Inverse Trigonometric Functions. September 5, 2018
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1 Inverse Trigonometric Functions September 5, 08 / 7
2 Restricted Sine Function. The trigonometric function sin x is not a one-to-one functions Π 6, 5Π 6, Π Π Π Π 0.5 We still want an inverse, so what to do? We must restrict the domain of the sin function. We want to choose some interval (as large as we can find) so that [ sin(x) is one-to-one on that interval. From the graph we see that π, π ] looks likes a good choice. To be super cautious note d sin(x) [ ] = cos(x) 0 on dx sin(x) is increasing on π, π ]. [ π, π and 0 only at the end points so / 7
3 The restricted sine function Define the restricted sine function by sin x π x π f(x) = undefined otherwise We have Domain(f) = [ π, π ] and Range(f) = [, ] and graph Π Π 4 4 Π Π / 7
4 Inverse Sine Function (arcsin x = sin x). The restricted sine function is one-to-one and hence has an inverse, shown in red in the diagram below..5, Π.0 Π, 0.5 Π Π 4 4 Π Π Π, 4 Π, This inverse function, f (x), is denoted by.5 f (x) = sin x or arcsin x. sin is terrible notation since you will confuse it with the cosecant function. 4 / 7
5 Properties of arcsin(x). Domain(arcsin) = [, ] and Range(arcsin) = [ π, π ]. Since f (x) = y if and only if f(y) = x, we have: arcsin x = y if and only if sin(y) = x and π y π. Since f(f (x)) = x f (f(x)) = x we have: sin(arcsin(x)) = x for x [, ] [ arcsin(sin(x)) = x for x π, π ]. From the graph: arcsin(x) is an odd function and so arcsin( x) = arcsin(x). 5 / 7
6 Evaluating arcsin(x). Example Evaluate arcsin We see that the point Therefore arcsin ( ) (, π 4 ( ) = π 4. using the graph above. ) is on the graph of y = arcsin(x). 6 / 7
7 ( ) ( ) 3 3 Example Evaluate arcsin and arcsin. ( ) 3 arcsin = y is the same statement as: y is an angle between π and π 3 with sin y =. Consulting our unit circle, we see that y = π 3. ( ) ( ) 3 3 arcsin = arcsin = π 3 7 / 7
8 More Examples For arcsin(x) Example Evaluate arcsin(sin π). It is tempting to write π until you realize the answer is between π and π. We have sin π = 0, hence arcsin(sin π) = arcsin(0) = 0. Example Evaluate cos(arcsin( 3/)). ( ) 3 We saw above that arcsin = π 3. ( ( )) 3 ( π ) Therefore cos arcsin = cos = 3. 8 / 7
9 Preparation for the method of Trigonometric Substitution Example Give a formula in terms of x for tan ( arcsin(x) ). We draw a right angled triangle with θ = arcsin(x). x θ - x From this we see that tan ( arcsin(x) ) = tan(θ) = x x. 9 / 7
10 Derivative of arcsin(x). d dx arcsin(x) = x, < x <. Proof We have arcsin(x) = y if and only if sin(y) = x and π/ y π/. Using implicit differentiation, we get cos(y) dy dx = or dy dx = cos(y). Now we know that cos (y) + sin (y) =, hence we have that cos (y) + x = and cos(y) = ± x For y between π and π, cos(y) 0 so cos(y) = x and d dx arcsin(x) = x 0 / 7
11 Derivative of arcsin(x): Example Example Find the derivative d dx arcsin( cos(x) ). We have d dx arcsin( cos(x) ) = = d ( cos(x) ) cos x dx sin x cos x cos x = sin x cos x cos x / 7
12 Inverse Cosine Function Inverse Cosine Function We can define the function cos x = arccos(x) similarly. The details are given at the end of your lecture notes. Domain(arccos) = [, ] and Range(arccos) = [0, π]. arccos(x) = y if and only if cos(y) = x and 0 y π. cos(arccos(x)) = x for x [, ] arccos(cos(x)) = x for x [ 0, π ]. / 7
13 Derivative of Inverse Cosine Function Inverse Cosine Function Also by implicit differentiation, we can show that (see end of the lecture notes) d dx arccos(x) = d dx arcsin(x) = x Note that since both have the same derivative, we must have arccos(x) = arcsin(x) + C for some constant C. Letting x = 0, we get that arcsin(x) + arccos(x) = π. This means that the arccos is not an odd function. In fact arccos( x) = π arccos(x). 3 / 7
