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Inverse Trigonometric Functions Table of domain and range of inversetrigonometric function Inverse-forward identities are Forward-inverse identities are
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Example -- "the angle ."1. Evaluate sin -1 2 whose sine is 2
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Solution. is the sine of what angle? 2 π = sin 2 4That is, π Sin-1 = . 2 4 The range of y = arcsin x=sin-1x is not the only π . is the sine of But angle whose sine 4 2 2 every 1st and is 2nd quadrant angle whose corresponding πacute angle is 4 3π sin = . 4 2 πsin ( + 2π) = . 4
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2 And so on. For the function y = sin-1x to be single-valued, then, we must restrict the values of y. How will we do that? We will restrict them tothose angles that have the smallest absolutevalue.In that same way, we will restrict the range ofeach inverse trigonometric function. (Topic 3of Precalculus.) is the angle ofπ smallest .4 absolute value 2 whose sine is Example2. Evaluate sin- )1 (− 2Solution. Angles whose sines are negative fallin the 3rd and 4th quadrants. The angle ofsmallest absolute value is in the 4th
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πquadrant. It is the angle − . 4 -1 π sin (− )=− . 2 4For angles whose sine is negative, we alwayschoose a 4th quadrant angle. In fact, sin-1 (−x) = − sin-1x sin-1(−½) = − sin-1½ π =− . 6 *To see that sin-1 (−x) = − sin-1x, look here:
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= θ. = −θ. Here, then, is the range of the function y =sin-1x
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To restrict the range of arcsin xis equivalent to restricting thedomain of sin x to those samevalues. This will be the casewith all the restricted rangesthat follow. .) Problem 2. Evaluate the following in radians. a) sin-1 0 = 0.
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b) sin-1 1 = π/2.c) sin-1 (−1) = −π/2. π/3. −π/3. −π/6. The range of y = tan-1 xSimilarly, we must restrict the range of y =tan-1 x. Like y = sin-1 x, y = tan-1 x has itssmallest absolute values in the 1st and 4thquadrants.
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For angles whose tangent is postive, wechoose a 1st quadrant angle. For angleswhose tangent is negative, we choose a 4thquadrant angle. Like sin-1 (−x), tan-1 (−x) = −tan -1x.
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= −θ. = θ.Problem 3. Evaluate the following. -1 π -1 πa) tan 1 = b) tan (−1) = − 4 4
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π π −1 c) tan = −1 d) tan (− ) = −3 3 -1 π e) tan 0 = 0 f) =− 6 The range of y = cos-1 xThe values of y = Cos-1 x will have theirsmallest absolute values when y falls withinthe 1st or 2nd quadrants. Example 3. Evaluate a) cos-1 ½ Solution. The radian angle whose π (60°).cosine is ½ is 3
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b) cos-1 (−½)Solution. An angle x whose cosine isnegative falls in the 2nd quadrant.And the cosine of a 2nd quadrant angle isthe negative of the cosine of its supplement. This implies: An angle θ whose cosine is −x is the supplement of the angle whose cosine is x. cos-1 (−x) = π – cos-1 x. Therefore, cos-1 (−½) = π – cos-1 ½ = 2π = 3Problem 4. Evaluate the following.
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a) cos-1 1 = 0 b) cos-1 (−1) = π −1 π −1 3π c) cos = d) cos (− )= 2 4 2 4 π 5π e) cos-1 0 = f) = 2 6 The range of y = sec-1 x In calculus, sin−1x, tan−1x, and cos−1x are the most important inverse trigonometric functions. Nevertheless, here are the ranges that make the rest single-valued.The Inverse Sine Function (sin-1 x)We define the inverse sine function as
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y = sin-1 x forwhere y is the angle whose sine is x. Thismeans thatx = sin yThe graph of y = sin-1 xLets see the graph of y = sin x first and thenderive the curve of y = sin-1 x.As we did previously , if we reflect theindicated portion of y = sin x through the liney = x, we obtain the graph of y = sin-1 x:
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Once again, what you see is what you get.The graph does not extend beyond theindicated boundaries of x and y.The domain (the possible x-values) of sin-1x is-1 ≤ x ≤ 1The range (of y-values for the graph) for sin-1 x is-π/2 ≤ sin-1 x ≤ π/2The Inverse Tangent Function (tan-1x)
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As a reminder, here is the graph of y = tan x,that we met before in Graphs of tanx, cotx,secx and cscx.Reflecting this portion of the graph in theline y = x, we obtain the graph of y = tan-1 x:
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This time the graph does extend beyondwhat you see, in both the negative andpositive directions of x.The domain (the possible x-values) of tan-1x isAll values of xThe range (of y-values for the graph) forarctan x is-π/2 < tan-1 x < π/2
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The graph of the inverse of cosine x isfound by reflecting the graph of cos xthrough the line y = x.We now reflect every point on this portion ofthe cos x curve through the line y = x.
