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  = $\frac{\text{Adjacent side to x}}{\text{Hypotenuse}}$   = $\frac{\text{a}}{\text{1}}$    =  a sin x  = $\frac{\text{Opposite side to x}}{\text{Hypotenuse}}$   =     =  b cos(- x) = cos x and sin(- x) = - sin x A = (1, 0) [âˆµ 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 Î”BOD and Î”AOC, OB = OC [âˆµ Radii of the circle are equal] OD = OA [âˆµ Radii of the circle are equal] Also, âˆ BOD = y + âˆ COD = âˆ AOC Î”BOD â‰… Î”AOC                  (By SAS congruence) â‡’ BD = AC â‡’ BD2 = AC2 â‡’ (cos x - cos y)2 + (sin x + sin y)2 = [cos(x + y) â€“ 1]2 + [sin(x + y) â€“ 0]2    (âˆµ Distance between two points is $\sqrt{{\text{(}{\text{x}}_{\text{2}}\text{-}{\text{x}}_{\text{1}}\text{)}}^{\text{2}}\text{+}{\text{(}{\text{y}}_{\text{2}}\text{-}{\text{y}}_{\text{1}}\text{)}}^{\text{2}}}$ ) â‡’ cos2x + cos2y - 2 cos x cos y + sin2x + sin2y + 2 sin x sin y â‡’ cos2(x + y) - 2 cos (x + y) + 1 + sin2(x + y) â‡’ -2 cos(x + y) = -2(cos x cos y â€“ sin x sin y)     (Since cos2 Î¸ + sin2 Î¸ = 1) â‡’ cos(x + y) = cos x cos y â€“ sin x sin y cos(x + y) = cos x cos y â€“ sin x sin y   cos(x - y) = cos x cos y + sin x sin y Replacing y by â€“ y, cos(x - (â€“y)) = cos x cos (â€“y) + sin x sin (â€“y) Since cos(â€“ x) = cos x and sin(â€“ x) = â€“ sin x cos(x - y) = cos x cos y + sin x sin y  ______ (2) cos ($\frac{Ï€}{\text{2}}$  - y) = sin x cos ($\frac{Ï€}{\text{2}}$  - x) = cos $\frac{Ï€}{\text{2}}$ cos x + sin $\frac{Ï€}{\text{2}}$ sin x = 0 Ã— cos x + 1 Ã— sin x cos ($\frac{Ï€}{\text{2}}$  - x) = sin x sin ($\frac{Ï€}{\text{2}}$  - x) = cos x sin ($\frac{Ï€}{\text{2}}$  - x) = cos ($\frac{Ï€}{\text{2}}$  â€“ ($\frac{Ï€}{\text{2}}$  â€“ x)) = cos x   sin(x + y) = sin x cos y + cos x sin y sin(x + y) = cos ($\frac{Ï€}{\text{2}}$  â€“ ( x + y ))                 = cos (($\frac{Ï€}{\text{2}}$  â€“ x ) â€“ y ))                 =  cos($\frac{Ï€}{\text{2}}$  â€“ x ) cos y + sin($\frac{Ï€}{\text{2}}$  â€“ x ) sin y â‡’ sin (x + y) = sin x cos y + cos x sin y      sin (x â€“ y) = sin x cos y â€“ cos x sin y Replacing y by â€“ y in the equation sin(x + (-y)) = sin x cos (â€“ y) + cos x sin (â€“ y) since cos (â€“ x) = cos x and sin (â€“ x) = â€“ sin x sin(x â€“ y) = sin x cos y â€“ cos x sin y cos($\frac{Ï€}{\text{2}}$  + x ) = â€“ sin x We know that  cos($\frac{Ï€}{\text{2}}$  â€“ x ) = sin x Replacing x by â€“ x, we get cos($\frac{Ï€}{\text{2}}$  â€“ (â€“ x) ) = sin (â€“ x) â‡’ cos($\frac{Ï€}{\text{2}}$  + x ) = â€“ sin x sin($\frac{Ï€}{\text{2}}$  + x ) = cos x We know that sin($\frac{Ï€}{\text{2}}$  â€“ x ) = cos x                 Replacing x by â€“ x, we have sin($\frac{Ï€}{\text{2}}$  â€“ (â€“ x) ) = cos (â€“ x)  â‡’ sin($\frac{Ï€}{\text{2}}$  + x ) = cos x Similarly, we can prove that cos (Ï€ â€“ x) = â€“ cos x sin (Ï€ â€“ x) =  sin x cos (Ï€ + x) = â€“ cos x sin (Ï€ + x) = â€“ sin x cos (2Ï€ â€“ x) = cos x cos (2Ï€ â€“ x) = â€“ 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.  $\frac{Ï€}{\text{2}}\text{- x}$  $\frac{Ï€}{\text{2}}\text{+ x}$    $\frac{3Ï€}{\text{2}}\text{- x}$  $\frac{3Ï€}{\text{2}}\text{+ x}$      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 Î¸ = 0 implies that Î¸ = (2n+1)  Ï€ 2  , where n is any integer. tan Î¸ = $\frac{sin\mathrm{Î¸}\text{}}{cos\mathrm{Î¸}\text{}}$ is defined if Î¸ â‰  (2n+1) Ï€/2 $\text{tan(x + y) =}\frac{\text{tan x + tan y}}{\text{1 - tan x tan y}}$ if x, y, x + y â‰  (2n+1) Ï€/2 tan(x+y) = $\frac{\text{sin (x + y)}}{\text{cos (x + y)}}$                              = $\frac{\text{sin x cos y + cos x sin y}}{\text{cos x cosy - sin x siny}}$               = $\frac{\frac{\text{sin x cos y}}{\text{cos x cos y}}\text{+}\frac{\text{cos x sin y}}{\text{sin x sin y}}}{\frac{\text{cos x cos y}}{\text{cos x cos y}}\text{-}\frac{\text{sin x sin y}}{\text{cos x cos y}}}$ $\text{tan(x + y) =}\frac{\text{tan x + tan y}}{\text{1 - tan x tan y}}$ $\text{tan(x - y) =}\frac{\text{tan x - tan y}}{\text{1 + tan x tan y}}$ Replacing y by â€“ y, we have, $\text{tan(x + (-y)) =}\frac{\text{tan x + tan (-y)}}{\text{1 - tan x tan (-y)}}$ â‡’ $\text{tan(x - y) =}\frac{\text{tan x - tan y}}{\text{1 + tan x tan y}}$ Sin Î¸ = 0 implies that Î¸ =nÏ€, where n is any integer. cot Î¸ = $\frac{cos\mathrm{Î¸}\text{}}{sin\mathrm{Î¸}\text{}}$ is defined if Î¸ â‰  nÏ€ $\text{cot (x + y) =}\frac{\text{cot x cot y - 1}}{\text{cot y + cot x}}$ if x, y, x + y â‰  nÏ€ $\text{}$$\text{cot (x + y) =}\frac{\text{cos (x + y)}}{\text{sin (x + y)}}$   = $\frac{\text{cos x cos y - sin x sin y}}{\text{sin x cos y + cos x sin y}}$ Dividing both numerator and denominator by sin x sin y, $\text{}$$\text{cot (x + y) =}\frac{\frac{\text{cos x cos y}}{\text{sin x sin y}}\text{-}\frac{\text{sin x sin y}}{\text{sin x sin y}}}{\frac{\text{sin x cos y}}{\text{sin x sin y}}\text{+}\frac{\text{cos x sin y}}{\text{sin x sin y}}}$ â‡’ $\text{cot (x + y) =}\frac{\text{cot x cot y - 1}}{\text{cot y + cot x}}$      $\text{cot (x - y) =}\frac{\text{cot x cot y + 1}}{\text{cot y - cot x}}$ Replacing y by â€“ y, $\text{cot (x - (-y)) =}\frac{\text{cot x cot ( -y) - 1}}{\text{cot (-y) + cot x}}$ $\text{cot (x - y) =}\frac{\text{- cot x cot y - 1}}{\text{- cot y + cot x}}$ â‡’ $\text{cot (x - y) =}\frac{\text{cot x cot y + 1}}{\text{cot y - cot x}}$

