19 ## INVERSE TRIGONOMETRIC FUNCTIONSTHE ANGLES in calculus will be in radian measure. Thus if we are given a radian angle, for example, then we can evaluate a function of it.
(Topic 13.) Inversely, if we are given that the sin "The sine of what angle is equal to ½?" We write however: Evaluate
"The The function is called the inverse of the funtion
arcsin Strictly, Now there are many angles whose sine is ½. It wll be any angle whose corresponding acute angle is . Therefore we must restrict the range of How will we do that? We will restrict them to those angles that have the smallest absolute value. They are called the principal values of Thus arcsin ½ = . The first quadrant angle is the angle with the smallest absolute value whose sine is ½. Example 1. Evaluate arcsin (−½).
The angle whose sine is − arcsin (−½) = −arcsin (½) = −. The range, then, of the function Angles whose sines are positive will be 1st quadrant angles. Angles whose sines are negative will fall in the 4th quadrant. To restrict the sin Another notation for arcsin Problem 1. Evaluate the following in radians. To see the answer, pass your mouse over the colored area. a) sin b) sin c) sin
Corresponding to each trigonometric function, there is its inverse function. arcsin arccos arctan arccsc arcsec arccot In each one, we are given the In each case, we must retstrict its range so that the function will be single-valued. The range of Like Note that Angles whose tangents are positive will be 1st quadrant angles. Angles whose tangents are negative will fall in the 4th quadrant. That is exactly the same as with arcsin (− The angle whose tangent is −
Problem 2. Evaluate the following.
The range of Example 2. Evaluate arccos ½.
Problem 3. Why is this not true? arccos (−½) = −. − is a 4th quadrant angle. And in the 4th quadrant, the cosine is positive. An angle whose cosine is negative will fall in the 2nd quadrant, where it will have its smallest absolute value. (Topic 15.) The cosine of a 2nd quadrant angle is the negative of the cosine of its corresponding acute angle, which is its supplement. In other words: The angle θ whose cosine is − arccos (− Example 3. Evaluate arccos (−½).
arccos ½ = . Therefore, arccos (−½) is the supplement of —which is the angle we must add to to equal π. + θ = π. Now, is one-third of π. Therefore, its supplement will be two- thirds of π: . θ =arccos (−½) = . The range, then, of An angle whose cosine is positive will be a 1st quadrant angle; an angle whose cosine is negative will fall in the 2nd. It will be the supplement of the 1st quadrant angle. Problem 4. Evaluate the following.
* The inverse relation is as follows: arccos For example,
This in general is the case. Problem 5. a) arctan b) arcsec c) arccos 1 = 0 if and only if 1 = cos 0.
The range of In calculus, sin If The only inverse function below in which Again, we restrict the values of The inverse relations If we put
and
then according to the definition of inverse functions (Topic 19 of Precalculus):
sin(arcsin In particular, if
By taking the inverse function of both sides, we have extracted, or freed, the argument
Example 4. Solve for
Therefore,
Problem 6. Solve for tan (
Problem 7. Solve for cos
Problem 8. Solve for
Theorem. If
then the product sec For, if When angle If Therefore, that product is never negative. (This theorem is referenced in the proof of the derivative of Next Topic: Trigonometric identities Please make a donation to keep TheMathPage online. Copyright © 2021 Lawrence Spector Questions or comments? E-mail: themathpage@yandex.com |