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Inverse-trignometric-function Formula Sheet

This page will help you to revise formulas and concepts of Inverse-trignometric-function instantly for various exams.
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Inverse trigonometric functions reverse the usual ratios, allowing you to find angles when given a ratio. They include functions like arcsin, arccos, and arctan, which map values from the ratio back to an angle in specific ranges. Think of them as the GPS that gets you back to the angle when you’ve wandered too far into sine, cosine, or tangent territory!

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Neetesh Kumar

Neetesh Kumar | September 06, 2024                                       \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space Share this Page on: Reddit icon Discord icon Email icon WhatsApp icon Telegram icon



1. Domain, Range & Graph of Inverse Trignometric Functions: (#toc1)

(a) f1:[1,1][π2,π2],{f}^{-1}:[-1, 1] \to [-\frac{\pi}{2}, \frac{\pi}{2}],
    f1(x)=sin1(x)\space \space \space \space {f}^{-1}{(x)} = sin^{-1}(x)

(b) f1:[1,1][0,π],{f}^{-1}:[-1, 1] \to [0, \pi],
    f1(x)=cos1x\space \space \space \space {f}^{-1}(x) = cos^{-1}x

(c) f1:R(π2,π2),{f}^{-1}:R \to (-\frac{\pi}{2}, \frac{\pi}{2}),
    f1(x)=tan1x\space \space \space \space {f}^{-1}(x) = tan^{-1}x

(d) f1:R(0,π),{f}^{-1}:R \to (0, \pi),
    f1(x)=cot1x\space \space \space \space {f}^{-1}(x) = cot^{-1}x

(e) f1:(,1][1,){f}^{-1}:(-\infin, -1] \cup [1, -\infin)
    [0,π2)(π2,π],\space \space \space \space \to [0,\frac{\pi}{2}) \cup (\frac{\pi} {2},\pi],
    f1(x)=sec1x\space \space \space \space {f}^{-1}(x) = sec^{-1}x

(f) f1:(,1][1,){f}^{-1}:(-\infin, 1] \cup [1, -\infin)
    [π2,0)(0,π2],\space \space \space \space \to [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}],
    f1(x)=cosec1x\space \space \space \space {f}^{-1}{(x)} = cosec^{-1}x


2. Properties of Inverse Circular Functions: (#toc2)

(a)

(i) y=sin(sin1x)=x,x[1,1],y[1,1],y = sin(sin^{-1}x) = x, x \in [-1,1], y \in [-1,1], y is aperiodic

(ii) y=cos(cos1x)=x,x[1,1],y[1,1],y = cos(cos^{-1}x) = x, x \in [-1,1], y \in [-1,1], y is aperiodic

(iii) y=tan(tan1x)=x, xR,yR,y = tan(tan^{–1}x) = x, \ x \in R, y \in R, y is aperiodic

(iv) y=cot(cot1x)=x,xR,yR,y = cot(cot^{-1}x) = x, x \in R, y \in R, y is aperiodic

(v) y=cosec(cosec1x)=x,x1,y1,y = cosec(cosec^{-1}x) = x, |x| \geq 1, |y| \geq 1, y is aperiodic

(vi) y=sec(sec1x)=x,x1;y1,y = sec(sec^{-1}x) = x, |x| \geq 1; |y| \geq 1, y is aperiodic


(b)

(i) y=sin1(sin x),xR,y[π2,π2].y = sin^{-1} (sin \ x), x \in R, y \in [-\frac{\pi}{2}, \frac{\pi}{2}]. Periodic with period 2π.2\pi.

    sin1(sin x)={πx    , 3π2xπ2    x          ,  π2xπ2 πx      ,   π2x3π2x2π     ,  3π2x5π23πx     ,  5π2x7π2x4π     ,  7π2x9π2\space \space \space \space sin^{-1} (sin \ x) = \begin{cases} -\pi-x \space \space \space \space, \space -\frac{3\pi}{2} \leq x \leq -\frac{\pi}{2} \\ \space \space \space \space x \space \space \space \space \space \space \space \space \space \space, \space \space -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \\ \space \pi-x \space \space \space \space \space \space, \space \space \space \frac{\pi}{2} \leq x \leq \frac{3\pi}{2} \\ x-2\pi \space \space \space \space \space, \space \space \frac{3\pi}{2} \leq x \leq \frac{5\pi}{2} \\ 3\pi-x \space \space \space \space \space , \space \space \frac{5\pi}{2} \leq x \leq \frac{7\pi}{2} \\ x-4\pi \space \space \space \space \space, \space \space \frac{7\pi}{2} \leq x \leq \frac{9\pi}{2} &\text\\ \end{cases}

