1 / 11

The inverse trigonometric functions

The inverse trigonometric functions. The inverse trigonometric functions The reciprocal trigonometric functions Trigonometric identities Examination-style question. Contents. 1 of 35. © Boardworks Ltd 2006. The inverse of the sine function.

terrence
Download Presentation

The inverse trigonometric functions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. The inverse trigonometric functions • The inverse trigonometric functions • The reciprocal trigonometric functions • Trigonometric identities • Examination-style question Contents 1 of 35 © Boardworks Ltd 2006

  2. The inverse of the sine function This is not the same as (sinx)–1 which is the reciprocal of sinx, . Suppose we wish to find θ such that sinθ= x In other words, we want to find the angle whose sine is x. This is written as θ= sin–1x or θ= arcsinx In this context, sin–1x means the inverse of sinx. Is y = sin–1x a function?

  3. The inverse of the sine function y x y We can see from the graph of y = sinx between x = –2π and x = 2π that it is a many-to-one function: y = sinx x The inverse of this graph is not a function because it is one-to-many: y = sin–1x

  4. The inverse of the sine function So, if we restrict the domain of f(x) = sinx to – ≤ x ≤ we have a one-to-one function: y 1 x –1 However, remember that if we use a calculator to find sin–1x (or arcsinx) the calculator will give a value between –90° and 90° (or between – ≤ x ≤ if working in radians). There is only one value of sin–1x in this range, called the principal value. y = sinx

  5. The graph of y = sin–1x y y y = sin–1x y = sin–1x 1 y = sinx x x –1 –1 1 1 –1 – ≤ sin–1x≤ Therefore the inverse of f(x) = sinx,– ≤ x ≤ , is also a one-to-one function: f–1(x) = sin–1x The graph of y = sin–1xis the reflection of y = sinxin the line y = x: (Remember the scale used on the x- and y-axes must be the same.) The domain of sin–1x is the same as the range of sinx : –1 ≤ x ≤ 1 The range of sin–1x is the same as the restricted domain of sinx :

  6. The inverse of cosine and tangent if f(x) = cosx for 0 ≤ x ≤ π then f–1(x) = cos–1x for –1 ≤ x ≤ 1. And if f(x) = tanx for – < x < then f–1(x) = tan–1x for x We can restrict the domains of cosx and tanx in the same way as we did for sin x so that The graphs cos–1xand tan–1xcan be obtained by reflecting the graphs of cosxand tanxin the line y = x.

  7. The graph of y = cos–1x y y = cos–1x y = cos–1x 1 x 0 –1 –1 1 1 y x 0 y = cosx –1 The domain of cos–1x is the same as the range of cosx : –1 ≤ x ≤ 1 The range of cos–1x is the same as the restricted domain of cosx : 0 ≤ cos–1x≤ π

  8. The graph of y = tan–1x y y = tan x y = tan–1x y = tan–1x x x – < tan–1x< y y = tanx x The domain of tan–1x is the same as the range of tanx : The range of tan–1x is the same as the restricted domain of tanx :

  9. Problems involving inverse trig functions radians 0 degrees 0° 30° 45° 60° 90° sin cos tan sin–1 = Find the exact value of sin–1 in radians. To solve this, remember the angles whose trigonometric ratios can be written exactly: 0 1 1 0 0 1 From this table

  10. Problems involving inverse trig functions cosθ = – We know that cos = = 1 From the graph, cos = – θ 0 So, cos–1 = –1 Find the exact value of sin–1 in radians. This is equivalent to solving the trigonometric equation for 0 ≤θ≤ π this is the range of cos–1x Sketching y = cosθfor 0 ≤θ≤ π:

  11. Problems involving inverse trig functions sin θ= Let sin–1 = θ But sin–1 = θso 4 θ cosθ= cos(sin–1 ) = Find the exact value of cos(sin–1 ) in radians. so Using the following right-angled triangle: 7 + a2 = 16 a = 3 3 The length of the third side is 3 so

More Related