trig problems 3

Question

1. Derive a formula for cos (A/2) if cos 2 A = 2 (cos^2) A-1. Use the formula to calculate the value of cos 15 degrees without a calculator 2. Solve for x if: (Sec^2)x + tan x -3 =0 ; 0 degrees less than equal 0 less than equal 360. 3. Calculate the value of sin 120degrees without the use of a calculator. 4. Prove that: Tan(90-B)=cot B

Answer 1

1)

cos 2 A = 2 (cos^2) A-1

Substitute A/2 for A.

cos A = 2 (cos^2) (A/2) -1

Add 1 to each side .

cos A + 1 = 2 (cos^2) (A/2)

Divide each side by 2

(cos^2) (A/2) = (cos A + 1) / 2

Square root on both sides.

cos (A/2) = √ [(cos A + 1) / 2]

Substitute A = 30° in above formula.

cos (30°/2) = √ [(cos 30° + 1) / 2]

cos (15°) = √ [(cos 30° + 1) / 2]

cos (15°) = √ [(√3/2) + 1) / 2]

                = √ [(√3 + 2) / 4]

                = √ [(√3 + 2)]/ 2

Answer 2

2)

(sec^2)x + tan x -3 =0 

(1+tan^2 (x)) + tan x -3 =0 

tan^2 (x) + tan x  - 2 =0

tan^2 (x) + 2tan x - tan x  - 2 =0

tanx (tanx + 2) - 1(tanx + 2) =0

(tanx + 2) ( tan x - 1) =0

tanx + 2 =0

tanx = -2

tanx = tan (- 63.43°)

The general solution is x = nπ - 63.43°, n ∈ Z.

At n = 1, x = π - 63.43° = 116.6°

At n = 2, x = 2π - 63.43° = 296.6°

tanx - 1 =0

tanx = 1

tanx = tan (45°)

The general solution is x = nπ + 45°, n ∈ Z.

At n = 1, x = π + 45° = 225°

The solutions are  x = 116.6°, 225°, and 296.6°

Answer 3

3)

sin (120) = sin (90 + 30)

Using the trignometric sum formula: sin (A + B) = sin A cos B + cos A sin B

= sin 90 cos 30 + cos 90  sin 30

= 1* cos 30 + 0* sin 30

= cos 30

= √3/2

Answer 4

4)

tan (90 - B)

Recall the formula tan (A - B) = (tan A - tan B )/ (1+ tan A tan B)

tan (90 - B) =( tan 90 - tan B ) (1 + tan 90 tan B)

                  =  ( (1/0) - tan B ) (1 + (1/0) tan B)

                 = 1 / tan B

                = cot B