Lecture 6 : Inverse Trigonometric Functions Inverse Sine Function (arcsin x = sin x) The trigonometric function sin x is not one-to-one functions, hence in orer to create an inverse, we must restrict its omain. The restricte sine function is given by sin x π x π f(x) = unefine otherwise We have Domain(f) = [ π, π ] an Range(f) = [, ]..0.0 0.5 6, 5 6, 0.5 0.5 0.5 We see from the graph of the restricte sine function (or from its erivative) that the function is one-to-one an hence has an inverse, shown in re in the iagram below..5,.0, 0.5,, 0.5.0.5 This inverse function, f (x), is enote by f (x) = sin x or arcsin x. Properties of sin x. Domain(sin ) = [, ] an Range(sin ) = [ π, π]. Since f (x) = y if an only if f(y) = x, we have:
sin x = y if an only if sin(y) = x an π y π. Since f(f )(x) = x f (f(x)) = x we have: sin(sin (x)) = x for x [, ] sin (sin(x)) = x for x [ π, π ]. from the graph: sin x is an o function an sin ( x) = sin x. ( ) Example Evaluate sin using the graph above. Example Evaluate sin ( /), sin ( /), Example Evaluate sin (sin π). Example Evaluate cos(sin ( /)). Example Give a formula in terms of x for tan(sin (x)) Proof or x sin x = Derivative of sin x., x. x We have sin x = y if an only if sin y = x. Using implicit ifferentiation, we get cos y y x = y x = cos y. Now we know that cos y + sin y =, hence we have that cos y + x = an cos y = x
an x sin x = x. If we use the chain rule in conjunction with the above erivative, we get x sin (k(x)) = k (x), x Dom(k) an k(x). (k(x)) Example. Fin the erivative x sin cos x Inverse Cosine Function We can efine the function cos x = arccos(x) similarly. The etails are given at the en of this lecture. Domain(cos ) = [, ] an Range(cos ) = [0, π]. cos x = y if an only if cos(y) = x an 0 y π. cos(cos (x)) = x for x [, ] cos (cos(x)) = x for x [ 0, π ]. It is shown at the en of the lecture that x cos x = x sin x = x an one can use this to prove that sin x + cos x = π. Inverse Tangent Function The tangent function is not a one to one function, however we can also restrict the omain to construct a one to one function in this case. The restricte tangent function is given by tan x π < x < π h(x) = unefine otherwise We see from the graph of the restricte tangent function (or from its erivative) that the function is one-to-one an hence has an inverse, which we enote by h (x) = tan x or arctan x.
6,, 5 5 6 Properties of tan x. Domain(tan ) = (, ) an Range(tan ) = ( π, π ). Since h (x) = y if an only if h(y) = x, we have: tan x = y if an only if tan(y) = x an π < y < π. Since h(h (x)) = x an h (h(x)) = x, we have: ( tan(tan (x)) = x for x (, ) tan (tan(x)) = x for x π, π ). Frpm the graph, we have: Also, since tan ( x) = tan (x). lim tan x = an lim tan x =, x ( π ) x ( π + ) we have lim x tan x = π an lim x tan x = π Example Fin tan () an tan ( ). Example Fin cos(tan ( )). Derivative of tan x. x tan x =, < x <. x +
Proof or We have tan x = y if an only if tan y = x. Using implicit ifferentiation, we get sec y y x = y x = sec y = cos y. Now we know that cos y = cos (tan x) = +x. proving the result. If we use the chain rule in conjunction with the above erivative, we get x tan (k(x)) = k (x) + (k(x)), x Dom(k) Example Fin the omain an erivative of tan (ln x) Domain = (0, ) x tan (ln x) = x + (ln x) = x( + (ln x) ) Integration formulas Reversing the erivative formulas above, we get x x = sin x + C, x + x = tan x + C, Example x = x 9 Let u = x, then x = u an x = 9 x 9 x x = x = x 9 x x 9 u u = sin u + C = sin x + C Example / 0 + x x Let u = x, then u = x, u(0) = 0, u(/) = an / 0 + x x = 0 + u x = tan u 0 = [tan () tan (0)] [π 0] = π 8. 5
The restricte cosine function is given by cos x g(x) = unefine We have Domain(g) = [0, π] an Range(g) = [, ]. 0 x π otherwise We see from the graph of the restricte cosine function (or from its erivative) that the function is one-to-one an hence has an inverse, g (x) = cos x or arccos x f( x) = cos - ( x) 6
Domain(cos ) = [, ] an Range(cos ) = [0, π]. Recall from the efinition of inverse functions: g (x) = y if an only if g(y) = x. cos x = y if an only if cos(y) = x an 0 y π. g(g (x)) = x g (g(x)) = x cos(cos (x)) = x for x [, ] cos (cos(x)) = x for x [ 0, π ]. Note from the graph that cos ( x) = π cos (x). cos ( /) = an cos ( /) = You can use either chart below to fin the correct angle between 0 an π.: tan(cos ( /)) = tan(cos (x)) = Must raw a triangle with correct proportions: -x θ x θ cos θ = x cos - x = θ cos θ = x x tan(cos - x) = tan θ = -x x 7
Proof or x cos x =, x. x We have cos x = y if an only if cos y = x. Using implicit ifferentiation, we get sin y y x = y x = sin y. Now we know that cos y + sin y =, hence we have that sin y + x = an sin y = x an x cos x = x. Note that x cos x = x sin x. In fact we can use this to prove that sin x + cos x = π. If we use the chain rule in conjunction with the above erivative, we get x cos (k(x)) = k (x), x Dom(k) an k(x). (k(x)) 8