Setting trig functions equal to each other

[Clandestiny]

Hey there math forum. First off, thank you guys again so much for the help you've already given me! It's been invaluable.

Okay, so we're working on area between curves. Seeing as how we just completed an entire section on integrating trig functions, I imagine this next section will also primarily focus on trig functions. The first step I've taken is memorizing the actual graphs of trig functions (as well as cos and sin squared). Beyond that, I'm wondering if my approach will work in all, or at least most cases, and if there are any rules I should be aware of?

When approaching problems where I need to set two trig functions equal to each other, the approach that has worked on the sample problems for me is to follow these steps:

1. If the equality isn't immediately obvious, I find an identity so that the functions on both sides are the same function (for instance, changing sec(x) = cos(x) I'd change to 1/cos(x)=cos(x)).

2. Get the trig functions on the same side, dividing/multiplying as necessary, followed by getting the constants on the other side.

3. Reduce the constant by raising it to the power of 1/p, where p is the power the trig function is raised to.

4. Simplify, and find the angle x based on the given the information.

So in 'almost' all cases, this method has worked wonders. I'm concerned though that it only works for a particular type of equality, as one problem on my worksheet could not be solved that way. The problem was simply sin(x)=cos(x). The answer was obviously x = pi/4 just by knowing the graphs and the unit circle, but I have no idea how to mathematically derive that (if it's important).

I've done some google searches and have found examples for specific problems that seem to follow my procedure, but I'm wondering if there's a system of rules/guides/methods that I'm unaware of?

 

[Deleted member 4993]

Hey there math forum. First off, thank you guys again so much for the help you've already given me! It's been invaluable.

Okay, so we're working on area between curves. Seeing as how we just completed an entire section on integrating trig functions, I imagine this next section will also primarily focus on trig functions. The first step I've taken is memorizing the actual graphs of trig functions (as well as cos and sin squared). Beyond that, I'm wondering if my approach will work in all, or at least most cases, and if there are any rules I should be aware of?

When approaching problems where I need to set two trig functions equal to each other, the approach that has worked on the sample problems for me is to follow these steps:

1. If the equality isn't immediately obvious, I find an identity so that the functions on both sides are the same function (for instance, changing sec(x) = cos(x) I'd change to 1/cos(x)=cos(x)).

2. Get the trig functions on the same side, dividing/multiplying as necessary, followed by getting the constants on the other side.

3. Reduce the constant by raising it to the power of 1/p, where p is the power the trig function is raised to.

4. Simplify, and find the angle x based on the given the information.

So in 'almost' all cases, this method has worked wonders. I'm concerned though that it only works for a particular type of equality, as one problem on my worksheet could not be solved that way. The problem was simply sin(x)=cos(x). The answer was obviously x = pi/4 just by knowing the graphs and the unit circle, but I have no idea how to mathematically derive that (if it's important).

I've done some google searches and have found examples for specific problems that seem to follow my procedure, but I'm wondering if there's a system of rules/guides/methods that I'm unaware of?

sin(x)=cos(x)

cos(π/2 - x) = cos(x)

π/2 - x = x

x = π/4

 

[stapel]

...The first step I've taken is memorizing the actual graphs of trig functions (as well as cos and sin squared).

Good! In my experience, having a good grasp of the basic graphs (for me, it was sine mostly, and also some cosine) is a fabulous help in working through just about anything that has trig stuff in it.

When approaching problems where I need to set two trig functions equal to each other, the approach that has worked on the sample problems for me is to follow these steps....

So in 'almost' all cases, this method has worked wonders. I'm concerned though that it only works for a particular type of equality....

The method they've given you is a helpful guide. But you're in calculus now and, as you've no doubt noticed, there is a lot more flexibility, now that you've passed beyond simple algebra. Now you have to come up with stuff on your own.

In this topic, that's going to mean "fiddling with stuff, until you notice something 'working' for you" and also "asking experienced people for hints, because they've seen lots of helpful 'tricks' down the years".

...one problem on my worksheet could not be solved that way. The problem was simply sin(x)=cos(x).

There are various ways of solving things; you've been provided with one. Another is to divide through by the cosine:

. . . . .$\displaystyle \sin(x)\, =\, \cos(x)$

. . . . .$\displaystyle \dfrac{\sin(x)}{\cos(x)}\, =\, \dfrac{\cos(x)}{\cos(x)}$

. . . . .$\displaystyle \tan(x)\, =\, 1$

Can you solve it now?

 

[Ishuda]

Or another way, although getting the result tan(x) = 1 is better IMO:
sin(x) = cos(x)
square both sides
sin²(x)=cos²(x) = 1 - sin²(x)
so
2 sin²(x) = 1
or
sin(x) = $\displaystyle \pm\frac{1}{\sqrt{2}}$
The plus sign is obvious, but what about the minus sign?

The result above also brings up the matter of what domain are you working in? The $\displaystyle \frac{\pi}{4}$ result is fine if your domain is $\displaystyle [-\frac{\pi}{2},\, \frac{\pi}{2}]$ but what if it is $\displaystyle [-\pi,\, \pi]$.

Strange - I though I had already posted this.

 

[stapel]

Or another way, although getting the result tan(x) = 1 is better IMO:
sin(x) = cos(x)
square both sides
sin²(x)=cos²(x) = 1 - sin²(x)

Warning: If one squares both sides (which sometimes is the best step to take), one must remember to check all solutions. Squaring is an "irreversible step", and thus can generate extraneous solutions.

 

[Ishuda]

Warning: If one squares both sides (which sometimes is the best step to take), one must remember to check all solutions. Squaring is an "irreversible step", and thus can generate extraneous solutions.

Thanks for pointing that out. I had had that in the original but something got lost.