2The inverse of the sine function Suppose we wish to find θ such thatsin θ = xIn other words, we want to find the angle whose sine is x.This is written asθ = sin–1 xorθ = arcsin xIn 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?
3The 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:yy = sin xxxyThe inverse of this graph is not a function because it is one-to-many:y = sin–1 xThe inverse graph can be obtained by reflecting the original graph in the line y = x, effectively interchanging the role of x and y.
4The 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:y1–1xy = sin x
5The graph of y = sin–1 xTherefore the inverse of f(x) = sin x, – ≤ x ≤ , is also aone-to-one function:f –1(x) = sin–1 x1–1xyxy1–1y = sin–1 x1–1y = sin–1 xThe graph of y = sin–1 x is the reflection ofy = 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 ≤ 1The range of sin–1 x is the same as the restricted domain ofsin x :– ≤ sin–1 x ≤
6The 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 thatiff(x) = cos xfor0 ≤ x ≤ πf –1(x) = cos–1 xthenfor–1 ≤ x ≤ 1.– < x <And iff(x) = tan xforf –1(x) = tan–1 xthenforxRemind 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.
7The graph of y = cos–1 xxy1–1y = cos–1 xy1–1y = cos–1 x1x–1y = cosxThe domain of cos–1 x is the same as the range of cos x :–1 ≤ x ≤ 1The range of cos–1 x is the same as the restricted domain of cos x :0 ≤ cos–1 x ≤ π
8The graph of y = tan–1 xxyy = tan xy = tan–1 xyy = tanxy = tan–1 xxThe domain of tan–1 x is the same as the range of tan x :xThe range of tan–1 x is the same as the restricted domain oftan x :– < tan–1 x <
9Problems 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:tancossin90°60°45°30°0°degreesradians11Remind 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 √3.1From this tablesin– =
10Problems involving inverse trig functions Find the exact value of sin– in radians.This is equivalent to solving the trigonometric equationcos θ = –for 0 ≤ θ ≤ πthis is the range of cos–1xWe 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.–1θ1From the graph,cos =–So, cos– =
11Problems involving inverse trig functions Find the exact value of cos (sin–1 ) in radians.Let sin– = θsosin θ =θ4Using the following right-angled triangle:7 + a2 = 16a = 33The length of the third side is 3 soAlternatively, 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 socos (sin–1 ) =