Question

Question concerning pre-calculous/trig equation on unit circle.

Hello! I'm currently struggling with understanding { how to write general equation for all solutions given either sin, tan or cos. }

For instance: sin θ = 1/2

How do I determine the angle and derive an equation from this?

Answer 1

@Michele_Laino @ganeshie8 @anyone

Answer 2

I've isolated that sin must be in the first and second quadrant because of y.

Answer 3

Another that is confusing is cos θ = 0

Answer 4

Help is greatly appreciated!

Answer 5

Could you provide a link by chance? Unable to find the video.

Answer 6

I am already familiar with the unit circle and not struggling with its understanding. It's determining sin theta. 1/2 corresponds to 30 degrees I've figured and the other is 150. Only problem is creating an equation.

Answer 7

I don't mind if you're unable to answer. I'll await for another. Thank you, nonetheless.

Answer 8

To solve the equation, you'd have to take the inverse sin of both sides

(sin^-1)sinx=(sin^-1)1/2

The sin^-1 & and sinx on the left cancels. Then you're just left with
x=(sin^-1)1/2
That's where referring to the unit circle comes in x=30° and x=150°

Answer 9

Or is the question, derive the other trig functions if sinx=1/2? (I just reread the question.)

Answer 10

Sine and cosine both have a range of 2*pi, so for any value of x, ... = sin(x - 4*pi) = sin(x - 2*pi) = sinx = sin(x+2*pi) = sin(x+4*pi) = ..., and same for cosine. In addition, for any value of y, sinx = y and cosx = y will both have two solutions equidistant from the x axis for x within the range [0,2pi]. Sine will have solutions in either quadrants 1 & 2 or 3 & 4, and cosine will have solutions in quadrants 1 & 4 or 2 & 3.

So then for example, working in degrees and given sinx = 0.5, the simple solution is x = 30. Since this is in quadrant 1, the other solution is going to be in quadrant 2 and 30 degrees from the x axis, or x = 150. However, adding or subtracting any multiple of 2pi radians, or 360 degrees, from either solution will also be a solution. So then the general equation will be:
x = 30 + 360k or x = 150 + 360k, k E I
(If you're not familiar, \[k \in I\] means "for any k in the set of integers")

Similarly, let's look at cosx = -0.5. Then x = 120 is one solution, given by the inverse cos function. This is in quadrant 2. Then the other solution will also be 60 degrees from the x axis and in quadrant 3, or x = 240. Then the general equation will be:
x = 120 + 360k or x = 240 + 360k, k E I

Tan is a little different for two reasons. First, Tan has a range of pi, not 2pi. Second, for any given value of y, tanx = y only has one solution in the range [0,pi]. So creating a general solution for tan is easy - it's just x + k*pi.

For example, let's use tanx = 1. The solution to this in [0, pi] is x = 45. Then the general solution is x = 45 + k*pi, k E I