Revision: Sets and Functions >> Trigonometry Mathematics ISC (Commerce) Class 11 CISCE

Angle: An angle consists of two rays that originate from a single initial point.  An angle is represented by the symbol ∠.

Arms of an Angle: The two rays forming the angle are called the arms or sides of the angle.

Vertex: The vertex of the angle is the common initial point of two rays.
Example:

The vertex of the angle is the common initial point of two rays.

 The two rays forming the angle are called the arms or sides of the angle.

An angle consists of two rays that originate from a single initial point.  An angle is represented by the symbol ∠.

\[sineA=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\]

\[cosineA=\frac{\mathrm{Base}}{\text{Hypotenuse}}\]

\[tangentA=\frac{\text{Perpendicular}}{\mathrm{Base}}\]

\[cotangent A = \frac{\text{Base}}{\text{Perpendicular}}\]

\[secantA=\frac{\text{Hypotenuse}}{\mathrm{Base}}\]

\[cosecantA=\frac{\text{Hypotenuse}}{\text{Perpendicular}}\]

\[\sin\mathrm{A}=\frac{1}{\mathrm{cosec~A}}\quad\mathrm{and}\quad\mathrm{cosec~A}=\frac{1}{\sin\mathrm{A}}\]

\[\cos\mathrm{A}=\frac{1}{\sec\mathrm{A}}\quad\mathrm{and}\quad\mathrm{sec}\mathrm{A}=\frac{1}{\cos\mathrm{A}}\]

\[\tan\mathrm{A}=\frac{1}{\cot\mathrm{A}}\quad\mathrm{and}\quad\cot\mathrm{A}=\frac{1}{\tan\mathrm{A}}\]

• sin⁡θ⋅cosec⁡θ = 1

• cos⁡θ⋅sec⁡θ = 1

• tan⁡θ⋅cot⁡θ = 1

\[tanA=\frac{\sin A}{\cos A}\]

\[cotA=\frac{\cos A}{\sin A}\]

Given that: `(sin(x + y))/(sin(x - y)) = (a + b)/(a - b)`

⇒ `(sin(x + y) + sin(x - y))/(sin(x + y) - sin(x - y)) = (a + b + a - b)/(a + b - a + b)`  .....(Using componendo and dividendo theorem)

⇒ `(2sin((x + y + x - y)/2) cos  ((x + y - x + y)/2))/(2cos((x + y + x - y)/2) sin((x + y - x + y)/2)) = (2a)/(2b)`

⇒ `(sinx . cos y)/(cosx . sin y) = a/b`

⇒ tan x.cot y = `a/b`

⇒ `tanx/tany = a/b`

Hence proved.

Given that: tanθ = `(sinalpha - cosalpha)/(sinalpha + cosalpha)`

⇒ tanθ = `(tanalpha - 1)/(tan alpha + 1)`

= `(tanalpha - tan  pi/4)/(1 + tan  pi/4  tan alpha)` 

⇒ tanθ = `tan(alpha - pi/4)`

∴  θ =  `alpha - pi/4`

⇒ cosθ = `cos(alpha - pi/4)`

⇒ cosθ = `cos alpha cos  pi/4 + sin alpha sin  pi/4`

⇒ cosθ = `cos alpha . 1/sqrt(2) + sin alpha . 1/sqrt(2)`

⇒ `sqrt(2) cos theta` = cosα + sinα

⇒ sinα + cosα = `sqrt(2) cos theta`

Hence proved.

Given that: cos(θ + Φ) = m cos(θ – Φ)

⇒ `(cos(theta + phi))/(cos(theta - phi)) = m/1`

Using componendo and dividendo theorem, we get

`(cos(theta + phi) + cos(theta - phi))/(cos(theta + phi) - cos(theta - phi)) = (m + 1)/(m - 1)`

⇒ `(2cos((theta + phi + theta - phi)/2).cos((theta+ phi - theta + phi)/2))/(-2sin((theta + phi + theta - phi)/2)*sin((theta + phi - theta + phi)/2)) = (m + 1)/(m - 1)`

⇒ `(costheta.cosphi)/(-sintheta.sinphi) = (m + 1)/(m - 1)`

⇒ `- cot theta . cot phi = (m + 1)/(m - 1)`

⇒ `(-cot phi)/(tantheta) = (m + 1)/(m - 1) - (1 + m)/(1 - m)`

⇒ tan θ  = `(1 - m)/(1 + m) cot phi`

Hence proved.

a cosθ + b sinθ = m  ......(i)

a sinθ - b cosθ = n  ......(ii)

Squaring and adding equations 1 and 2, we get,

(a cosθ + b sinθ)² + (a sinθ - b cosθ)² = m² + n²

⇒ a²cos²θ + b²sin²θ + 2ab sin θ cos θ + a²sin²θ + b²cos²θ - 2ab sin θ cos θ = m² + n²

⇒ a²cos²θ + b²sin²θ + a²sin²θ + b²cos²θ = m² + n²

⇒ a²(sin²θ + cos²θ) + b²(sin²θ + cos²θ) = m² + n²

Using, sin²θ + cos²θ = 1

We get,

⇒ a² + b² = m² + n²

Theorem 1: For any real numbers x and y,                 
sin x = sin y implies x = nπ + (–1)n y, where n ∈ Z
Proof :If sin x = sin y, then
sin x – sin y = 0

 or  `2cos  (x+y)/2 sin  (x-y)/2= 0`

which gives `cos   (x+y)/2= 0` or `sin  (x-y)/2= 0`

Therefore `(x+y)/2= (2n+1) π/2` or `(x-y)/2= nπ`, where n ∈ Z

i.e. x = (2n + 1) π – y  or x = 2nπ + y, where n∈Z
Hence x = (2n + 1)π + (–1)2n + 1 y or x = 2nπ +(–1)2n y, where n ∈ Z.
Combining these two results, we get x = nπ + (–1)n y, where n ∈ Z.

Theorem 2: For any real numbers x and y, cos x = cos y, implies x = 2nπ ± y, where n ∈ Z
Proof: If cos x = cos y, then
cos x – cos y = 0   i.e.,

-2 sin `(x+y)/2 sin  (x-y)/2= 0`

Thus, `sin  (x+y)/2= 0` or `sin  (x-y)/2= 0`

Therefore, `(x+y)/2= nπ`  or `(x-y)/2= nπ`, where n ∈ Z
i.e. x = 2nπ – y  or x = 2nπ + y, where n ∈ Z Hence x = 2nπ ± y, where n ∈ Z

Theorem 3:Prove that if x and y are not odd mulitple of  `π/2`, then
tan x = tan y implies x = nπ + y, where n ∈ Z
Proof : If tan x = tan y,   then   tan x – tan y = 0
or `(sin x cos y- cos x sin y)/(cos x cos y)= 0`
which gives sin (x – y) = 0
Therefore x – y = nπ, i.e., x = nπ + y, where n ∈ Z.

For an acute angle A in a right-angled triangle:

• Hypotenuse is the side opposite the right angle.

• Perpendicular is the side opposite angle A.

• Base is the side adjacent to angle A.