# General solutions problems

1. Jul 3, 2012

### jackscholar

I have two problems that were posed to me and without having any knowledge on the subject I cannot understand what they ask of, nor do I know the answers. These questions are:
Solve sinθ=cosθ for -2pi<θ<2pi

and find and show the general solution to sinθ=a.

Any help or explanations would be highly appreciated.

2. Jul 3, 2012

### chiro

Hey jackscholar and welcome to the forums.

The first problem relates to having a function f(x) = sin(x) and g(x) = cos(x) and finding when these lines 'cross' when x > -2pi and x < 2pi.

The way to algebraically solve this is to use:

sin(x) = cos(x).
sin(x)/cos(x) = 1 (cos(x) <> 0). But sin(x)/cos(x) = tan(x) So
tan(x) = 1.

Now what we do here is use what is called an inverse function. Inverses basically undo a function to get the original argument which in this case is x. So lets say we calculate f(x). If we then do f^-1(f(x)) (i.e. we take the value of f(x) and we put this value in our inverse), then if its a proper inverse function we get back x. From this, we apply the inverse function to both sides and we get:

tan^-1(tan(x)) = tan^-1(1)

x = tan^-1(1).

Now this has a solution of pi/4, 3pi/4, 7pi/4 for positive x as well -3pi/4 -7pi/4. for negative x. The reason for this is that sin(pi/4) = 1/SQRT(2) and cos(pi/4) = 1/SQRT(2). You then need to use other properties to show for other values.

In terms of why you need to find it, there are many reasons why you would and it's not only a one reason answer: it depends on the problem.

3. Jul 3, 2012

### jackscholar

Thank you for your help, it is highly appreciated. For the second problem, would it be similar to the general solution for tanθ=a because they both only have one solution between -pi/2 and pi/2?

4. Jul 3, 2012

### chiro

It's not the same for the second problem sin(x) = a.

For this problem you need to apply the inverse sine function (call it sin^-1(x)) to both sides. The reason I did a tan is because I divided both sides by cos(x) since the RHS had a cos(x) in it so that I got tan(x) = 1 (remember tan(x) = sin(x)/cos(x)).

So by applying inverse function to both sides I get:

sin^-1(sin(x)) = sin^-1(a) which means
x = sin^-1(a).

Since you haven't supplied a value for a, the solution will depend on what value of a you pick. If for example a = 1, the solution would be x = pi/2. If it was -1 it would be -pi/2. If it was 0 then x would be 0.

If you supply a general number, you tend to use a calculator or a computer to get the answer that is approximate, but close enough to use for most purposes unless you doing say an engineering computation or something serious where you would use a lot more accuracy, and possibly complex math to make sure the answers you get are correct to so many decimal places.

5. Jul 3, 2012

### jackscholar

Well thank you for all your help, it is easier to interpret your workings than the block of text i was presented with from my teacher.

6. Jul 3, 2012

### haruspex

I think the key word in the second problem is 'general'.

jackscholar,
Many common functions can give the same output for two different inputs. -2 and +2 have the same square. sin(π/6) = sin(5π/6). When you try to define an inverse function for these, there is an ambiguity. Is the square root of four 2 or -2?
To get around this, certain conventions are adopted. By definition, the function √ returns the positive value. If, in the algebra, you know it's the negative value that's wanted you write -√, or if it could be either then ±√. Similarly with trig functions. I believe the standard range for arcsin is (-π/2, +π/2].
The question asks for the general solution. One solution is arcsin(a), a number greater than -π/2 and less than or equal to +π/2. What you have to do is state what all the (infinity of) valid answers are to the equation.

Last edited by a moderator: May 6, 2017
7. Jul 4, 2012

### chiro

Good point haruspex! jackscholar, what haruspex is saying is important to be aware of so I'd take a geezer at the above post.

8. Jul 5, 2012

### jackscholar

Thank you both for your contributions.