Trigonometry Help!!?

Question

A.Use special triangles to show these are true.
sin (225°) = sin (-45°)
cos (30°) = cos (-30°)
tan (-60°) = tan (120°)

B.For 120°, show that tan T = sin T/cos T.

C.Solve the below for 0° < t < 360°

2cost+√3=0
cost=0
2tant=2
2sin3t-1=0

D. Prove identities:
cotT sinT = cosT
sin2T sec2T = tan2T

Answer 1

B.

The angle θ = T = 120^o lies in second quadrant (90^o ≤ θ ≤ 180^o) and sin(θ) is positive and cos(θ) and tan(θ) are negative in second quadrant.

Left hand side expression = tan(T)

                                                = tan(120^o)

                                                = tan(90^o + 30^o)

                                                = - cot(30^o)                                      [tan(90^o + θ^o) = - tan(θ)]

                                                = - √3.

Right hand side expression = sin(T)/cos(T)

                                                = sin(120^o) / cos(120^o)

                                                = sin(90^o + 30^o) / cos(90^o + 30^o)

Since sin(90^o + θ^o) = cos(θ) and cos(90^o + θ^o) = - sin(θ).

                                                = cos(30^o) / [- sin(30^o)]

                                                = (√3/2) / (- 1/2)

                                                = - √3.

Therefore, tan(T) = sin(T) / cos(T).

 

Answer 2

A.

Statement : sin(225^o) = sin(-45^o).

Left hand side expression = sin(225^o).

The angle θ = 225^o lies in third quadrant (180^o ≤ θ ≤ 270^o) and sin(θ) is negative in third quadrant.

sin(225^o) = sin(180^o + 45^o)

               = - sin(45^o)                                       [sin(180^o + θ) = - sin (θ)]

               = sin(- 45^o)                                       [- sin(θ) = sin(- θ)]

              = Right hand side expression.

Therefore, the above statement is true.

 

Statement : cos(30^o) = cos(-30^o).

Right hand side expression = cos(-30^o).

                                          = cos(30^o)              [cos(-θ) = cos(θ)]

                                          = Left hand side expression.

Therefore, the above statement is true.

 

Statement : tan(-60^o) = tan(120^o).

Right hand side expression = tan(120^o).

The angle θ = 120^o lies in second quadrant (90^o ≤ θ ≤ 180^o) and sin(θ) is negative in second quadrant.

tan(120^o) =tan(180^o - 60^o)

               = - tan(60^o)                                       [tan(90^o + θ) = - cot(θ)]

               = tan(- 60^o)                                       [- tan(θ) = tan(- θ)]

               = Left hand side expression.

Therefore, the above statement is true.

Answer 3

C.

Equation : 2 cos(t) + √3 = 0 and 0^o < t < 360^o.

⇒ cos(t) = - (√3)/2

cos(t) is negative in second and third quadrant.

In second quadrant (90^o < t < 180^o),

cos(t) = - (√3)/2

           = - cos(30^o)

           = cos(180^o - 30^o)                           [ - cos(θ) = cos(180^o - θ) ]

cos(t) = cos(150^o)

⇒ t = 150^o.

In third quadrant (180^o < t < 270^o),

cos(t) = - (√3)/2

           = - cos(30^o)

           = cos(180^o - 30^o)                           [ - cos(θ) = cos(180^o + θ) ]

cos(t) = cos(210^o)

⇒ t = 210^o.

The solutions in the interval (0^o, 360^o) are 150^o and 210^o.

 

Equation : cos(t) = 0 and 0^o < t < 360^o.

⇒ cos(t) = 0

cos(t) is positive in first and fourth quadrant.

In second quadrant (0^o < t < 90^o),

cos(t) = cos(90^o)

⇒ t = 90^o.

In fourth quadrant (270^o < t < 360^o),

cos(t) = 0

           = cos(90^o)

           = cos(360^o - 90^o)                           [ cos(θ) = cos(360^o - θ) ]

cos(t) = cos(270^o)

⇒ t = 270^o.

The solutions in the interval (0^o, 360^o) are 90^o and 270^o.

 

Equation : 2 tan(t) = 2 and 0^o < t < 360^o.

⇒ tan(t) = 1

tan(t) is positive in first and third quadrant.

In first quadrant (0^o < t < 90^o),

tan(t) = 1

tan(t) = tan(45^o)

⇒ t = 45^o.

In third quadrant (180^o < t < 270^o),

tan(t) = 1

           = tan(45^o)

           = tan(180^o + 45^o)                           [ tan(θ) = tan(180^o + θ) ]

tan(t) = tan(225^o)

⇒ t = 225^o.

The solutions in the interval (0^o, 360^o) are 45^o and 225^o.

 

Equation : 2 sin(3t) - 1 = 0 and 0^o < t < 360^o.

⇒ sin(3t) = 1/2

Let 3t = θ.

The sin(θ) is positive in first and second quadrant.

In first quadrant (0^o < θ < 90^o),

sin(θ) = 1/2

sin(θ) = sin(30^o)

⇒ θ = 30^o ⇒ 3t = 30^o ⇒ t = 10^o.

In second quadrant (90^o < θ < 180^o),

sin(θ) = 1/2

            = sin(30^o)

           = sin(180^o - 30^o)                           [ sin(θ) = sin(180^o - θ) ]

sin(θ) = sin(150^o)

⇒ θ = 150^o ⇒ 3t = 150^o ⇒ t = 50^o.

The solutions in the interval (0^o, 360^o) are 10^o and 50^o.

Answer 4

(D).

Identity : cot(T) * sin(T) = cos(T).

Left hand side expression = cot(T) * sin(T)

                                        = [ cos(T) / sin(T) ] * sin(T) [ Reciprocal identity : cot(θ) = cos(θ) / sin(θ) ]

                                        = cos(T)

                                        = Left hand side expression.

Therefore, Identity is proved.

 

Identity : sin(2T) * sec(2T) = tan(2T).

Left hand side expression = sin(2T) * sec(2T)

                                        = sin(2T) / cos(2T)  [ Reciprocal identity : sec(θ) = 1/cos(θ) ]

                                        = tan(2T)                [ Reciprocal identity : sin(θ) / cos(θ) = tan(θ) ]

                                        = Right hand side expression.

Therefore, Identity is proved.