Inverse functions and substitution

**terrorsquid** · August 17th 2011, 07:13 AM

Hi, I am having trouble finding resources on how to apply the properties of inverse functions arcsin, arccos and arctan, as well as how to use substitution to solve equations such as:

$5cos(x) + 12cos(x) = 3~$

$~0\leq x \leq 2\pi$

Everything I come across seems to go over the properties of the inverse functions and graph them etc. which I already understand; however, I am still confused as to how I would apply the properties to solve trig equations.

How should I go about solving the above problem and can anyone direct me to some more examples?

Thanks.

**TheChaz** · August 17th 2011, 07:15 AM

Originally Posted by terrorsquid

Hi, I am having trouble finding resources on how to apply the properties of inverse functions arcsin, arccos and arctan, as well as how to use substitution to solve equations such as:

$5cos(x) + 12cos(x) = 3~$

$~0\leq x \leq 2\pi$

Everything I come across seems to go over the properties of the inverse functions and graph them etc. which I already understand; however, I am still confused as to how I would apply the properties to solve trig equations.

How should I go about solving the above and can anyone direct me to some more examples?

Thanks.

Solve for cos(x), then solve for x!
Since 5 + 12 = 17, we have
17cos(x) = 3
cos(x) = 3/17
x = arccos(3/17) and 2pi - arccos(3/17)

Now if the problem had been
5cos(x) + 12sin(x) = 3, then we would take a different (and more interesting) approach...

**Siron** · August 17th 2011, 02:20 PM

Are you also searching for other types of goniometric equations? ...

**terrorsquid** · August 17th 2011, 09:03 PM

Originally Posted by TheChaz

Now if the problem had been
5cos(x) + 12sin(x) = 3, then we would take a different (and more interesting) approach...

This is what I meant to write and is what I have been trying to find examples of but I can only seem to find definitions of inverse functions or trig substitution used to integrate and not for solving trig equations.

The only thing I seem to spot is that the squares of the coefficients on the LHS add up to a clean 169 so if you divided both sides by 13, you would you have the squares of the coefficients = 1.

How would you apply substitution or the properties of inverse functions to the eqation:

$5cos(x) + 12sin(x) = 3$

Never thought I'de like trig, but this has got me interested again

**terrorsquid** · August 17th 2011, 09:07 PM

Originally Posted by Siron

Are you also searching for other types of goniometric equations? ...

If 5cos(x) + 12sin(x) = 3 is a goniometric equation then yes

**Siron** · August 18th 2011, 04:05 AM

I think you can try something with the t-formulas, let $t=\tan\left(\frac{x}{2}\right)$ .
Are you familiar with this? ...

**TheChaz** · August 18th 2011, 04:57 AM

Originally Posted by terrorsquid

This is what I meant to write and is what I have been trying to find examples of but I can only seem to find definitions of inverse functions or trig substitution used to integrate and not for solving trig equations.

The only thing I seem to spot is that the squares of the coefficients on the LHS add up to a clean 169 so if you divided both sides by 13, you would you have the squares of the coefficients = 1.

How would you apply substitution or the properties of inverse functions to the eqation:

$5cos(x) + 12sin(x) = 3$

Never thought I'de like trig, but this has got me interested again

You are on the right track! And while (5, 12, 13) is a pythagorean triple, you can perform this with any values of A and B for

$A\cos(x) + B\sin(x) = D$ . Just divide by $\sqrt{(A^2 + B^2)}$

So here we go.
Dividing by 13, we have

$\frac{5}{13} \cos(x) + \frac{12}{13} \sin(x) = \frac{3}{13}$

Let t be the angle with sine 5/13 (and hence cosine of 12/13).

Then our equation becomes

$\sin(t) \cos(x) + \cos(t) \sin(x) = \frac{3}{13}$

The left hand side is the sine angle sum...

$\sin(t + x) = \frac{3}{13}$

arcsine
find the value of t
etc.

**terrorsquid** · August 18th 2011, 05:37 AM

Ah, ok. I got to the point where I could see that with the new coefficients, the equation was obviously on the unit circle; however, when I drew in the coordinates 5/13 and 12/13, I didn't know which was cos(x) and which was sin(x) (x,y) on the circle. I tried once just guessing and put +5/13 on the x axis and +12/13 on the y axis. Therefore, I had:

$cos(t) = \frac{5}{13}$

$sin(t) = \frac{12}{13}$

$cos(t)cos(x) + sin(t)sin(x) = \frac{3}{13}$

LHS = cos angle difference:

$cos(t - x) = \frac{3}{13}$

$arccos\left(\frac{3}{13}\right) = t - x$

$x = t - arccos\left(\frac{3}{13}\right) = arccos\left(\frac{5}{13}\right) - arccos\left(\frac{3}{13}\right)$

Is it that the same answer? how do I know which coefficient to use on which axis?

