 # Trigonometry Formulas For Class 12

Trigonometry is the study between the relationships dealing with angles, heights and lengths of triangles and also the relationships between the different circle parts and other geometric figures. In trigonometry class 12, we study trigonometry which finds its application in the field of astronomy, engineering, architectural design, and physics.Trigonometry Formulas for class 12 contains all the essential trigonometric identities which can fetch some direct questions in competitive exams on the basis of formulae.

Trigonometric identities given in the 12th trigonometry formula are very useful and help to solve the problems better. There are huge numbers of fields in which these trigonometry identities and trigonometric equations are used.

### The Difference Between Trigonometric Identities And Trigonometric Ratios:

Trigonometric Identities: Formula of trigonometry class 12 involves trigonometric functions and trigonometric identities. These identities of trigonometry are accurate for all variable’s values.

Trigonometric Ratio: The relationship of the angle measurement and the right-angle triangle side length is known for its trigonometric ratio.

### All Trigonometry Formulas For Class 12

Sum And Difference Of Angles:

1. Sin A Sin B = ½  [cos(A-B)-cos(A+B)]

2. cos A cos B = ½ [cos(A-B)+ cos(A+B)]

3. Sin A cos B = ½ [sin(A+B)+ sin(A-B)]

4. cos A sin B = ½ [cos(A+B)+ sin(A-B)]

Double Angle Formulas:

1. sin2θ= 2sinθcosθ

2. cos2θ= Cos²θ - Sin²θ

= 2Cos²θ - 1

= 1-2Sin²θ

1. tan2θ= $\frac{2Tanθ}{1-Tan^{2}θ}$

Triple Angle Formula:

1. sin3θ= 3Sinθ - 4$Sin^{3}θ$

2. cos3θ= 4$Cos^{3}θ$ - 3Cosθ

3. tan3θ= (3tanθ-tan³θ) / (1-3tan²θ)

4. cot3θ= $\frac{Cot^{3}θ - 3Cotθ}{3Cot^{2}θ - 1}$

Converting Products Into Sums And Difference:

1. sinA sinB = ½ [cos(A-B) - cos(A+B)]

2. cosA cosB = ½ [cos(A-B)+cos(A+B)]

3. sinA cosB = ½ [sin(A+B)+sin(A-B)]

4. cosA sinB = ½ [cos(A+B)+sin(A-B)]

Half Angle Identities:

1. Sin($\frac{x}{2}$) = ±$\sqrt{\frac{1-2cos(x)}{2}}$

2. Tan($\frac{x}{2}$) = ±$\sqrt{\frac{1-cos(x)}{1+Cos(x)}}$ = $\frac{1-Cos(x)}{Sin(x)}$ = $\frac{Sin(x)}{1+Cos(x)}$

3. Cos($\frac{x}{2}$) = ±$\sqrt{\frac{1+cos(x)}{2}}$

On squaring the above identities we can re-state the above equations as following-

1. $Sin^{2}(x)$ = $\frac{1}{2}$[1-Cos(2x)]

2. $Cos^{2}(x)$ = $\frac{1}{2}$[1+Cos(2x)]

3. $Tan^{2}(x)$ = $\frac{1-Cos(2x)}{1+Cos(2x)}$

Complex Relations:

1. Sinθ = $\frac{e^{iθ} - e^{-iθ}}{2i}$

2. Cosθ = $\frac{e^{iθ} + e^{-iθ}}{2i}$

3. Tanθ = $\frac{e^{iθ} - e^{-iθ}}{(e^{iθ} + e^{-iθ})i}$

4. Cosecθ = $\frac{2i}{e^{iθ} - e^{-iθ}}$

5. Secθ =$\frac{2i}{e^{iθ} - e^{-iθ}}$

6. Cotθ =$\frac{(e^{iθ} + e^{-iθ})i}{e^{iθ} - e^{-iθ}}$

### Inverse Trigonometric Functions:

Definition:

θ = $Sin^{−1}(x)$ is equivalent to x = Sin θ

θ = $Cos^{−1}(x)$ is equivalent to x = Cos θ

θ = $Tan^{−1}(x)$ is equivalent to x = Tan θ

Inverse Properties:

These properties hold for x in the domain and θ in the range

sin(sin−1 (x)) = x

cos(cos−1 (x)) = x

tan(tan−1 (x)) = x

sin−1 (sin(θ)) = θ

cos−1 (cos(θ)) = θ

tan−1 (tan(θ)) = θ

Other Notations

sin−1 (x) = arcsin(x)

cos−1 (x) = arccos(x)

tan−1 (x) = arctan(x)

## Domain and Range:

 Function Domain Range θ = sin−1 (x) −1 ≤ x ≤ 1 − π /2 ≤ θ ≤ π/2 θ = cos−1 (x) −1 ≤ x ≤ 1 0 ≤ θ ≤ π θ = tan−1 (x) −∞ ≤ x ≤ ∞ − π/ 2 < θ < π/2

## Inverse Trigonometric Functions:

 Name Usual Notation Definition Domain of x for real number Range of usual principal values(Radians) Range of usual principal values(Degrees) arcsine y=arcsin(x) x=sin(y) −1 ≤ x ≤ 1 − π /2 ≤ y ≤ π/2 −90⁰ ≤ y ≤ 90⁰ arccosine y=arcos(x) x=cos(y) −1 ≤ x ≤ 1 0 ≤ y ≤ π 0⁰ ≤ y ≤ 180⁰ arctangent y=arctan(x) x=tan(y) All real numbers − π /2 ≤ y ≤ π/2 −90⁰ ≤ y ≤ 90⁰ arccotangent y=arccot(x) x=cot(y) All real numbers 0 ≤ y ≤ π 0⁰ ≤ y ≤ 180⁰ arcsecant y=arcsec(x) x=sec(y) x ≤ -1 or 1 ≤ x 0 ≤ y < π/2 or π/2 < y ≤π 0 ≤ y < 90⁰ or 90⁰ < y ≤180⁰ arccosecant y=arccosec(x) x=cosec(x) x ≤ -1 or 1 ≤ x - π/2 ≤ y < 0 or 0 < y ≤ π/2 - 90⁰ ≤ y < 0 or 0 < y ≤ 90⁰

### Solved Examples:

Question 1) Find the principal values of $Sin^{-1}$ ($\frac{1}{\sqrt{2}}$)

Solution) Let $Sin^{-1}$($\frac{1}{\sqrt{2}}$) = α; then sin α =$\frac{1}{\sqrt{2}}$= sin 45 degree

α = 45 degree or $\frac{π}{4}$, which is the required principal value.

Question 2) Show that $Tan^{-1}$$\frac{1}{2}$ + $Tan^{-1}$$\frac{1}{3}$ = $\frac{π}{4}$

Solution) L.H.S. = $Tan^{-1}$ $\frac{1}{2}$ + $Tan^{-1}$ $\frac{1}{3}$ = $Tan^{-1}$ = $\frac{\frac{1}{2} + \frac{1}{3}}{1- \frac{1}{2} X \frac{1}{3}}$ = $Tan^{-1}$ ($\frac{\frac{5}{6}}{\frac{5}{6}}$) = $Tan^{-1}$ = $\frac{π}{4}$

### Fun Facts

• The word "Trigonometry" is taken from the word "Triangle Measure".

• Trigonometry is used by the engineers to figure out the angles of the sound waves and how to design rooms.

• Trigonometry is connected with music and architecture.