Notes On Trigonometric Functions of Sum and Difference of Two Angles (Part I) - CBSE Class 11 Maths

Consider a unit circle such that its centre is at the origin. Let A, B, C and D be points on the circle such that angle AOB is x, angle BOC is y and angle AOD is -y.
 
P(a, b) = P(cos x, sin x)
 
cos x  = Adjacent side to x Hypotenuse    =  a 1    =  a
sin x  = Opposite side to x Hypotenuse    =  b 1    =  b

cos(- x) = cos x and sin(- x) = - sin x

A = (1, 0) [∵ Radius of the circle is 1 unit]

B = (cos x, sin x)

C = [cos(x + y), sin (x + y)]

D = [cos(-y , sin (-y))] = (cos y, -sin y)

In ΔBOD and ΔAOC,

OB = OC [∵ Radii of the circle are equal]

OD = OA [∵ Radii of the circle are equal]

Also, ∠BOD = y + ∠COD = ∠AOC

ΔBOD ≅ ΔAOC                  (By SAS congruence)

⇒ BD = AC

⇒ BD2 = AC2

⇒ (cos x - cos y)2 + (sin x + sin y)2 = [cos(x + y) – 1]2 + [sin(x + y) – 0]2    (∵ Distance between two points is ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2  )
⇒ cos2x + cos2y - 2 cos x cos y + sin2x + sin2y + 2 sin x sin y

⇒ cos2(x + y) - 2 cos (x + y) + 1 + sin2(x + y)

⇒ -2 cos(x + y) = -2(cos x cos y – sin x sin y)     (Since cos2 θ + sin2 θ = 1)

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

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

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

Replacing y by – y, cos(x - (–y)) = cos x cos (–y) + sin x sin (–y)

Since cos(– x) = cos x and sin(– x) = – sin x

cos(x - y) = cos x cos y + sin x sin y  ______ (2)

cos ( π 2   - y) = sin x

cos ( π 2   - x) = cos  π 2 cos x + sin π 2 sin x

= 0 × cos x + 1 × sin x

cos ( π 2   - x) = sin x

sin ( π 2   - x) = cos x

sin ( π 2   - x) = cos ( π 2   – ( π 2   – x)) = cos x  

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

sin(x + y) = cos ( π 2   – ( x + y ))

                = cos (( π 2   – x ) – y ))

                =  cos( π 2   – x ) cos y + sin( π 2   – x ) sin y

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

     sin (x – y) = sin x cos y – cos x sin y

Replacing y by – y in the equation

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

since cos (– x) = cos x and sin (– x) = – sin x

sin(x – y) = sin x cos y – cos x sin y

cos( π 2   + x ) = – sin x

We know that  cos( π 2   – x ) = sin x

Replacing x by – x, we get

cos( π 2   – (– x) ) = sin (– x)

⇒ cos( π 2   + x ) = – sin x

sin( π 2   + x ) = cos x

We know that sin( π 2   – x ) = cos x
               
Replacing x by – x, we have sin( π 2   – (– x) ) = cos (– x)

 ⇒ sin( π 2   + x ) = cos x

Similarly, we can prove that

cos (π – x) = – cos x
sin (π – x) =  sin x
cos (π + x) = – cos x
sin (π + x) = – sin x
cos (2π – x) = cos x
cos (2π – x) = – sin x

Similar results for tan x, cot x, sec x and cosec x can be obtained from the results of sin x and cos x.  These results are shown in the following table.
  π 2 - x   π 2 + x   π  - x π  + x   2 - x   2 + x   2π  - x   2π  + x
 sin  cos x  cos x  sin x  - sin x  - cos x  - cos x  - sin x  sin x
 cos  sin x  - sin x  - cos x  - cos x  - sin x  sin x  cos x  cos x
 tan  cot x  - cot x  - tan x  tan x  cot x  - cot x  - tan x  tan x
 cot  tan x  - tan x  - cot x  cot x  tan x  - tan x  - cot x  cot x
 sec  cosec x  - cosec x  - sec x  - sec x  - cosec x  sec x  sec x  sec x
 cosec  sec x  sec x  cosec x  - cosec x  - sec x  - sec x  - cosec x cosec x

 Tan and Cot functions of the sum and differences of two angles:
 
Cos θ = 0 implies that θ = (2n+1)  π 2  , where n is any integer.

tan θ =  sin θ cos θ is defined if θ ≠ (2n+1) π/2

tan(x + y) = tan x + tan y 1 - tan x tan y  if x, y, x + y ≠ (2n+1) π/2

tan(x+y) = sin (x + y) cos (x + y)
              
              = sin x cos y + cos x sin y cos x cosy - sin x siny

              = sin x cos y cos x cos y + cos x sin y sin x sin y cos x cos y cos x cos y - sin x sin y cos x cos y

tan(x + y) = tan x + tan y 1 - tan x tan y

tan(x - y) = tan x - tan y 1 + tan x tan y

Replacing y by – y, we have, tan(x + (-y)) = tan x + tan (-y) 1 - tan x tan (-y)

tan(x - y) = tan x - tan y 1 + tan x tan y

Sin θ = 0 implies that θ =nπ, where n is any integer.

cot θ = cos θ sin θ  is defined if θ ≠ nπ

cot (x + y) = cot x cot y - 1 cot y + cot x

if x, y, x + y ≠ nπ

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

Dividing both numerator and denominator by sin x sin y,

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

⇒  cot (x + y) = cot x cot y - 1 cot y + cot x

     cot (x - y) = cot x cot y + 1 cot y - cot x

Replacing y by – y, cot (x - (-y)) = cot x cot ( -y) - 1 cot (-y) + cot x

cot (x - y) = - cot x cot y - 1 - cot y + cot x

cot (x - y) = cot x cot y + 1 cot y - cot x

Summary


Consider a unit circle such that its centre is at the origin. Let A, B, C and D be points on the circle such that angle AOB is x, angle BOC is y and angle AOD is -y.
 
