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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.
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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 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?
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
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
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 :
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.
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
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 :
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
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:
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