Trigonometry help?

Question

1. a) Given that cos⁡θ=-3/4 and θ is in the third quadrant, find sinθ and cosθ exact value no decimal. Include a diagram.

b) If sin⁡θ= 3/8 and θ is in the first quadrant, what is sin⁡2θ. Include a diagram.

2. Simplify cos y (tan y – sec y).

3. Solve cos⁡2x + sinx = 1 in [0,2π).

Answer 1

1).

a).

cos θ = - 3/4 and θ is in third quadrant.

So, cos (- θ) = cos θ and sin (- θ) = - sin θ.

Therefore, exact value of cos θ = 3/4, and,

Exact value of sin θ = √ (1 - cos² θ) = √ (1 - 9/16) = √[(16 - 9)/16] = √7/4.

b).

sin θ = 3/8 and θ is in first quadrant.

cos θ = √ (1 - sin² θ) = √ (1 - 9/64) = √ [(64 - 9)/64 ] = √55/8.

sin 2θ = 2 sin θ cos θ

= 2 (3/8)(√55/8)

= 3√55/32.

2).

cos y(tan y - sec y)

= cos y[ (sin y/cos y) - (1/cos y))                        (∴ tan y = sin y/cos y and sec y = 1/cos y)

= cos y[ (sin y - 1)/cos y ]

= sin y - 1.

Therefore, cos y(tan y - sec y) = sin y - 1.

Comments

The above solution for 1. (a) is wrong,

See the correct solution, answered by the steve.

Answer 2

  

3).

The trigonometric equation is cos 2x + sin x = 1.

1 - 2sin² x + sin x - 1 = 0                                           (∴ cos 2x = 1 - 2sin² x)

2sin² x - sin x = 0

sin x(2 sin x - 1) = 0

sin x = 0 and 2sin x - 1 = 0

sin x = 0 and sin x = 1/2

x = sin^-1 (0) and x = sin^-1 (1/2)

The function sin(x) has a period of 2π, first find all solutions in the interval (0, 2π).

The function sin(x) ispositive in first and second quadrant.

In first Quadrant, 0 ≤ x ≤ π/2.

x = sin^-1 (0) = sin^-1 (sin (0)) = 0, and

x = sin^-1 (1/2) = sin^-1 (sin (π/6)) = π/6

In second Quadrant, π/2 ≤ x ≤ π.

  x = sin^-1 (0) = sin^-1 (sin (π/2 + π/2)) = 2π/2 = π, and  

x = sin^-1 (1/2) = sin^-1 (sin (π - π/6)) = sin^-1 (sin (5π/6)) = 5π/6.

The solutions x = 0, x = π, x = π/6 and x = 5π/6 in the interval (0, 2π).

  

Answer 3

Definitions of Trigonometric Functions of Any Angle :

Let θ be an angle in standard position with (x, y) a point on the terminal side of θ and r = √(x² + y²) ≠ 0.

sin(θ) = y/r,                               cos(θ) = x/r,

tan(θ) = y/x,   x ≠ 0                   cot(θ) = x/y,   y ≠ 0,

sec(θ) = r/x,   x ≠ 0                   csc(θ) = r/y,   y ≠ 0.

1 . (a)

The trigonometric function is cos(θ) = - 3/4 and θ lies in the third quadrant.

cos(θ) = x/r = - 3/4.

The value of x is negative in Quadrant  III, so x = - 3 and r = 4.

To find the value of y, substitute the values of  x = - 3 and r = 4 in r = √(x² + y²).

(4) = √[ (- 3)² + y² ]

16 = 9 + y²

7 = y²

y = ± √7.

Here θ lies in the third quadrant, so the value of y is negative in in Quadrant  III,  → y = - √7.

sin(θ) = y/r = - √7/4.

1 . (b)

The trigonometric function is sin(θ) = 3/8 and θ lies in the first quadrant.

The double angle formula of sine function : sin(2θ) = 2 sin(θ) cos(θ).

First find the cos(θ) :

Using the Pythagorean identity sin² θ + cos² θ = 1.

(3/8)² + cos² θ = 1

9/64 + cos² θ = 1

cos² θ = 1 - 9/64 = (64-9)/64 = 55/64.

Because cos θ > 0 in Quadrant I, so here positive root to obtain cos θ = √55/√64 = (√55)/8.

Next find sin(2θ) :

sin(2θ) = 2 sin(θ) cos(θ) = 2 (3/8) (√55)/8 = (3√55)/32.