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Trigonometry Help!!?

0 votes
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
asked Jul 24, 2014 in TRIGONOMETRY by anonymous

4 Answers

0 votes

B.

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

Left hand side expression = tan(T)

                                                = tan(120o)

                                                = tan(90o + 30o)

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

                                                = - √3.

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

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

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

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

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

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

                                                = - √3.

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

 

answered Jul 28, 2014 by casacop Expert
0 votes

A.

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

Left hand side expression = sin(225o).

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

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

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

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

              = Right hand side expression.

Therefore, the above statement is true.

 

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

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

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

                                          = Left hand side expression.

Therefore, the above statement is true.

 

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

Right hand side expression = tan(120o).

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

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

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

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

               = Left hand side expression.

Therefore, the above statement is true.

answered Jul 29, 2014 by casacop Expert
edited Jul 29, 2014 by casacop
0 votes

C.

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

⇒ cos(t) = - (√3)/2

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

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

cos(t) = - (√3)/2

           = - cos(30o)

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

cos(t) = cos(150o)

⇒ t = 150o.

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

cos(t) = - (√3)/2

           = - cos(30o)

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

cos(t) = cos(210o)

⇒ t = 210o.

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

 

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

⇒ cos(t) = 0

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

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

cos(t) = cos(90o)

⇒ t = 90o.

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

cos(t) = 0

           = cos(90o)

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

cos(t) = cos(270o)

⇒ t = 270o.

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

 

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

⇒ tan(t) = 1

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

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

tan(t) = 1

tan(t) = tan(45o)

⇒ t = 45o.

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

tan(t) = 1

           = tan(45o)

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

tan(t) = tan(225o)

⇒ t = 225o.

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

 

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

⇒ sin(3t) = 1/2

Let 3t = θ.

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

In first quadrant (0o < θ < 90o),

sin(θ) = 1/2

sin(θ) = sin(30o)

⇒ θ = 30o ⇒ 3t = 30o ⇒ t = 10o.

In second quadrant (90o < θ < 180o),

sin(θ) = 1/2

            = sin(30o)

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

sin(θ) = sin(150o)

⇒ θ = 150o ⇒ 3t = 150o ⇒ t = 50o.

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

answered Jul 29, 2014 by casacop Expert
0 votes

(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.

answered Jul 29, 2014 by casacop Expert

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