3.1 The Inverse Sine, Cosine, and Tangent Functions

Transcription

1 3.1 The Inverse Sine, Cosine, and Tangent Functions Let s look at f(x) = sin x The domain is all real numbers (which will represent angles). The range is the set of real numbers where -1 sin x 1. However, in order for the sine function to have an inverse function, it has to be 1-to-1. y = sin x Page (π/, 1) (5π/, 1) (-π/, -1) passes Horizontal Line Test (0,0) {-π/ x π/} (π, 0) (3π/, -1) (π, 0) Entire domain of sin x does not pass Horizontal Line Test π π If we restrict the domain of y = sin x to the interval, then it will have an inverse function. The inverse sine function is denoted f -1 (x) = sin -1 x. The input of the inverse function is a real number between -1 and 1, and the output of inverse sine is a real number that is an angle in radians between π/ and π/. The domain is the set of real numbers {x -1 x 1}. π 1 π The range is the set of real numbers such that sin x By the property of inverse functions, for any x in the domain ofsin x, sin -1 (sin (x)) = x and sin(sin -1 (x)) = x Notice that the input of the sine function is an angle, and the output of the inverse sine function is an angle (in radians). The inverse sine function also called the arcsinfunction. NOTE: sin -1 x 1 / sin x

2 Page Example on p. Find the exact value of sin -1 ( -½ ) What this is asking is, for what angle, θ, (where -π/ θ π/) does sin θ = ½? Let s look at the unit circle, x + y = 1. Remember each point on the unit circle is (cos θ, sin θ). So find a point on the unit circle with y-coordinate = -½ sin θ = - ½ when θ= -π/6 We can t choose θ = 7π/6 because that is not within the range of sin -1 Therefore sin -1 (-½) = -π/6 You try Exercise #19 on p.30 What is sin -1?

3 The Inverse Cosine Function In order for the cosine function to be one-to-one we must restrict its domain to {x 0 x π}. Page 3 y = cos x , π/6, π/6, {0 x π} passes Horizontal Line Test π, π/6, π/6, π/3, π/3, π/3, π/3, π/, π/, π/, π/, π/3, π/3, π/3, π/3, π/6, π/6, π/6, π, π/6, π, The inverse cosine function is denoted f -1 (x) = cos -1 x. The output of the inverse cosine function is a real number that is an anglein radians. The domain (input) is the set of real numbers {x -1 x 1}. The range (output) is the set of real numbers such that 1 0 cos x π { } By the property of inverse functions, for any x in the domain ofcos x, cos -1 (cos (x)) = x and cos(cos -1 (x)) = x The inverse cosine function is also called the arccos function. Example 5 Find the exact value of cos 1 Look on the unit circle of an angle between 0 and π that gives a cosine value of Which quadrant within that range yields a positive cos value?

4 The Inverse Tangent Function In order for the tangent function to be one-to-one we must restrict π π its domain to < x < π. Notice that x cannot equal ± because tan x is undefined at those x-values. Page 4 {-π/ <x < π/} passes Horizontal Line Test 1 (π/4, 1) (5π/4, 1) (9π/4, 1) -π/ 0 π/ π 3π/ π 5π/ (-π/4, -1) -1 (3π/4, -1) (7π/4, -1) - Notice that the range of the tangent function is all real numbers. The graph extends to - when close to π/ and to + when close to π/. Therefore, for tan -1 x The domain is all real numbers. The range is the set of real numbers such that π By the property of inverse functions, for any x in the domain oftan x, tan -1 (tan (x)) = x and tan(tan -1 (x)) = x < tan x < 1 π Example 8 on p.9 Find the exact value of Hint: Since range is negative tan value? tan ( 3 ) 1 π < tan x < 1 π, which quadrant will yield a Now you do Exercise #17 on p.30

5 3. More Inverse Functions

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7 Example 1 Find exact value of sin 1 sin 5π 4 Page 7 This is a trick problem, because we were taught that f -1 (f(x))=x However, the x in this example, 5π/4, is not in the range of sin -1 (x). So we must choose an angle within the range of sin -1 (x) {- π/ θ π/} that yields the same value as sin(5π/4). Use your unit circle.

8 Example Find the exact value of sin tan θ 1 1 Output of tan -1 is an angle, θ, that has a tan θ of ½. What if we can t find an angle on the unit circle that yields a tan value of ½? That s OK. Draw a triangle. tan θ = opp/adj = 1/ We are looking for sin θ, which is opp/hyp Hypotnuse= 1 sin θ = opp/hyp = + = = Notice that we didn t actually have to find θ to find sin θ! 1

9 Example 3 Find the exact value of cos sin Page 9 Pay attention to the output of sin-1. The output must be an angle in Quadrant I or IV, and -1/3 means it must be in Quadrant IV. Will cosine be positive? Yes. adjacent = 3 ( 1) = 3 8 = -1 cos θ = 3

10 Remember that range of sec-1 is 0 θ π, excluding π/. [Quadrants I and II] Remember that range of cot-1 is 0< θ<π [Quadrants I and II]

11 Page 11 HOMEWORK p. 9 Concepts and Vocab: #7-1 Exercises p #15-57 ETP p.36 #9-53 EOO Extra Credit (+ pts) p.31 #64 (all work must be shown) Hint for (c) (calculator graphing): Since x is the number of feet from the screen, and the first row is 5 feet from the screen, set Xmin at 5 and set Xmax at 50 (a theater is probably that long.) Since Y represents the angle in degrees, set Ymin =0 and Ymax = 90.