Class 12 : MATHS : SOLUTIONS:Chapter #3 : Trigonometric Functions

SOLUTIONS : CHAPTER 3 # TRIGONOMETRIC FUNCTIONS

Subject : Maths                                                         Class XII

Total Marks: 15                                                Time : 40 Min

Note: Following are the solution for The Question Bank   https://spectrum6.blogspot.in/2017/08/trigonometry-problems-simple-level-v.html

ANS : Q 1 Answer the following questions                       (5M)

a) The solutions of a trigonometric equation of an unknown angle x, between 0 to 2π are called (ii)Principle Solution.

b) Which of the following is not a trigonometric equation.

    iii) ⎷3x = 17.

c) 1 + Cot²Ө = i) Sec²Ө. 

d) If Cos Ө = 0 then Ө = ii) (2n+1)π/2.

e) For Sin x = ½ then the general solution of x is iii) π/6,5π/6,13π/6

ANS : Q2 Prove the following equations (any three)       (6M)

a) If tan ²x =  tan ²β then x = nπ  β

           As tan ²x =  tan ²β             

           Cos 2x  = Cos 2β

    using result Cos x  = Cos β , x=2nπ 土 β

                  2x  =  2nπ 土 2β

              ∴   x  =   nπ 土 β   

   ∴   tan ²x =  tan ²β  implies  x  =   nπ 土 β  where n∈Z        

b) Find the Principle solution of Sin x = 1/⎷2

       Sin x = 1/⎷2 = Sin π /4

       also ,Sin (π-x) = Sin x

 ∴   Sin π /4   = Sin (x- π /4) = Sin 3π /4      

 ∴   Sin π /4  =  Sin 3π /4 = 1/⎷2

       such that 0 < π /4  < 2π  and 0 < 3π /4 < 2π

  ∴  The required solutions are 

         x = π /4 and x = 3π /4.

c) Find the general Solution of CosӨ = 0

        here using CosӨ = 0 = Cos π /2 = Cos 3π /2 

        also using the periodicity, we get

        Cos π /2 = Cos (2π + π /2) =  Cos (4 π + π /2) = ......

   ∴  Cos (2nπ + π /2) = 0   for  n∈Z

d) Find the Principle solution of Cot x = 1/⎷3

                                        ∴  tan x  = ⎷3

                                        ∴  tan x  = ⎷3 = tan π /3 by using allied    

                                                                                        angles

                              ∴   tan (π + x) =  tan x

                             ∴   tan (2π + x)=  tan x

                              ∴   tan (π/3) =  tan (π + π/3) = ⎷3            

                            ∴   tan (π/3) =  tan (2π + π/3) = ⎷3

                                ∴   tan (4π/3) =  tan (7π/3) = ⎷3

Hence x = 4π/3 and 7π/3 are the required Principle solutions of the given equation Cot x = 1/⎷3.

ANS : Q3 Solve the following  (any three)                         (9M)

a) If   ⎷3 Cosec x = 2 then find the general Solution of equation. 

                 Cosec x = 2/⎷3

                           ∴  sin x = ⎷3/2 = sin π/3
                    ∴  sin x = sin π/3

         ∴Using the result that  sin x = sin α implies x= nπ + (-1)ⁿα

        We get,                        x = nπ + (-1)ⁿ π/3

∴ The required General Solution is

                    x = {nπ + (-1)ⁿ π/3} , where n ϵ Z

b) Find the Principle solution of tan x = -1/⎷3   

     tan π /6 = -1/⎷3   and by using allied angles

     tan (π - x) = - tan x  and

     tan (2π - x) = - tan x

     - tan π /6 = tan (π - π/6) = -1/⎷3
   - tan π /6 = tan (2π - π/6) = -1/⎷3

     tan 5π/6 π /6π /6 π /6= -1/⎷3 =   tan 11π/6

  Hence x = 5π/6 and 11π/6 are the required principal solutions   of given equation.

c) Convert 40⁰20’10’’ into radian measurement.

   Angle in Radian 
         = (π / 180) X Angle in Degree

         = (π / 180) X 40⁰20’10’’

         = (π / 180) X [40⁰ + (20’/ 60)⁰ + (10’’/60)’ ]

         = (π / 180) X [40⁰ + (20’/ 60)⁰ + (10’’/60 X 60)⁰ ]

         = (π / 180) X [40 + 1/3 + 1/360]⁰

         = (π / 180) X [40 + 0.33 + 0.003]

         = (π / 180) X [40.333]

         = (0.22 π)^c
Hence 40⁰20’10’’ = (0.22 π)^c