trig problems1

Question

1. Determine the value of cos 105 degrees, without the use of a calculator 2. Prove that: [(2 sin y+1)/(sin 2+cos y)]=sec y 3. Simplfy: Cosec y + cot y 4. If sin P=(12/13) and cos Q=(3/5) and both P and Q are acute angles, determine without the use of a calculator, the value of cos (P+Q). 5. Solve for P if: 4 (sin^2) P-1=0, where 0 less than equal to P less than equal to 360.

Answer 1

1)

cos (105°)

 105° can be written as 60 + 45

= cos (60° + 45°)

Use trignometric sum rule

cos (A + B) = cos A cos B - sin A sin B

= cos 60°cos 45° - sin 60° sin 45°

= (1/2) (1/√2 ) - (√3 /2) (1/√2 )

= (1/2√2)  - (√3 /2√2)

= (1 - √3 ) / 2√2

Answer 2

3)

 

 cosec y + cot y 

Recall the formula cosec y  = 1/ sin y and cot y = cos y / sin y

 = 1/ sin y  + cos y / sin y

  = (1 +  cos y)/ sin y

   = 2 cos² (y/2) / 2 sin (y/2) cos (y/2) 

   = cos (y/2) / cos (y/2)

   = cot (y/2)

 

Answer 3

4)

sin P=(12/13) and cos Q=(3/5)

sin P=(12/13)

cos P = √ (1- sin² P)

         = √ [1- (12/13)²]

        = √ 25/169

        = 5 / 13

cos Q=(3/5)

sin Q = √ (1- cos² P)

         = √ [1- (3/5)²]

        = √ 16/25

        = 4 / 5

cos (P +Q) = cos P cos Q - sin P sin Q

                 =cos P cos Q - sin P sin Q

                = (5/13)(3/5) - (12/13) (4/5)

                =  15/65 - 48/ 65

                = -33/65   

Answer 4

5)

The trignometric equation

sin² P - 1 =0

sin² P  = 1

sin P = ± 1

sin P = 1

sin P = sin π/2

The general solution is P = nπ + (-1)^n π/2   n ∈ Z.

At n =1 , P = π -π/2 = π/2

sin P = - 1

sin P = sin 3π/2

The general solution is P = nπ + (-1)^n 3π/2   n ∈ Z.

The solution is π/2    , 0 < P ≤ 360°