i hate trig...?

Question

idk how to solve these type of questions...


sin(theta) = - square root 3/2, tan(theta) > 0


can someone please give me step by step on how to solve this?

Answer 1

I'm assuming you meant

sin(A) = -sqrt(3)/2, tan(A) > 0
A = (5/3)pi or (4/3)pi

(4/3)pi falls under (-x,-y)
(5/3)pi falls under (x,-y)

since we know that tan(A) = x/y

for tan(A) to be greater than 0, (4/3)pi is your answer

ANS : (4/3)pi

just try it, tan((4/3)pi) = sqrt(3), tan((5/3)pi) = -sqrt(3)

Answer 2

Easy.

The first thing to know is the standard triangles. A 60/30 right triangle has sides in the ration of 1:root3:2, and the sine of 60 degrees is therefore root 3 / 2.

The second thing to know is the quadrants of the cartesian plane, and which quadrants have sine and cosine positive or negative.

Sine and cosine have the same sign in quadrants 1 (both positive) and 3 (both negative). Remember that sine is the y-coordinate and cosine is the x-coordinate of the intersection with the unit circle. Therefore, tan is positive in quadrants 1 and 3.

If the sine of the angle is negative and the tangent is positive, we are in quadrant 3. If you extend a 60 degree angle into quadrant 3, you are adding 180 degrees to it, and the result is 240 degrees.

Answer 3

This is a standard value
if sin x = -root(3) / 2
x = -60 degrees and -120 degrees.

But the only one of those that give a positive tan is -120.
This is the same as 240 degrees.

Answer 4

you opt to apply some trig identities like tan(x) = sin(x)/cos(x) so in case you remedy for cos and replace on your equation you get sin^2 + sin^2/tan^2 = a million or sin^2(a million + a million/tan^2) = a million or sin^2 = a million/(a million + a million/tan^2) so sin = sqrt( a million/ (a million + a million/ tan^2))

Answer 5

If you hate trig, you're screwed ☺

And your problem is screwed as well, because the value of the sine function never exceeds ±1 so sin(Θ) = -√(3/2) = -1.2247 has no value for Θ (except a complex number value, and you do *not* want to go there just yet ☺)

Doug