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Find the all Trig Ratios

+2 votes
1) tan a= 2, 0 < a < π/2
2) cos x= -1/3, π < x < 3π/2
asked Feb 23, 2013 in TRIGONOMETRY by angel12 Scholar

2 Answers

+3 votes

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.

answered Feb 23, 2013 by richardson Scholar
+1 vote

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.

answered Feb 23, 2013 by richardson Scholar
edited Feb 23, 2013 by moderator

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