Finding Real Solutions for Trigonometric Equations

[FaraDazed]

The question is "Find the real solutions θ for both of the following, a. sinθ=1/2, b. sin(7θ)=sin(5θ)"

On part a, I know that theta must be 30° or π/6 but I know that I need to mathematically show it and not sure how. And part b just has me head spinning, don't know where to start on that one.

 

[haruspex]

The question is "Find the real solutions θ for both of the following, a. sinθ=1/2, b. sin(7θ)=sin(5θ)"

On part a, I know that theta must be 30° or π/6 but I know that I need to mathematically show it and not sure how. And part b just has me head spinning, don't know where to start on that one.

Anyway, for a., it's reasonable to assume π/6 is a solution; your problem is to find all the others. If you're not sure what the other solutions are, I suggest sketching a graph.
For b., I haven't tried it, but my approach would be to write 7 as 6+1 and 5 as 6-1, then use the usual trig formulae to expand the sines.

 

[mfb]

sin(a)=sin(b) has two classes of solutions:
1) both points are in a region where the sine is increasing OR both are in a region where the sine is falling.
2) one point is in a region where the sine is increasing, the other is in a region where the sine is falling.
(solutions with sin(a)=sin(b)=1 or -1 are in both categories)

Those cases can be reduced to conditions for a-b and a+b respectively, and then you have a simple linear equation.

 

[HallsofIvy]

"On part a, I know that theta must be 30° or π/6 but I know that I need to mathematically show it and not sure how."

What, exactly to you feel would show that? Would putting "30 degrees" into a calculator and determining that "sin(30)" is equal to 0.5 be enough?

How about if you construct an equilateral triangle, so that all three angles are 60 degrees and all three sides have length 1. If you drop a perpendicular from one vertex to the opposite side, it divides the equilateral triangle into two right triangles with angles 30 and 60 degrees, hypotenuse of length 1 and leg opposite the 30 degree angle of length 1/2. The sine of 30 degrees is (1/2)/1= 1/2.

But I doubt that your teacher would require that. (It is not nearly so "nice" for other angles.) It should be enough to assert that "sin(60)= 1/2" or use a calculator to show that "arcsin(1/2)= 60 degrees".

But what I would consider the important part, you are not even considering! "sin(t)" is the y-component of the a point distance "t" around the unit circle. Drawing the horizontal line, y= 1/2, across the unit circle, it is easy to see that it crosses twice, once at "x" and the other at "180- x". So in addition to 60 degrees, another solution is 180- 60= 120. But sine is periodic with period 360 degrees to all solutions are of the form 60+ 360n or 120+ 360n degrees.

(You gave "60 degrees" first and then "$\pi/3$ radians". In fact, "degrees" are only used in problems that actually have angles measured in degrees. Most applications of "trig function" problems have little or no application to actual angles and the arguments are given in radians. I would consider the best answer to this problem $\pi/3+ 2\pi n$ and $2\pi/3+ 2\pi n$.)

 

[FaraDazed]

(You gave "60 degrees" first and then "$\pi/3$ radians". In fact, "degrees" are only used in problems that actually have angles measured in degrees. Most applications of "trig function" problems have little or no application to actual angles and the arguments are given in radians. I would consider the best answer to this problem $\pi/3+ 2\pi n$ and $2\pi/3+ 2\pi n$.)

Thank you for helping. :)

It is sine not cos mind you, and only +1/2 as a solution (i think). So that would be $\frac{\pi}{6} + 2 \pi n$ and $\frac{5\pi}{6} + 2 \pi n$ right?

 

[haruspex]

Fwiw, I have now tried what I suggested in post #2 for b and can report that it works extremely well.

 

[FaraDazed]

Fwiw, I have now tried what I suggested in post #2 for b and can report that it works extremely well.

Thanks, I gave what I think you were on about a go and have not got very far. Did you mean the general addition formumla ie for sin(A+B) etc?

I did...

$$sin(7\theta)=sin(5\theta) \\<br /> sin(6\theta+1\theta)=sin(6\theta-1\theta) \\<br /> sin(6\theta)cos(\theta)+sin(\theta)cos(6\theta)=sin(6\theta)cos(\theta)-sin(\theta)cos(6\theta) \\<br /> sin(\theta)cos(6\theta)=-sin(\theta)cos(6\theta) \\<br /> 2(sin(\theta)cos(6\theta))=0$$

I don't really know where I am going with that one :/

 

[haruspex]

Thanks, I gave what I think you were on about a go and have not got very far. Did you mean the general addition formumla ie for sin(A+B) etc?

I did...

$$sin(7\theta)=sin(5\theta) \\<br /> sin(6\theta+1\theta)=sin(6\theta-1\theta) \\<br /> sin(6\theta)cos(\theta)+sin(\theta)cos(6\theta)=sin(6\theta)cos(\theta)-sin(\theta)cos(6\theta) \\<br /> sin(\theta)cos(6\theta)=-sin(\theta)cos(6\theta) \\<br /> 2(sin(\theta)cos(6\theta))=0$$

I don't really know where I am going with that one :/

If the product of two expressions is zero, what can you say about the two expressions?

 

[FaraDazed]

If the product of two expressions is zero, what can you say about the two expressions?

That at least one of them must equal zero??

 

[mfb]

That at least one of them must equal zero??

Sure. What about the next step?

 

[FaraDazed]

Sure. What about the next step?

I am really not sure sorry. Maybe, if sinθ=0 then θ must equal 0 and/or π,. If cos(6θ)=0 then θ must equal (π/2)*6 = 3π and/or (3π/2)*6= 9π.

 

[haruspex]

I if sinθ=0 then θ must equal 0 and/or π,.

It can't be 0 and pi, but yes, it could be 0 or pi or ...?

If cos(6θ)=0 then θ must equal (π/2)*6 = 3π and/or (3π/2)*6= 9π.

Again, it's or, not and, and there is an infinity of solutions.