P(a, b) = P(cos x, sin x)
 
cos x  = Adjacent side to x Hypotenuse    =  a 1    =  a
sin x  = Opposite side to x Hypotenuse    =  b 1    =  b

cos(- x) = cos x and sin(- x) = - sin x

A = (1, 0) [∵ Radius of the circle is 1 unit]

B = (cos x, sin x)

C = [cos(x + y), sin (x + y)]

D = [cos(-y , sin (-y))] = (cos y, -sin y)

In ΔBOD and ΔAOC,

OB = OC [∵ Radii of the circle are equal]

OD = OA [∵ Radii of the circle are equal]

Also, ∠BOD = y + ∠COD = ∠AOC

ΔBOD ≅ ΔAOC                  (By SAS congruence)

⇒ BD = AC

⇒ BD2 = AC2

⇒ (cos x - cos y)2 + (sin x + sin y)2 = [cos(x + y) – 1]2 + [sin(x + y) – 0]2    (∵ Distance between two points is ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2  )
⇒ cos2x + cos2y - 2 cos x cos y + sin2x + sin2y + 2 sin x sin y

⇒ cos2(x + y) - 2 cos (x + y) + 1 + sin2(x + y)

⇒ -2 cos(x + y) = -2(cos x cos y – sin x sin y)     (Since cos2 θ + sin2 θ = 1)

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

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

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

Replacing y by – y, cos(x - (–y)) = cos x cos (–y) + sin x sin (–y)

Since cos(– x) = cos x and sin(– x) = – sin x

cos(x - y) = cos x cos y + sin x sin y  ______ (2)

cos ( π 2   - y) = sin x

cos ( π 2   - x) = cos  π 2 cos x + sin π 2 sin x

= 0 × cos x + 1 × sin x

cos ( π 2   - x) = sin x

sin ( π 2   - x) = cos x

sin ( π 2   - x) = cos ( π 2   – ( π 2   – x)) = cos x  

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

sin(x + y) = cos ( π 2   – ( x + y ))

                = cos (( π 2   – x ) – y ))

                =  cos( π 2   – x ) cos y + sin( π 2   – x ) sin y

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

     sin (x – y) = sin x cos y – cos x sin y

Replacing y by – y in the equation

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

since cos (– x) = cos x and sin (– x) = – sin x

sin(x – y) = sin x cos y – cos x sin y

cos( π 2   + x ) = – sin x

We know that  cos( π 2   – x ) = sin x

Replacing x by – x, we get

cos( π 2   – (– x) ) = sin (– x)

⇒ cos( π 2   + x ) = – sin x

sin( π 2   + x ) = cos x

We know that sin( π 2   – x ) = cos x
               
Replacing x by – x, we have sin( π 2   – (– x) ) = cos (– x)

 ⇒ sin( π 2   + x ) = cos x

Similarly, we can prove that

cos (π – x) = – cos x
sin (π – x) =  sin x
cos (π + x) = – cos x
sin (π + x) = – sin x
cos (2π – x) = cos x
cos (2π – x) = – sin x

Similar results for tan x, cot x, sec x and cosec x can be obtained from the results of sin x and cos x.  These results are shown in the following table.
  π 2 - x   π 2 + x   π  - x π  + x   2 - x   2 + x   2π  - x   2π  + x
 sin  cos x  cos x  sin x  - sin x  - cos x  - cos x  - sin x  sin x
 cos  sin x  - sin x  - cos x  - cos x  - sin x  sin x  cos x  cos x
 tan  cot x  - cot x  - tan x  tan x  cot x  - cot x  - tan x  tan x
 cot  tan x  - tan x  - cot x  cot x  tan x  - tan x  - cot x  cot x
 sec  cosec x  - cosec x  - sec x  - sec x  - cosec x  sec x  sec x  sec x
 cosec  sec x  sec x  cosec x  - cosec x  - sec x  - sec x  - cosec x cosec x

 Tan and Cot functions of the sum and differences of two angles:
 
Cos θ = 0 implies that θ = (2n+1)  π 2  , where n is any integer.

tan θ =  sin θ cos θ is defined if θ ≠ (2n+1) π/2

tan(x + y) = tan x + tan y 1 - tan x tan y  if x, y, x + y ≠ (2n+1) π/2

tan(x+y) = sin (x + y) cos (x + y)
              
              = sin x cos y + cos x sin y cos x cosy - sin x siny

              = sin x cos y cos x cos y + cos x sin y sin x sin y cos x cos y cos x cos y - sin x sin y cos x cos y

tan(x + y) = tan x + tan y 1 - tan x tan y

tan(x - y) = tan x - tan y 1 + tan x tan y

Replacing y by – y, we have, tan(x + (-y)) = tan x + tan (-y) 1 - tan x tan (-y)

tan(x - y) = tan x - tan y 1 + tan x tan y

Sin θ = 0 implies that θ =nπ, where n is any integer.

cot θ = cos θ sin θ  is defined if θ ≠ nπ

cot (x + y) = cot x cot y - 1 cot y + cot x

if x, y, x + y ≠ nπ

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

Dividing both numerator and denominator by sin x sin y,

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

⇒  cot (x + y) = cot x cot y - 1 cot y + cot x

     cot (x - y) = cot x cot y + 1 cot y - cot x

Replacing y by – y, cot (x - (-y)) = cot x cot ( -y) - 1 cot (-y) + cot x

cot (x - y) = - cot x cot y - 1 - cot y + cot x

cot (x - y) = cot x cot y + 1 cot y - cot x

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