  x/2)/(2sin  x/2 cos  x/2)`

= `cot  x/2`

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

= `cot^(-1) (cot  x/2)`

= `x/2` = R.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
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