14 Restricted Tangent Function The tangent function is not a one to one function. The restricted tangent function is given by tan x π < x < π h(x) = undefined otherwise We see from the graph of the restricted tangent function (or from its derivative) that the function is one-to-one and hence has an inverse, which we denote by h (x) = tan x or arctan x. 4 / 7
15 6 4 Π 4, Π Π 4 Π 4 4 Π Π 4, Π Π / 7
16 Properties of arctan(x). Domain(arctan) = (, ) and Range(arctan) = ( π, π ). Since h (x) = y if and only if h(y) = x, we have: arctan(x) = y if and only if tan(y) = x and π < y < π. Since h(h (x)) = x and h (h(x)) = x, we have: tan(arctan(x)) = x for x (, ) ( arctan(tan(x)) = x for x π, π ) 6 / 7
17 Π 4 From the graph, we have: arctan( x) = arctan(x). Also, since lim tan x = and lim tan x =, x ( π ) x ( π )+ we have lim x arctan(x) = π and lim x arctan(x) = π Π Π 4, Π / 7
18 Evaluating arctan(x) ( Example Find arctan() and arctan 3 ). arctan() is the unique angle, θ, between π and π with tan(θ) = sin(θ) cos(θ) =. By inspecting the unit circle, we see that θ = π 4. ) ( arctan is the unique angle, θ, between π 3 and π with tan(θ) = sin(θ) cos(θ) =. By inspecting the unit circle, we see that 3 θ = π 6. 8 / 7
19 Example ( ( Find cos arctan 3 )). ( ( )) ( π ) cos arctan 3 = cos = / 7
20 Derivative of arctan(x). Using implicit differentiation, we get d dx arctan(x) = x, < x <. + tan(y) = x; y sec (y) = so y ( + tan (y) ) = so y = + x. We can use the chain rule in conjunction with the above derivative. Example Find the domain and derivative of arctan ( ln(x) ) Domain = Domain of ln x = (0, ) d dx arctan( ln(x) ) = + (ln x) x = x( + (ln x) ). 0 / 7
21 Integration Formulas Reversing the derivative formulas above, we get dx = arcsin(x) + C, x dx = arctan(x) + C, + x Example / 0 + 4x dx We use substitution. Let u = x, then du = dx, u(0) = 0, u(/) =. / 0 + 4x dx = 0 + u du = arctan(u) 0 = ( ) arctan() arctan(0) = ( π ) 4 0 = π 8. / 7
22 Integration Example 9 x dx dx = 9 x dx = 3 x 3 9 dx x 9 Let u = x, then dx = 3du 3 dx = 9 x 3 3 ( x ) du = arcsin(u) + C = arcsin + C u 3 / 7
23 arcsec(x) (on 09/07) Here is a graph of the secant function. There are vertical asymptotes at π + kπ, k any integer. There is disagreement in the literature as to how to restrict it. The [ book uses 0, π ) [ π, 3π ) so we will too, but some authors prefer [ 0, π ) ( π ], π. 3 / 7
24 We are using the red curves, but some authors prefer the red in the first quadrant union the blue in the fourth. Our domain is [ 0, π ) [ π, 3π ) and the range is (, ] [, ). (The other choice has the same range.) 4 / 7
25 The red curve is the graph y = arcsec(x). The domain [ is(, ] [, ) and the range is 0, π ) [ π, 3π ). There are two limits (or horizontal asymptotes) lim x arcsec(x) = 3π and lim x arcsec(x) = π 5 / 7
26 d dx arcsec(x) = x x y = arcsec(x); sec y = x; y sec(y) tan(y) =. Now sec (y) = + tan (y) so tan (y) = sec y and hence tan(y) = ± sec y = ± x. Hence y = ± x x. From the graph, when x < 0 the curve is decreasing so the derivative is negative and the sign is +. Similarly, when x > 0 the curve is increasing so the derivative is positive and the sign is +. 6 / 7
27 The other choice for the arcsec gives a function which is always increasing but has derivative x. Since we will only use the x arcsec to integrate, we prefer the definition we chose. There are also inverse trig functions arccot and arccsc. These do not allow us to integrate anything new so we will not discuss them. 7 / 7
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