4.7 Solving Problems with Inverse Trig Functions

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1 4. Solving Problems with Inverse Trig Functions 4.. Inverse trig functions create right triangles An inverse trig function has an angle (y or θ) as its output. That angle satisfies a certain trig expression and so we can draw a right triangle that represents that expression. One can always draw a right triangle with an inverse trig function and think of the output as a certain angle in that triangle. For example, the equation arcsin(z) = θ implies that sin θ = z and so it can be viewed as corresponding to a right triangle with hypotenuse, with θ one of the acute angles and z the length of the side opposite θ. (See Figure 34, below.) z θ z Figure 34. The triangle that appears with the equation arcsin(z) = θ. We will practice this idea with some worked problems... Some Worked Problems. Draw a right triangle with the appropriate lengths and use that triangle to find the sine of the angle θ if (a) cos(θ) = 3 (b) cos(θ) = (c) cos(θ) = 0.8. (d) cos(θ) = 0.6. Partial solutions. (a) If cos(θ) = 3 then draw a triangle with legs of length, and hypotenuse of length 3. If the cosine of θ is 3 then the sine of θ is then 3. (b) If cos(θ) = then draw a triangle with legs of length, and hypotenuse of length. The sine of θ is then. (c) If cos(θ) = 0.8. then draw a triangle with legs of length 3, 4 and hypotenuse of length. The sine of θ is then 3. (d) If cos(θ) = 0.6. then draw a triangle with legs of length 3, 4 and hypotenuse of length. The sine of θ is then 4. When we work with inverse trig functions it is especially important to draw a triangle since the output of the inverse trig function is an angle of a right triangle. Indeed, one could think of inverse trig functions as creating right triangles. 84

2 The angle θ in the drawing in Figure 34 is arcsin(z). Notice that the Pythagorean theorem then gives us the third side of the triangle (written in blue); its length is z. This allows us to simplify expressions like cos(arcsin z), recognizing that cos(arcsin z) = cos(θ) = z. In a similar manner, we can simplify tan(arcsin z) to tan(arcsin(z)) = z z. Some worked problems.. Simplify (without use of a calculator) the following expressions (a) arcsin[sin( π 8 )]. (b) arccos[sin( π 8 )]. (c) cos[arcsin( 3 )]. (a) Since arcsin is the inverse function of sine then arcsin[sin( π 8 )] = π 8. (b) If θ is the angle π 8 then the sine of θ is the cosine of the complementary angle π π 8, which, after getting a common denominator, simplifies to 3π 8. In other words, the sine of π 8 is the cosine of 3π 8 so arccos[sin( π 8 )] = 3π. (Notice that I ve solved this problem this without ever 8 having to figure out the value of sin( π 8 ). (c) To simplify cos[arcsin( 3 )] we draw a triangle with hypotenuse of length 3 and one side of length, placing the angle θ so that sin(θ) = 3. The other short side of the triangle must have length 8 = by the Pythagorean theorem so the cosine of θ is 3. So cos[arcsin( 3 )] =. 3. Simplify (without the use of a calculator) the following expressions: (a) arccos(sin(θ)), assuming that θ is in the interval [0, π ]. (b) arccos(y) + arcsin(y). (a) To simplify arccos(sin(θ)), we draw a triangle (on the unit circle, say) with an acute angle θ and short sides of lengths x, y and hypotenuse. (See the figure below.) π θ y θ Figure 3. A right triangle with θ and its complementary angle π θ. x 8

