19 ## INVERSE TRIGONOMETRIC FUNCTIONSTHE ANGLES in theoretical work 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 could answer:
The algebraic abbreviation for that sentence is:
arcsin It is the Strictly, The inverse of the function
is
Their inverse relation is as follows: arcsin Thus,
Corresponding to each trigonometric function, there is its inverse function. arcsin arccos arctan arccsc arcsec arccot In each one, we are given the Problem 1. To see the answer, pass your mouse over the colored area. a) arctan b) arcsec c) arccos 1 = 0 if and only if 1 = cos 0.
Example 1. Evaluate arcsin -- "the angle whose sine is ."
The range of But is not the only angle whose sine is . is the sine of every first and second quadrant angle whose corresponding acute angle is .
And so on. For the function They are called the principal values of arcsin is the angle of smallest absolute value whose sine is . It is the first quadrant angle between 0 and . Example 2. Evaluate arcsin (−).
The range, then, of the function To restrict the Angles whose sines are positive will be 1st quadrant angles, while angle whose sines are negative will fall in the 4th quadrant. If fact,
The angle whose sine is − To see that, look here:
sin Another notation for arcsin Problem 2. Evaluate the following in radians. a) sin b) sin c) sin
The range of Similarly, we must restrict the range of Note that For an angle whose tangent is positive, we choose a 1st quadrant angle. For an angle whose tangent is negative, then, as with arcsin (− arctan (−
Problem 3. Evaluate the following.
The range of Example 3. Evaluate arccos ½.
Problem 4. 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 4. 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 5. Evaluate the following.
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,
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
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 Copyright © 2018 Lawrence Spector Questions or comments? E-mail: themathpage@yandex.com Private tutoring available. |