Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,435 questions

17,804 answers

1,438 comments

774,936 users

Trigonometry help?

0 votes
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).
asked May 13, 2014 in TRIGONOMETRY by anonymous

3 Answers

0 votes

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.

answered May 13, 2014 by lilly Expert
The above solution for 1. (a) is wrong,

See the correct solution, answered by the steve.
0 votes

 

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).

 

answered May 13, 2014 by lilly Expert
0 votes

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.

graph cos(x) = - 3/4

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.

graph cos(x) = - 3/4

answered May 23, 2014 by steve Scholar

Related questions

asked Jul 16, 2014 in TRIGONOMETRY by anonymous
asked May 13, 2014 in TRIGONOMETRY by anonymous
asked Apr 23, 2014 in TRIGONOMETRY by anonymous
asked Jul 22, 2014 in TRIGONOMETRY by anonymous
asked Jun 24, 2014 in TRIGONOMETRY by anonymous
asked Sep 25, 2014 in ALGEBRA 2 by anonymous
asked Feb 9, 2015 in TRIGONOMETRY by anonymous
asked Jul 28, 2014 in TRIGONOMETRY by anonymous
asked Jul 16, 2014 in TRIGONOMETRY by anonymous
asked Apr 21, 2014 in TRIGONOMETRY by anonymous
...