trig help questions

Question

1)Given: tan(θ)=−√3 and cos(θ)<0. Which of the following can be the angle θ?

4π/3

5pi/6

7pi/6

2pi/3

5pi/3

2)Let P(x,y) denote the point where the terminal side of an angle θ meets the unit circle. If P is in Quadrant IV and x=1/4, find tan(θ).

3)Let P(x,y) denote the point where the terminal side of an angle θ meets the unit circle. If P is in Quadrant II and y=3/7, find tan(θ)+sec(θ)

Answer 1

(1)

ASTC

tangent and cosine are negative in quadrant 2

inverse_tangent ( sqrt(3)) = 60

so the angle is 120 degrees = 2*pi/3

(2)

x^2 + y^2 = 1

(1/4)^2 + y^2 = 1

(1/16) + y^2 = 1

y^2 = 15/16

y = - sqrt(15)/4 <--- in quadrant 4

then tangent is -sqrt(15)/16 divided by 1/4 = - sqrt(15)/4

incidentally the angle is ALMOST 316 degrees

(3)

x^2 + y^2 = 1

x^2 + (3/7)^2 = 1

x^2 + 9/49 = 1

x^2 = 40/49

x = -2*sqrt(10)/7

tangent is 3/7 divided by -2*sqrt(10)/7 =

3/7 * 7/(-2*sqrt(10)) =

3 / (-2 * sqrt(10)) =

-3*sqrt(10) / 2

which incidentally makes the angle ALMOST 102 degrees.

cosine is = -2*sqrt(10)/7

so the secant is -7 / (2*sqrt(10)) = -7*sqrt(10) / 20

tangent + secant is (-30 * sqrt(10) + -7*sqrt(10))/20 = -37*sqrt(10)/20

Answer 2

1) cos(θ)<0 in quadrants II and III. tan(θ) is negative in quadrant II and positive in quadrant III, so we are looking for θ in quadrant II where both conditions are met.

Let sin(θ)=x. Remember the trigonometric identity: sin^2(θ)+cos^2(θ)=1, thus cos(θ)=√(1-x^2)

tan(θ) = sin(θ)/cos(θ) = x/√(1-x^2)

Given tan(θ)=−√3, x/√(1-x^2) = −√3

Solve: x = sin(θ) = (√3)/2

Remember, we are looking for θ in quadrant II such that its sine is (√3)/2. That's 120' or 2pi/3.

2) The x-coordinate will be the cosine. Given cos(θ)=1/4 and using the trigonometric identity, sin^2(θ)+1/16=1 thus sin(θ) = -(√15)/4 (it's negative because sine in quadrant IV is negative), and tan(θ) = sin(θ)/cos(θ) = -√15

3) This question utilizes the same concepts as #2. Can you try to solve it? The answer is below.

The y-coordinate is the sine. Given sin(θ)=3/7 and using the trigonometric identity, 9/49+cos^2(θ)=1 thus cos(θ) = -[2√10]/7 (negative because cosine in quadrant II is negative). tan(θ) = sin(θ)/cos(θ) = -3/[2√10] and sec(θ) = -7/[2√10], thus tan(θ)+sec(θ) = -10/[2√10]. Rationalize: -10/[2√10] = -[10√10]/20 = -√10/2