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Thats it for the graph - it does not extendbeyond what you see here. (If it did, therewould be multiple values of y for each valueof x and then we would no longer have afunction.)The domain (the possible x-values) of cos-1x is-1 ≤ x ≤ 1The range (of y-values for the graph) forcos-1 x is0 ≤ cos-1 x ≤ πThe Inverse Tangent Function (arctan)As a reminder, here is the graph of y = tan x,that we met before in Graphs of tan, cot,sec and csc.
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Reflecting this portion of the graph in theline y = x, we obtain the graph of y = tan-1 x :
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This time the graph does extend beyondwhat you see, in both the negative andpositive directions of x.The domain (the possible x-values) of tan-1x isAll values of xThe range (of y-values for the graph) fortan-1 x is-π/2 < tan-1 x < π/2The Inverse Tangent FunctionLets begin by thinking about the graph of. If we want to solve , we may drawa horizontal line units above the axis andchoose one of the points which lies on theintersection of the graph and the horizontal line.
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From this demonstration, you can see that,as you vary the horizontal line, there arealways lots of solutions. However, there isalways a unique solution between and .For this reason, we have a well-definedfunction if define the inverse tangentfunction by saying if is the value between and such that .This is similar to the square root function:there are two values which satisfy but
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we agree, by convention, that the square rootof 4 is the positive value.Here are some famous values of the inversetangent function: -1 0 0 1
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In fact, we can sketch the graph of theinverse tangent as belowOther inverse trigonometric functionsIn the same way, we can build up the inversesine and cosine functions: if is the value between and such that .
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We can understand the graph of the inversesine functionAlso, if is the value between and such that .
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Notice that the domain of both of thesefunctions is restricted: if , there is noangle such that . This means thatwe require in both of these functions.You should verify for yourself the followingrelationships:
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Lets first recall the graph of y = cos x ,so we can see where the graph of y =cos-1 x comes from.We now choose the portion of this graphfrom x = 0 to x = π.
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The result is the graph y = cos-1 x:Thats it for the graph - it does notextend beyond what you see here. (If itdid, there would be multiple values of yfor each value of x and then we wouldno longer have a function.)The domain (the possible x-values) ofcos-1 x is-1 ≤ x ≤ 1The range (of y-values for the graph) forcos-1 x is
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0 ≤ cos-1 x ≤ πThe Inverse Secant Function (sec-1x)The graph of y = sec x, that we metbefore in Graphs of tanx, cotx, secx andcscx:The graph of y = sec-1x:
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The domain of sec-1x isAll values of x, except -1 < x < 1The range of sec-1x is0 ≤ arcsec x ≤ π, sec-1x ≠ π/2The Inverse Cosecant Function(arccsc)The graph of y = csc x, that we metbefore in Graphs of tanx, cotx, secx andcscx:
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The graph extends in the negative andpositive x-directions.The domain of csc-1 x isAll values of x, except -1 < x < 1The range of csc-1 x is-π/2 ≤ csc-1 x ≤ π/2, csc-1 x ≠ 0The Inverse Cotangent Function (cot-1 )The graph of y = cot x, that we metbefore in Graphs of tanx, cotx, secx andcscx is as follows:
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Taking the highlighted portion as above,and reflecting it in the line y = x, we havethe graph of y = cot-1 x:
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The graph extends in the negative andpositive x-directions.So the domain of cot-1 x is:All values of xThe range of cot-1 x is−π/2 < cot-1 x ≤ π/2For suitable values of x and y
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sin-1 x + sin-1 y= sin-1 (x√1-y2+ y√1-x2)sin-1 x - sin-1 y=sin-1 (x√1-y2- y√1-x2)cos-1 x + cos-1 y= cos-1 (xy- √1-x2√1-y2)cos-1 x - cos-1 y= cos-1 (xy+√1-x2√1-y2)tan-1 x + tan-1y = ; xy<1tan-1 x – tan-1 y= ; xy>-12tan-1 x= = =Trigonometry examplesExample 1:Solve the following equation:Suggested answer:
