1. ## Inverse Trigonometric Functions

When solving for a angle in a right triangle, one of the things I am confused about is we often have to use inverse trigonometric functions.

In the example below, the step outlined in blue, what is the algebra that allows us to move sin from the left hand side of the equation, to the right hand side and invert it using the -1 power, in order to isolate theta?

2. ## Re: Inverse Trigonometric Functions

First a note on notation: the "-1" exponent as it's used here is NOT "the negative one power" - it signifies the arcsin function. It might be easier to consider using notation "arcsin" instead of sin^-1, until you become comfortable with it. So, to get from the first step to the second, simply take the arcsin of both sides:

sin(theta) = 5/6
arcsin [sin (theta)] = arcsin(5/6)

Now recognize that the arc sin of the sin of an angle is that angle, and the left hand side becomes theta.

3. ## Re: Inverse Trigonometric Functions

Originally Posted by ChipB
First a note on notation: the "-1" exponent as it's used here is NOT "the negative one power" - it signifies the arcsin function. It might be easier to consider using notation "arcsin" instead of sin^-1, until you become comfortable with it. So, to get from the first step to the second, simply take the arcsin of both sides:

sin(theta) = 5/6
arcsin [sin (theta)] = arcsin(5/6)

Now recognize that the arc sin of the sin of an angle is that angle, and the left hand side becomes theta.
Thanks! Just to clarify on the second step, arcsin [sin (theta)] = arcsin(5/6)
the reason arcsin and sin cancel out is because it is a reciprocal? Sort of like if you had 2/3 x 3/2 its just equal to 1. So that leaves theta by itself. Is that right?

4. ## Re: Inverse Trigonometric Functions

Originally Posted by cmf0106
Thanks! Just to clarify on the second step, arcsin [sin (theta)] = arcsin(5/6)
the reason arcsin and sin cancel out is because it is a reciprocal? Sort of like if you had 2/3 x 3/2 its just equal to 1. So that leaves theta by itself. Is that right?
No that is a total misunderstanding. Look at this calculation. Study it very carefully by looking at this graph.

The function $\arcsin$ has domain $-1\le t\le 1$ and range $-\frac{\pi}{2}\le\arcsin(t)\le \frac{\pi}{2}$

5. ## Re: Inverse Trigonometric Functions

Not to complicate things too much (at first) - you do need to be careful to understand the proper range of inverse functions, or you get bad results. Perhaps a better example that you may be familiar with already is with powers. An example: if you take the square root of a number, then square the result you get the original number back: (x^1/2)^2 = x. However, the opposite may or may not be true: if you square a number, then take the square root of the result, you may end up with a number that is different sign from the original. Example: -2 squared is 4, and the square root of 4 is +/- 2. Another example using logarithms: the log base 10 of 100 is 2, and 10^2 = 100, so it seems that 10^(log x) = x. This is true for positive values of x, but falls apart for negative values of x. Getting back to your original question: you can see that for theta = 56.44 degrees you have sin(theta) = 5/6, and arcsin(sin(56.44 degrees)) = arcsin(5/6) = 56.44 degrees. But what if theta = 180-56.44 = 123.56 degrees? Then you would have arcsin(sin(123.56 degrees) = 56 degrees. Bottom line is you must be very careful to know which quadrant you are working on, or you run the risk of concluding that 56.44 degrees equals 123.56 degrees.

6. ## Re: Inverse Trigonometric Functions

Originally Posted by cmf0106
When solving for a angle in a right triangle, one of the things I am confused about is we often have to use inverse trigonometric functions.

In the example below, the step outlined in blue, what is the algebra that allows us to move sin from the left hand side of the equation, to the right hand side and invert it using the -1 power, in order to isolate theta?

It means the inverse of the ratio of opposite side and hypotenuse is equal to $\displaystyle 56.44^{\circ}$ From the ratio of sides you are getting the measurement of angle $\displaystyle \theta$ whereas when you use $\displaystyle \sin(\theta)$ then you are getting the opposite which is the ratio of the opposite side and hypotenuse.

7. ## Re: Inverse Trigonometric Functions

Originally Posted by x3bnm
It means the inverse of the ratio of opposite side and hypotenuse is equal to $\displaystyle 56.44^{\circ}$ From the ratio of sides you are getting the measurement of angle $\displaystyle \theta$ whereas when you use $\displaystyle \sin(\theta)$ then you are getting the opposite which is the ratio of the opposite side and hypotenuse.
Had a student of mine had written the above I would have assigned a grade of zero.
"the inverse of the ratio of opposite side and hypotenuse is equal to $\displaystyle 56.44^{\circ}$" what does that mean? Do you mean the of the opposite side and the hypotenuse is equal to $\displaystyle 56.44^{\circ}$ But lengths are real numbers what the heck are degrees( nobody knows)? It is true that is a right triangle with legs of length $x~\&~y$ if $\theta$ is the base angle opposite the side of length $y$ then $\sin(\theta)=\dfrac{y}{\sqrt{x^2+y^2}}$.

8. ## Re: Inverse Trigonometric Functions

Originally Posted by ChipB
Example: -2 squared is 4, and the square root of 4 is +/- 2.
I strongly disagree with that example. $4$ does have two square roots one is $2=\sqrt4$ and the other is $-2=-\sqrt4$.

My objection is basically to saying that $\sqrt4=\pm2$ I consistently required faculty to mark that wrong.
We can say that the square roots of four are $2~\&~-2$. If anyone taught complex variables as long as I did, knows the battle to get students not to use $\pm$ with radicals.