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:
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.
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:
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
. Just divide by
So here we go.
Dividing by 13, we have
Let t be the angle with sine 5/13 (and hence cosine of 12/13).
Then our equation becomes
The left hand side is the sine angle sum...
arcsine
find the value of t
etc.
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:
LHS = cos angle difference:
Is it that the same answer? how do I know which coefficient to use on which axis?
Thanks.
Ok, cool. Would the function still have period ? and therefore an infinite amount of answers at where (I'm a little confused on how to determine the solution set).
Also, with the other method I get:
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.
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?