# Possibly Evaluating a Trig Function of an Inverse Trig Function

• Apr 23rd 2011, 10:26 PM
ragnar
Possibly Evaluating a Trig Function of an Inverse Trig Function
I'm asked to find (root-3)sinx - cosx = -1 on the interval [0, 2pi). It comes in the middle of some questions which ask to evaluate trig functions of inverse trig functions, like cos(sin^-1(1/4)) so that tells me I'm supposed to do something like this at some point, but I can't see how.

I came up with a solution though somewhat inelegant, and I'm wondering if anyone sees a better solution. Mine went: Move the -1 and -cosx to the other sides of the equation, square both sides, factor sinx, solve each factor for 0, test each solution in the interval, obtain only 0 and 4pi/5.
• Apr 24th 2011, 12:53 AM
Opalg
Quote:

Originally Posted by ragnar
I'm asked to find (root-3)sinx - cosx = -1 on the interval [0, 2pi). It comes in the middle of some questions which ask to evaluate trig functions of inverse trig functions, like cos(sin^-1(1/4)) so that tells me I'm supposed to do something like this at some point, but I can't see how.

Divide both sides by 2, to get (√3/2)sin x – (1/2)cos x = –1/2. Then notice that √3/2 = cos(π/6) and 1/2 = sin(π/6), and use the formula for sin(x–y).
• Apr 25th 2011, 09:33 AM
HallsofIvy
Quote:

Originally Posted by ragnar
I'm asked to find (root-3)sinx - cosx = -1 on the interval [0, 2pi). It comes in the middle of some questions which ask to evaluate trig functions of inverse trig functions, like cos(sin^-1(1/4)) so that tells me I'm supposed to do something like this at some point, but I can't see how.

I'm not sure how you would go from cos(sin^(-1)(1/4)) to that! You know that cos(x)= sqrt(1- sin^2(x)) and that sin(sin^(-1)(1/4)= 1/4. So cos(sin^(-1)(1/4))= sqrt(1- (1/4^2)= sqrt(1- 1/16)= sqrt(15)/4. (The plus sign is assuming the principle value for sin^(-1).)
Another say to do this is to the angle, t, such that sin(t)= 1/4 would be, in a right triangle, the angle opposite a leg of length 1 while the hypotenuse had length 4. By the Pythagorean theorem, the other leg has length sqrt(4^2- 1)= sqrt(15) so cos(sin^(-1)(1/4))= cos(t)= sqrt(15)/4.

Quote:

I came up with a solution though somewhat inelegant, and I'm wondering if anyone sees a better solution. Mine went: Move the -1 and -cosx to the other sides of the equation, square both sides, factor sinx, solve each factor for 0, test each solution in the interval, obtain only 0 and 4pi/5.
To solve the equation you give, since I am not as clever as Opalg at recognizing trig functions like that, I would do this:
First, write cos(x) as sqrt(1- sin^2(x)) so the equation becomes sqrt(3) sin(x)- sqrt(1- sin^2(x))= -1. Now, to reduce the writing- and the confusion, let y= sin(x). The equation is now sqrt(3)y- sqrt(1- y^2)=-1. Isolate that square root- sqrt(3)y+ 1= sqrt(1- y^2)- and square both sides: 3y^2+ 2sqrt(3)y+ 1= 1- y^2 or 4y^2+ 2sqrt(3)y= 2y(2+ sqrt(3)y)= 0.

One solution is y= sin(x)= 0 so x= 0 or x= pi. Checking back into the original equation, 0 is a solution, but pi is not (when we square both sides of an equation or multiply both sides of an equation by something involving the unknown, we may introduce "spurious" solutions). Another is y= sin(x)= -2/sqrt(3) which is impossible because it is less than -1.
• Apr 25th 2011, 04:04 PM
ragnar
Right, I did the earlier cosine-arcsine problem the second way you listed: By a right triangle and got high with a little help from my friend Pythagoras. I guess the location of the problem within these trig-functions-of-inverse-functions was misleading. I like Opalg's solution a whole lots because I saw the root-3 and knew I should be seeing something about that, which I just wasn't seeing. And it doesn't seem like too much of a stretch to realize that if I divide out by 2 I can use one of my common trig values.

I think my solution went very much the way that your solution went. Even though I didn't represent anything as a square-root, I basically rearranged things to make squaring the thing to do. It's good enough for government work. Many thanks!
• Apr 26th 2011, 04:24 AM
sa-ri-ga-ma
Another method using half angles.
given problem can be written as
sqrt(3)*sin(x) = cos(x) - 1
sqrt(3)*2sin(/2)cos(x/2) = -(2sin^2(x/2)
sqrt(3)*2sin(/2)cos(x/2) + (2sin^2(x/2) = 0
That gives you sin(x/2) = 0 or tan(x/2) = - sqrt(3)
Hence x = 0 or x/2 = 2π/3 or x = 4π/3.
• Apr 26th 2011, 06:08 AM
topsquark
Quote:

Originally Posted by sa-ri-ga-ma
Another method using half angles.
given problem can be written as
sqrt(3)*sin(x) = cos(x) - 1
sqrt(3)*2sin(/2)cos(x/2) = -(2sin^2(x/2)
sqrt(3)*2sin(/2)cos(x/2) + (2sin^2(x/2) = 0
That gives you sin(x/2) = 0 or tan(x/2) = - sqrt(3)
Hence x = 0 or x/2 = 2π/3 or x = 4π/3.

Clever. I'll have to remember that one.

-Dan