3 The sine of θ is then y and the arccosine of y must be the complementary angle π θ. So arccos(sin(θ)) = π θ. (b) Notice in the triangle in Figure 3 that the sine of θ is y and the cosine of π arcsin(y) = θ and arccos(y) = π θ. Therefore θ is y. So arccos(y) + arcsin(y) = θ + ( π θ) = π. Indeed, the expression arccos(y) + arcsin(y) merely asks for the sum of two complementary angles! By definition, the sum of two complementary angles is π! 4.. Drawing triangles to solve composite trig expressions Some problems involving inverse trig functions include the composition of the inverse trig function with a trig function. If the inverse trig function occurs first in the composition, we can simplify the expression by drawing a triangle. Here are some worked problems. Worked problems.. Do the following problems without a calculator. Find the exact value of (a) sin(arccos( 3 4 )) (b) tan(arcsin( 3 4 )) (c) sin( arctan( 4 3 )) (Use the trig identity sin θ = sin θ cos θ.) (a) To compute sin(cos ( 3 4 )) draw a triangle with legs 3, and hypotenuse 4. The angle θ needs to be in the second quadrant so the sine will be positive. So the sine of the angle θ should be 4. (b) To compute tan(sin ( 3 4 )) draw a triangle with legs 3, and hypotenuse 4. The tangent of the angle θ should be 3. But the angle θ is in the fourth quadrant so the final answer is 3. (c) To compute sin( tan ( 4 3 )) = sin θ cos θ where tan(θ) = 4. draw a triangle with legs 3, 4 3 and hypotenuse. The cosine of the angle θ is 3 and the sine of the angle θ is 4. Since the original problem has a negative sine in it, we must be working with an angle in the fourth quadrant, so the sine is really 4. Now we just plug these values into the magical 6 identity given us: sin(θ) = sin θ cos θ = ( 4 )(3 ) = 4. 6 See section.3 of these notes for an explanation of the double angle identity for sine. 86

4 . Draw a right triangle with short legs of length and x and then compute (a) sin(arctan(x)) (b) tan(arctan(x)) (c) cot(arctan(x)) (d) sec(arctan(x)) (a) To compute sin(arctan(x)) draw a right triangle with sides, x and hypotenuse + x. The x sine of the angle θ is. + x (b) To compute tan(arctan(x)) just recognize that tan x and arctan x are inverse functions and so tan(arctan(x)) = x. (c) To compute cot(arctan(x)) draw a right triangle with sides, x and hypotenuse + x. The cotangent of the angle θ is x. (d) To compute sec(arctan(x)) draw a right triangle with sides, x and hypotenuse + x. The secant of the angle θ should be + x More on inverting composite trig functions Just like other functions, we can algebraically manipulate expressions to create an inverse function. Some worked problems. Find the inverse function of the following functions.. y = sin( x) +. y = sin( x + ) 3. y = sin( x + ) 4. y = e sin( x+). y = sin(arccos x). To find the inverse function of y = sin( x) +, let s exchange inputs and outputs: x = sin( y) + and then solve for y by subtracting from both sides x = sin( y), applying the arcsin to both sides, and then squaring both sides arcsin(x ) = y (arcsin(x )) = y so that the answer is is y = (arcsin(x )). 8

5 . The inverse function of y = sin( x + ) is y = (arcsin(x) ). 3. The inverse function of y = sin( x + ) is y = (arcsin x). 4. The inverse function of y = e sin( x+) is y = (arcsin(ln x) ).. The inverse function of y = sin(cos x) is the inverse function of y = x. It happens that the inverse function of y = x obeys the equation x = y so x = y so y = x so y = x. (That is y = x is its own inverse function!) 4..4 Other resources on sinusoidal functions (Hard copy references need to be fixed.) In the free textbook, Precalculus, by Stitz and Zeager (version 3, July 0, available at stitz-zeager.com) this material is covered in section 0.6. In the free textbook, Precalculus, An Investigation of Functions, by Lippman and Rassmussen (Edition.3, available at this material is covered in section 6.4 and 6.. In the textbook by Ratti & McWaters, Precalculus, A Unit Circle Approach, nd ed., c. 04 this material appears in section??. In the textbook by Stewart, Precalculus, Mathematics for Calculus, 6th ed., c. 0 (here at Amazon.com) this material appears??. Homework. As class homework, please complete Worksheet 4., More Inverse Trig Functions available through the class webpage. 88