Revision: Sets and Functions >> Trigonometric Functions Maths Commerce (English Medium) Class 11 CBSE

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 ∠.

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

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

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

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

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

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

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: `(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: 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.

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.

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

Given that: cosα + cosβ = 0

And sinα + sinβ = 0

∴ (cosα + cosβ)² - (sinα + sinβ)² = 0

⇒ (cos²α + cos²β + 2cosα cosβ) - (sin²α + sin²β + 2sinα sinβ) = 0

⇒ cos²α + cos²β + 2 cosα cosβ - sin²α - sin²β - 2 sinα sinβ = 0

⇒ (cos²α - sin²α) + (cos²β - sin²β) + 2(cosα cosβ - sinα sinβ) = 0

⇒ cos2α + cos2β + 2cos(α + β) = 0

Hence, cos2α + cos2β = -2cos(α + β).

Hence proved

Given that: cos(α + β) = `4/5`

∴ tan(α + β) = `3/4`

And sin(α – β) = `5/13`

∴ tan(α – β) = `5/12`

Now tan 2α = tan[α + β + α – β]

= tan[(α + β) + (α – β)]

= `(tan(alpha + beta) + tan(alpha - beta))/(1 - tan(alpha + beta).tan(alpha - beta))`

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

= `((9 + 5)/12)/((48 - 15)/48)`

= `14/12 xx 48/33`

= `56/33`

Hence, tan 2α = `56/33`.

Given that: m sinθ = n sin(θ + 2α)

⇒ `(sin(theta + 2alpha))/sintheta = m/n`

Using componendo and dividendo theorem, we get,

⇒ `(sin(theta + 2alpha) + sintheta)/(sin(theta + 2alpha) - sintheta) = (m + n)/(m - n)`

⇒ `(2sin((theta + 2alpha + theta)/2).cos((theta + 2alpha - theta)/2))/(2cos((theta + 2alpha + theta)/2).sin((theta + 2alpha - theta)/2)) = (m + n)/(m - n)`    .......`[(because sinA + sinB = 2sin  (A + B)/2 . cos  (A - B)/2),(sinA - sinB = 2cos  (A + B)/2 . sin  (A - B)/2)]`

⇒ `(sin(theta + alpha).cosalpha)/(cos(theta + alpha).sinalpha) = (m + n)/(m - n)`

⇒ `tan(theta + alpha)cotalpha = (m + n)/(m - n)`

Hence proved.

Given that: tan(A + B) = p, tan(A – B) = q

tan 2A = tan(A + B + A – B)

= tan[(A + B) + (A – B)]

= `(tan(A + B) + tan(A - B))/(1 - tan(A + B).tan(A - B))`

= `(p + q)/(1 - pq)`

Hence proved.

acos2θ + bsin2θ = c

α and β are the roots of the equation.

Using the formula of multiple angles,

We know that,

cos2θ = `(1 - tan^2theta)/(1 + tan^2theta)` and sin2θ = `(2tantheta)/(1 + tan^2theta)`

∴ `a((1 - tan^2theta)/(1 + tan^2theta)) + b((2tantheta)/(1 + tan^2theta)) - c` = 0

We know that,

The sum of roots of a quadratic equation,

ax² + bx + c = 0 is given by `((-b)/a)`.

Therefore,

tanα + tanβ = `(-2b)/(-(c + a)`

= `(2b)/((c + a))`

Hence, tanα + tanβ = `(2b)/((c + a))`

tanθ + sinθ = m   ......(i)

tanθ – sinθ = n  ......(ii)

Adding equations i and ii

2tanθ = m + n   ......(iii)

Subtracting equation ii from i

We get,

2sinθ = m – n  ......(iv)

Multiplying equations (iii) and (iv)

2sinθ(2tanθ) = (m + n)(m – n)

⇒ 4sinθ tanθ = m² – n²

Hence,

m² – n² = 4sinθ tanθ

sin4A = sin(2A + 2A)

We know that,

sin(A + B) = sinA cosB + cosA sinB

Therefore, sin4A = sin2A cos2A + cos2A sin2A

⇒ sin4A = 2sin2A cos2A

From T-ratios of multiple angles,

We get,

sin2A = 2sinA cosA and cos2A = cos²A – sin²A

⇒ sin4A = 2(2sinA cosA)(cos²A – sin²A)

⇒ sin4A = 4sinA cos³A – 4cosA sin³A

Hence, sin4A = 4sin A cos³A – 4cosA sin³A

Given that: x = sec Φ – tan Φ

And y = cosec Φ + cot Φ

xy + x – y + 1 = 0

L.H.S. xy + x – y + 1

= (sec Φ – tan Φ) (cosec Φ + cot Φ) + (sec Φ – tan Φ) – (cosec Φ + cot Φ) + 1

= `(1/cosphi - sinphi/cosphi) (1/sinphi + cosphi/sinphi) + (1/cosphi - sinphi/cosphi) - (1/sinphi - sinphi/cosphi) + 1`

= `((1 - sin phi)/cos phi) ((1 + cos phi)/sinphi) + (1 - sin phi)/cosphi - (1 + cosphi)/sinphi + 1`

= `(1 - sinphi + cosphi - sinphi cosphi)/(cosphisinphi) + (sinphi - sin^2 phi - cos phi - cos^2 phi)/(cos phi sin phi) + 1`

= `(1 - sin phi + cosphi - sinphi cosphi + sinphi - cosphi - (sin^2 phi + cos^2 phi) + sin phi cos phi)/(cosphi sin phi)`

= `(1 - 1)/(cos phi sin phi)`

= 0. R.H.S.

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

Hence proved.

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.