Find the all Trig Ratios

Question

1) tan a= 2, 0 < a < π/2

2) cos x= -1/3, π < x < 3π/2

Answer 1

1). Tan(a) = 2.

Pythagorean Identities: 1 + tan²θ = sec²θ

So, sec²(a) = 1 + [tan(a)]² = 1 + 2² = 1 + 4 = 5.

sec²(a) = 5.

Take square root each side.

√[sec²(a)] = √5.

sec(a) = √5

cos(a) = 1/sec(a) = 1/√5.

cot(a) = 1/tan(a) = 1/2

tan(a) = sin(a)/cos(a).

Multiply each side by cos(a).

sin(a) = [tan(a)][cos(a)]

Substitute tan(a) = 2 and cos(a) = 1/√5 in the equation.

sin(a) = (2)(1/√5) = 2/√5.

csc(a) = 1/sin(a) = 1/(2/√5) = √5/2.

Therefore sin(a) = 2/√5, cos(a) = 1/√5, tan(a) = 2, csc(a) = √5/2, sec(a) = √5 and cot(a) = 1/2.

Answer 2

2) cos(x) = -1/3, π < x < 3π/2

You are working with a reference angle and right triangle in quadrant III

cos(x) = -1/3 = adjacent / hypotenuse.

Opposite side = - √[(hypotenuse)² - (adjacent side)²]

Opposite side = - [(3)² - (-1)²] = - √[9 - 1] = - √(8) = - 2√2.

Trigonometric ratios:

sin(x) = - 2√2/3

cos(x) = - 1/3

tan(x) = 2√2

csc(x) = -3/2√2

sec(x) = - 3

cot(x) = 1/2√2.