Inverse Trigonometric Functions Math 1120 Added to Class 1 - - PowerPoint PPT Presentation

Math 1120 Added to Class 1

Inverse Trigonometric Functions

Exercise: d dx (Arccsc x) =? y = Arccos x, y = Arctan x, y = Arccot x, y = Arcsec x, y = Arccsc x are in section 3.9. The notation in the text is sin−1 x rather than Arcsin x, but I think this leads to the possible confusion that sin−1 x might be 1/ sin x. The function y = Arccsc x is defined as: Given x, y = Arccsc x is the unique value −π/2 ≤ y ≤ π/2 which satisfies csc y = x ⇔ sin y = 1/x. The domain is the set of x for which the equation csc y = x has a solution; so x ≤ −1 or x ≥ 1. For such x’s there are infinitely many solutions y; but only one satisfiying −π/2 ≤ y ≤ π/2.

Math 1120 Added to Class 1

Solution to Exercise

Rewrite the equation as 1/ sin y = x, or sin y = 1/x, and differentiate both sides (using the chain rule) cos y · y′ = −1/x2 ⇔ y′ = − 1 x2 cos y We need to express y′ in terms of x alone. Use cos2 y + sin2 y = 1, so solving for cos y we get cos y = ±

• 1 − sin2 y = ±
• 1 − 1/x2.

However −π/2 ≤ y ≤ π/2 so cos y ≥ 0. Since square roots are always positive quantities, the final answer is y′ = − 1 x2 1 − 1/x2

Math 1120 Added to Class 1

Solution to Exercise continued

IMPORTANT: − 1 x2 1 − 1/x2 = − 1 x2

• x2−1

x2

= − 1 x2

√ x2−1 √ x2

= − 1 x2

√ x2−1 |x|

because √ x2 = |x| NOT x! The final answer is (Arccsc x)′ = − 1 |x| √ x2 − 1 . If you use a right triangle with one angle equal to y, the hypotenuse of size x, and the opposite side equal to 1, to compute cos y, you are implicitly assuming that x > 0; this is not the case for the definition of Arccsc .