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Example 2:Prove the following equation:Suggested answer:
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The angles in theoretical work will be in radian measure. Thus if we are given a radian π for example, then we can angle, 6 evaluate a function of it.Problems – Solve Inverse TrigonometricFunctions Problems:Problem 1:Prove that + = , x<Solution:Let x = tan θ , then θ = tan-1 x. we have
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You will take R.H.S to prove the given expressionR.H.S = = tan-1 (tan 3θ) = 3 θ = 3 tan-1 x = tan-1 x + 2 tan-1 x = + = L.H.SHence, the given expression will be proved. OR We can take L.H.S. = + =By using tan-1 x + tan-1y = =R.H.S.Example problem 2:Solve tan-1 2x + tan-1 3x = π/4Solution:Given: tan-1 2x + tan-1 3x = π/4Ortan-1 ((2x + 3x)/(1 - 2x . 3x)) = π/4
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tan-1 ((5x)/(1 - 6x2)) = π/4 ∴ (5x)/(1 - 6x2) = tan π/4 = 1 or 6x2 + 5x – 1 = 0 That means, (6x – 1)(x + 1) = 0 Which gives x = 1/6 or x = -1 since x = -1 does not satisfy the equation ,the equation of the L.H.S is negative, so x = 1/6 is the only solution of the given equation. Practice Question – Solve Inverse Trigonometric FunctionsProblems: Practice problem 1: Prove the given expresssion 3sin-1 x =sin-1 (3x – 4x3 ), x ∈ (-1/2,1/2 ) Practice problem 2: Prove the given expression 3 cos-1 x =cos-1 (4x3 – 3x), x ∈ (1/2, 1) Practice problem 3: Find the value of tan-1 [ 2 cos(2 sin-1 (½))] [answer is п/4] ASSIGNMENT: Question.1 Evaluate: (i) sin-1(sin10) (ii) cos-1 (cos10) (iii)tan-1(tan(-6))
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Answer: (i) if –π/2 ≤x ≤π/2, then sin-1(sinx)=x but x= 10radians does not lie between –π/2 and π/2 3π – 10 lies between –π/2 and π/2 ∴ sin-1(sin(3π-10)) = 3π-10. Similarly for (ii) cos-1 (cos10) = cos-1 (cos(4π-10)) =4π-10. [10 radians does not lie between 0 and π. ∴ 0≤4π-10≤π] For (iii) tan-1(tan(-6)) = tan-1(tan(2π-6)) = 2π-6 . { -6 radiansdoes not lie in [ –π/2 , π/2]} Question.2 If x = cos-1(cos4) and y = sin-1(sin3), then whichholds? (give reason) (i) x=y=1 (ii) x+y+1=0 (iii) x+2y=2 (iv) tan(x+y) = -tan7. Question.3 if + = , then prove that -cos + = sin2 [Hint: + = - ]= ⇨ cos )2 = )2 Simplify it] Question.4 (i) sin-1x + sin-1y + sin-1z = π, then prove that X4+y4+z4+4x2y2z2 = 2(x2y2+y2z2+z2x2) (ii) If + + = π/2 ; prove thatxy+yz+xz = 1. (iii) If + + = π , prove thatx+y+z = xyz.
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[Hint: for (i) sin-1x + sin-1y = π - sin-1z ⇨ cos(sin-1x + sin-1y)=cos( π - sin-1z) Use cos(A-B) = cosAcosB – sinAsinB and cos(π – )= -cos It becomes - xy = - and simply it. [Hint: for (ii) tan-1 x + tan-1y = ] Question.5 Write the following functions in the simplestform: (i) ) (ii) ) (iii) ,-a<x<a [Hint: for (i) write cosx = cos2x/2 – sin2x/2 and 1+sinx=(cosx/2 +sinx/2)2 , then use tan(A-B), answer is π/4 – x/2 ] [ Hint: for (ii) write cosx = sin(π/2 – x) and sinx = cos(π/2 –x), then use formula of 1-cos(π/2 – x)= 2sin2(π/4 – x/2) and sin(π/2 –x) = 2 sin(π/4 – x/2) cos(π/4 – x/2) Same method can be applied for (i) part also. Answer is π/4+ x/2] [ for (iii) put x=a cos , then answer will be ½ ] Question.6 If y = ) - , prove thatsiny = tan2(x/2).
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[Hint: y = - 2 , use formula 2 = )] Question.7 (i) Prove that + + = π. (ii) Prove that ) + ) + ) = 0. [Hint: for (i) = - = , then useformula of tan-1 x + tan-1y = ] [Hint: write = ] Question.8 Solve the following equations: (i) + = . (ii) + = . [Hint: write = - , put = y] [Hint: use ] Question.9 Using principal values, evaluate ) + ). [answer is π] Question.10 Show that tan( )= and justify whythe other value is ignored?
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[ Hint: put =∅ ⇨ ¾ = sin2∅ = 2tan∅/(1+tan2∅), findtan∅] ** SOME HOT QUESTIONS: 1. Which is greater tan 1 or tan-11? 2. Find the value of sin(2 ) + cos( 3. Find the value of x which satisfies the equation + = . 4. Solve the equation: + ) = -π/2. 5. Show that tan ) = ). 6. If = - ), then find thegeneral value of . ANSWERS WITH HINTS: 1. Since 1> π/4 ⇨ tan1> 1> tan-11. 2. sin(2 ) + cos( = sin2x + cosy ⇨ + = + = . 3. + = sin( ) by usingsin(A+B)=sinA cosB + cosA sinB
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⇨ x + (1-x) = ∵sin( )= ⇨ 2x – x2 = 1 ⇨ x = 0 or ½. 4. =- - ) ⇨ 6x = sin[- - )] = -cos[ ]= -cos[ ]=- etc. 5. 1/2(2 tan )) , use formula 2 =and tan2x/2 = . 6. Put tan = t and use sin2 = and cos2 = then put t/3 = T,answer is = nπ, nπ+π/4. - = ½ ⇨ = ½ =½ = ½ (2T), then tan = 0 ,1. .
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