SOLUTION: These are actually trigonometry problems, but I can't figure out the right way to do this. I've done it before, but I can't seem to find it in my book for my class. cos(Sin^(-1

Question 717713: These are actually trigonometry problems, but I can't figure out the right way to do this. I've done it before, but I can't seem to find it in my book for my class.
cos(Sin^(-1)(1/3)-Cot^(-1)(3/4))
sin((1/2)(Arcsin(1/4))
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
Since these are Trig problems you should post them under that category. (I have change the category to Trig.) Tutors, like me, who like to help with Trig look for problems posted as Trig problems. Most days I would never have seen this problem since I rarely look for Geometry problems. So you will increase your chances of a quick response by posting in an appropriate category.

(Note: Algebra.com's formula drawing software mistakenly puts multiplication symbols between inverse trig functions and their arguments. Please ignore this.)
This problem would be a lot easier if you were comfortable with the idea that inverse Trig functions return angles. So this expression is the cos(one angle minus another angle). This is a job for the cos(A-B) formula. To help you see this let's say

and

Substituting these into our expression we literally end up with:
cos(A-B)
From the formula we that this is equal to:

Substituting back in for the A's and B's we get:

(Someday you will be able to see how to go directly from

and leave out the use of the A's and B's.)

You might very well be thinking: "How is better than ?" This answer is: Because we can figure out the longer expression. Anytime you have a-trig-function(an-inverse-trig-function) you can simplify it. Here's how:

Draw a right triangle.
Look at the inner inverse trig function and set up the two appropriate sides so that they form the right ratio with respect to angle A. (More on this as we do one later.)
Look at the outer Trig function and use the appropriate sides of our triangle to find the right ratio. This may require using the Pythagorean Theorem to find the third side of the triangle.

Let's try this on :
I'll leave it up to you to draw a triangle and choose angle A (steps 1 and 2). For the third step we look at the inner inverse trig function, . We want sin(A) to be 1/3. Since sin is opposite over hypotenuse, set up the sides of the triangle so that the opposite side is 1 and the hypotenuse is 3. Now we look at the outer trig function, cos. So we want to find cos(A). Cos is adjacent over hypotenuse. We have the hypotenuse of our triangle but not the adjacent side. So we use the Pythagorean Theorem to find it:

So the adjacent side is making the . So . Substituting this into our big expression we get:

Skipping ahead to because we don't even need a triangle to figure this out. This expression is "the sin of an angle whose sin is 1/3". Think about that statement. I hope that it is obvious that this would be 1/3. Now we have:

For we will need a new triangle. cot is adjacent over opposite so make the side adjacent to A be 3 and the side opposite to A be 4. cos is still adjacent over hypotenuse and we do not have the hypotenuse. So on to the Pythagorean Theorem:

So Substituting this into the expression we now have:

For we do not need a new triangle. The angle, , is the same angle as in the previous triangle. So we can reuse the sides of the last triangle. sin is opposite over hypotenuse so . Now the expression is:

Now we just simplify:

This is an exact expression for the simplified expression.

This is the sin of 1/2 of an angle. So we will be using the sin((1/2)x) formula. This formula has a + in front of the square root. Since arcsin returns angles between -90 and 90 degrees and since the angle's sin is positive, the angle arcsin will return will be between 0 and 90. And since half of an angle between 0 and 90 is still between 0 and 90, we know that we should use just the positive square root: Using this on we get:

Now use the procedure we used above to find . Insert that expression into and simplify. I'll leave that up to you.
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