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**The inverse trigonometric functions**

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

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**The inverse of the sine function**

Suppose we wish to find θ such that sin θ = x In other words, we want to find the angle whose sine is x. This is written as θ = sin–1 x or θ = arcsin x In this context, sin–1 x means the inverse of sin x. This is not the same as (sin x)–1 which is the reciprocal of sin x, Stress that while we do write sin2x to mean (sin x)2, sin–1x does not mean (sin x)–1. Is y = sin–1 x a function?

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**The inverse of the sine function**

We can see from the graph of y = sin x between x = –2π and x = 2π that it is a many-to-one function: y y = sin x x x y The inverse of this graph is not a function because it is one-to-many: y = sin–1 x The inverse graph can be obtained by reflecting the original graph in the line y = x, effectively interchanging the role of x and y.

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**The inverse of the sine function**

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

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The graph of y = sin–1 x Therefore the inverse of f(x) = sin x, – ≤ x ≤ , is also a one-to-one function: f –1(x) = sin–1 x 1 –1 x y x y 1 –1 y = sin–1 x 1 –1 y = sin–1 x The graph of y = sin–1 x is the reflection of y = sin x in the line y = x: y = sin x (Remember the scale used on the x- and y-axes must be the same.) Stress that the graph of an inverse function will only be a reflection of the original function in y = x if the scale used on the axes is the same. For this reason we label the axes using radians rather than degrees, where π/2 is just over 1.5 radians. The domain of sin–1 x is the same as the range of sin x : –1 ≤ x ≤ 1 The range of sin–1 x is the same as the restricted domain of sin x : – ≤ sin–1 x ≤

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**The inverse of cosine and tangent**

We can restrict the domains of cos x and tan x in the same way as we did for sin x so that if f(x) = cos x for 0 ≤ x ≤ π f –1(x) = cos–1 x then for –1 ≤ x ≤ 1. – < x < And if f(x) = tan x for f –1(x) = tan–1 x then for x Remind students that tan x is undefined for x = –π/2 and x = π/2 and so these values are not included when restricting the domain. The graphs cos–1 x and tan–1 x can be obtained by reflecting the graphs of cos x and tan x in the line y = x.

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

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

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**Problems involving inverse trig functions**

Find the exact value of sin– in radians. To solve this, remember the angles whose trigonometric ratios can be written exactly: tan cos sin 90° 60° 45° 30° 0° degrees radians 1 1 Remind students that these can be recalled quickly by sketching a right-angled triangle with acute angles equal to 45° and sides of length 1 and √2, or a right-angled triangle with acute angles equal to 30° and 60° and sides of length 1, 2 and √3. 1 From this table sin– =

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**Problems involving inverse trig functions**

Find the exact value of sin– in radians. This is equivalent to solving the trigonometric equation cos θ = – for 0 ≤ θ ≤ π this is the range of cos–1x We know that cos = = Sketching y = cos θ for 0 ≤ θ ≤ π : Alternatively, we could draw a quadrant diagram to show that cos 3π/4 is –√2/2 since it is in the second quadrant where cos θ is negative. –1 θ 1 From the graph, cos = – So, cos– =

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**Problems involving inverse trig functions**

Find the exact value of cos (sin–1 ) in radians. Let sin– = θ so sin θ = θ 4 Using the following right-angled triangle: 7 + a2 = 16 a = 3 3 The length of the third side is 3 so Alternatively, we could use the identity sin2 θ + cos2 θ = 1. This would give us 7/16 + cos2 θ = 1 cos2 θ = 9/16 cos θ = 3/4. cos θ = But sin– = θ so cos (sin–1 ) =

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8.3 Solving Right Triangles

8.3 Solving Right Triangles

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