#### 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  = $\frac{\text{Adjacent side to x}}{\text{Hypotenuse}}$   = $\frac{\text{a}}{\text{1}}$    =  a sin x  = $\frac{\text{Opposite side to x}}{\text{Hypotenuse}}$   =     =  b cos(- x) = cos x and sin(- x) = - sin x A = (1, 0) [âˆµ 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 Î”BOD and Î”AOC, OB = OC [âˆµ Radii of the circle are equal] OD = OA [âˆµ Radii of the circle are equal] Also, âˆ BOD = y + âˆ COD = âˆ AOC Î”BOD â‰… Î”AOC                  (By SAS congruence) â‡’ BD = AC â‡’ BD2 = AC2 â‡’ (cos x - cos y)2 + (sin x + sin y)2 = [cos(x + y) â€“ 1]2 + [sin(x + y) â€“ 0]2    (âˆµ Distance between two points is $\sqrt{{\text{(}{\text{x}}_{\text{2}}\text{-}{\text{x}}_{\text{1}}\text{)}}^{\text{2}}\text{+}{\text{(}{\text{y}}_{\text{2}}\text{-}{\text{y}}_{\text{1}}\text{)}}^{\text{2}}}$ ) â‡’ cos2x + cos2y - 2 cos x cos y + sin2x + sin2y + 2 sin x sin y â‡’ cos2(x + y) - 2 cos (x + y) + 1 + sin2(x + y) â‡’ -2 cos(x + y) = -2(cos x cos y â€“ sin x sin y)     (Since cos2 Î¸ + sin2 Î¸ = 1) â‡’ cos(x + y) = cos x cos y â€“ sin x sin y cos(x + y) = cos x cos y â€“ sin x sin y   cos(x - y) = cos x cos y + sin x sin y Replacing y by â€“ y, cos(x - (â€“y)) = cos x cos (â€“y) + sin x sin (â€“y) Since cos(â€“ x) = cos x and sin(â€“ x) = â€“ sin x cos(x - y) = cos x cos y + sin x sin y  ______ (2) cos ($\frac{Ï€}{\text{2}}$  - y) = sin x cos ($\frac{Ï€}{\text{2}}$  - x) = cos $\frac{Ï€}{\text{2}}$ cos x + sin $\frac{Ï€}{\text{2}}$ sin x = 0 Ã— cos x + 1 Ã— sin x cos ($\frac{Ï€}{\text{2}}$  - x) = sin x sin ($\frac{Ï€}{\text{2}}$  - x) = cos x sin ($\frac{Ï€}{\text{2}}$  - x) = cos ($\frac{Ï€}{\text{2}}$  â€“ ($\frac{Ï€}{\text{2}}$  â€“ x)) = cos x   sin(x + y) = sin x cos y + cos x sin y sin(x + y) = cos ($\frac{Ï€}{\text{2}}$  â€“ ( x + y ))                 = cos (($\frac{Ï€}{\text{2}}$  â€“ x ) â€“ y ))                 =  cos($\frac{Ï€}{\text{2}}$  â€“ x ) cos y + sin($\frac{Ï€}{\text{2}}$  â€“ x ) sin y â‡’ sin (x + y) = sin x cos y + cos x sin y      sin (x â€“ y) = sin x cos y â€“ cos x sin y Replacing y by â€“ y in the equation sin(x + (-y)) = sin x cos (â€“ y) + cos x sin (â€“ y) since cos (â€“ x) = cos x and sin (â€“ x) = â€“ sin x sin(x â€“ y) = sin x cos y â€“ cos x sin y cos($\frac{Ï€}{\text{2}}$  + x ) = â€“ sin x We know that  cos($\frac{Ï€}{\text{2}}$  â€“ x ) = sin x Replacing x by â€“ x, we get cos($\frac{Ï€}{\text{2}}$  â€“ (â€“ x) ) = sin (â€“ x) â‡’ cos($\frac{Ï€}{\text{2}}$  + x ) = â€“ sin x sin($\frac{Ï€}{\text{2}}$  + x ) = cos x We know that sin($\frac{Ï€}{\text{2}}$  â€“ x ) = cos x                 Replacing x by â€“ x, we have sin($\frac{Ï€}{\text{2}}$  â€“ (â€“ x) ) = cos (â€“ x)  â‡’ sin($\frac{Ï€}{\text{2}}$  + x ) = cos x Similarly, we can prove that cos (Ï€ â€“ x) = â€“ cos x sin (Ï€ â€“ x) =  sin x cos (Ï€ + x) = â€“ cos x sin (Ï€ + x) = â€“ sin x cos (2Ï€ â€“ x) = cos x cos (2Ï€ â€“ x) = â€“ 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.  $\frac{Ï€}{\text{2}}\text{- x}$  $\frac{Ï€}{\text{2}}\text{+ x}$    $\frac{3Ï€}{\text{2}}\text{- x}$  $\frac{3Ï€}{\text{2}}\text{+ x}$      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 Î¸ = 0 implies that Î¸ = (2n+1)  Ï€ 2  , where n is any integer. tan Î¸ = $\frac{sin\mathrm{Î¸}\text{}}{cos\mathrm{Î¸}\text{}}$ is defined if Î¸ â‰  (2n+1) Ï€/2 $\text{tan(x + y) =}\frac{\text{tan x + tan y}}{\text{1 - tan x tan y}}$ if x, y, x + y â‰  (2n+1) Ï€/2 tan(x+y) = $\frac{\text{sin (x + y)}}{\text{cos (x + y)}}$                              = $\frac{\text{sin x cos y + cos x sin y}}{\text{cos x cosy - sin x siny}}$               = $\frac{\frac{\text{sin x cos y}}{\text{cos x cos y}}\text{+}\frac{\text{cos x sin y}}{\text{sin x sin y}}}{\frac{\text{cos x cos y}}{\text{cos x cos y}}\text{-}\frac{\text{sin x sin y}}{\text{cos x cos y}}}$ $\text{tan(x + y) =}\frac{\text{tan x + tan y}}{\text{1 - tan x tan y}}$ $\text{tan(x - y) =}\frac{\text{tan x - tan y}}{\text{1 + tan x tan y}}$ Replacing y by â€“ y, we have, $\text{tan(x + (-y)) =}\frac{\text{tan x + tan (-y)}}{\text{1 - tan x tan (-y)}}$ â‡’ $\text{tan(x - y) =}\frac{\text{tan x - tan y}}{\text{1 + tan x tan y}}$ Sin Î¸ = 0 implies that Î¸ =nÏ€, where n is any integer. cot Î¸ = $\frac{cos\mathrm{Î¸}\text{}}{sin\mathrm{Î¸}\text{}}$ is defined if Î¸ â‰  nÏ€ $\text{cot (x + y) =}\frac{\text{cot x cot y - 1}}{\text{cot y + cot x}}$ if x, y, x + y â‰  nÏ€ $\text{}$$\text{cot (x + y) =}\frac{\text{cos (x + y)}}{\text{sin (x + y)}}$   = $\frac{\text{cos x cos y - sin x sin y}}{\text{sin x cos y + cos x sin y}}$ Dividing both numerator and denominator by sin x sin y, $\text{}$$\text{cot (x + y) =}\frac{\frac{\text{cos x cos y}}{\text{sin x sin y}}\text{-}\frac{\text{sin x sin y}}{\text{sin x sin y}}}{\frac{\text{sin x cos y}}{\text{sin x sin y}}\text{+}\frac{\text{cos x sin y}}{\text{sin x sin y}}}$ â‡’ $\text{cot (x + y) =}\frac{\text{cot x cot y - 1}}{\text{cot y + cot x}}$      $\text{cot (x - y) =}\frac{\text{cot x cot y + 1}}{\text{cot y - cot x}}$ Replacing y by â€“ y, $\text{cot (x - (-y)) =}\frac{\text{cot x cot ( -y) - 1}}{\text{cot (-y) + cot x}}$ $\text{cot (x - y) =}\frac{\text{- cot x cot y - 1}}{\text{- cot y + cot x}}$ â‡’ $\text{cot (x - y) =}\frac{\text{cot x cot y + 1}}{\text{cot y - cot x}}$

Previous
âž¤