Thanks.

**terrorsquid** · August 18th 2011, 05:44 AM

Originally Posted by Siron

I think you can try something with the t-formulas, let $t=\tan\left(\frac{x}{2}\right)$ .
Are you familiar with this? ...

I have something written down:

$sin(2\theta) = \frac{2t}{1 + t^2}$

$cos(2\theta) = \frac{1 - t^2}{1 + t^2}$

I don't know where they came from though and don't understand their use; the only identity I know for $sin(2\theta)$ and $cos(2\theta)$ would be the double angle identities.

**Siron** · August 18th 2011, 06:22 AM

That's ok, now you can wright the equation as:
$5\cdot \left(\frac{1-t^2}{1+t^2}\right)+12\cdot \left(\frac{2t}{1+t^2}\right)-3=0$

$\Leftrightarrow \frac{5(1-t^2)+24t-3(1+t^2)}{1+t^2}=0$

Solve this equation for $t$ and afterwards do the back-substitution.

**TheChaz** · August 18th 2011, 06:35 AM

Originally Posted by terrorsquid

$cos(t) = \frac{5}{13}$

$sin(t) = \frac{12}{13}$

$cos(t)cos(x) + sin(t)sin(x) = \frac{3}{13}$

LHS = cos angle difference:

$cos(t - x) = \frac{3}{13}$

$arccos\left(\frac{3}{13}\right) = t - x$

$x = t - arccos\left(\frac{3}{13}\right) = arccos\left(\frac{5}{13}\right) - arccos\left(\frac{3}{13}\right)$

Is it that the same answer? how do I know which coefficient to use on which axis?

Thanks.

I prefer sine, so I chose it as such.
They are the same, but don't forget that there are infinite (though coterminal) solutions - and that arccos(5/13) = arcsin(12/13) and vice versa.

**terrorsquid** · August 18th 2011, 08:08 AM

Ok, cool. Would the function still have period $\pi$ ? and therefore an infinite amount of answers at $\pi k$ where $k \in \mathbb{Z}$ (I'm a little confused on how to determine the solution set).

Also, with the other method I get:

$t = \frac{\pm\sqrt{10} + 3}{2}$

So, I subbed the two values in for cos(x) and solved for sin(x) and vice versa to get 4 different answers. approx ~ 9,-9, 35 and 144. Again, I don't know how to define the infinite set of answers or, if there was a given restriction, how to discern which are valid answers.

**Siron** · August 18th 2011, 09:48 AM

Is this the $t$ you become by using the t-formulas?
The only thing you've to do now is the back-substitution:
$t=\tan\left(\frac{x}{2}\right) \Leftrightarrow \frac{x}{2}=\arctan(t)+k\cdot \pi \Leftrightarrow x=2\cdot \arctan(t)+2k\pi$

Can you continue now? (as The Chaz noticed there are infinite solutions)

**terrorsquid** · August 19th 2011, 01:13 AM

Originally Posted by Siron

Is this the $t$ you become by using the t-formulas?
The only thing you've to do now is the back-substitution:
$t=\tan\left(\frac{x}{2}\right) \Leftrightarrow \frac{x}{2}=\arctan(t)+k\cdot \pi \Leftrightarrow x=2\cdot \arctan(t)+2k\pi$

Can you continue now? (as The Chaz noticed there are infinite solutions)

I'm ok solving the equations now, I am just having trouble visualising which quadrants the function appears in because there is a sin and a cos in it. If it was just cos(x) = 1/2, I could just think cos(x) is positive therefore it is in the 1st and 4th quadrants and then think it is either 60 or 360-60 (and 2pi k multiples of each). Or if it was sin(x) = 0 I can imagine the sin curve and know that it will be 0 at 0 and any integer multiple of pi. But how do you interpret a heterogeneous function like this one and know where it will reoccur?

**terrorsquid** · August 19th 2011, 01:49 AM

Originally Posted by terrorsquid

$cos(t) = \frac{5}{13}$

$sin(t) = \frac{12}{13}$

$cos(t)cos(x) + sin(t)sin(x) = \frac{3}{13}$

LHS = cos angle difference:

$cos(t - x) = \frac{3}{13}$

$arccos\left(\frac{3}{13}\right) = t - x$

$x = t - arccos\left(\frac{3}{13}\right) = arccos\left(\frac{5}{13}\right) - arccos\left(\frac{3}{13}\right)$

Say, for example, the above was between $-\pi$ and $\pi$ what would you be thinking when figuring out the solution set?

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