(ii) y=cos1(cos x),xR,y[0,π],y = cos^{-1} (cos \ x), x \in R, y \in [0,\pi], periodic with period 2π2\pi

     cos1(cos x)={x         , πx0   x         ,   0xπ2πx   ,   πx2πx2π   ,  2πx3π4πx   ,  3πx4π\space \space \space \space \space cos^{-1} (cos \ x) = \begin{cases} -x \space \space \space \space \space \space \space \space \space, \space -\pi \leq x \leq 0 \\ \space \space \space x \space \space \space \space \space \space \space \space \space, \space \space \space 0 \leq x \leq \pi \\ 2\pi-x \space \space \space, \space \space \space \pi \leq x \leq 2\pi \\ x-2\pi \space \space \space, \space \space 2\pi \leq x \leq 3\pi \\ 4\pi-x \space \space \space , \space \space 3\pi \leq x \leq 4\pi &\text\\ \end{cases}

(iii) y=tan1(tan x)y = tan^{-1} (tan \ x)
      xR{(2n1)π2,nI};yπ2,π2\space \space \space \space \space \space x \in R - \bigg \{(2n-1) \frac{\pi}{2}, n \in \Iota \bigg \}; y \in \bigg \lgroup -\frac{\pi}{2}, \frac{\pi}{2} \bigg \rgroup periodic with period π\pi

       tan1(tan x)={x+π   ,3π2<x<π2   x       ,  π2<x<π2xπ   ,   π2<x<3π2x2π ,  3π2<x<5π2x3π ,  5π2<x<7π2\space \space \space \space \space \space \space tan^{-1} (tan \ x) = \begin{cases} x+\pi \space \space \space, -\frac{3\pi}{2} < x < -\frac{\pi}{2} \\ \space \space \space x \space \space \space \space \space \space \space, \space \space -\frac{\pi}{2} < x< \frac{\pi}{2} \\ x - \pi \space \space \space, \space \space \space \frac{\pi}{2} < x < \frac{3\pi}{2} \\ x - 2\pi \space, \space \space \frac{3\pi}{2} < x < \frac{5\pi}{2} \\ x -3\pi \space, \space \space \frac{5\pi}{2} < x < \frac{7\pi}{2} &\text\\ \end{cases}

(iv) y=cot1(cot x),xR{n π},y(0,π), y = cot^{-1} (cot \ x),x \in R - \{n \ \pi\}, y \in (0, \pi), periodic with period π\pi

(v) y=cosec1(cosec x),xR{n π,nI}y[π2,00,π2], y = cosec^{-1} (cosec \ x), x \in R - \{n \ \pi, n \in \Iota \} y \in \bigg[ -\frac{\pi}{2},0 \bigg \rgroup \cup \bigg \lgroup 0, \frac{\pi}{2} \bigg ], y is periodic with period 2π.2\pi.

(vi) y=sec1(sec x),y = sec^{-1} (sec \ x), y is periodic with period 2π2\pi
     xR{(2n1)π2nI},y[0,π2π2,π]\space \space \space \space \space x \in R - \bigg \{(2n-1) \frac{\pi}{2} n \in \Iota \bigg \}, y \in \bigg [0, \frac{\pi}{2} \bigg \rgroup \cup \bigg \lgroup \frac{\pi}{2}, \pi \bigg]


(c)

(i) cosec1x=sin11x; x1 or x1 cosec^{-1} x = sin^{-1} \dfrac{1}{x}; \space x \leq -1 \ or \ x \geq 1

(ii) sec1x=cos11x; x1 or x1sec^{-1} x = cos^{-1} \dfrac{1}{x}; \space x \leq -1 \ or \ x \geq 1

(iii) cot1x=tan11x; x>0cot^{-1} x = tan^{-1} \dfrac{1}{x}; \space x > 0
                =π+tan11x; x<0\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space = {\pi} + tan^{-1} \dfrac{1}{x}; \space x < 0


(d)

(i) sin1(x)=sin1x,1x1sin^{-1}(-x) = -sin^{-1}x, -1 \leq x \leq 1

(ii) tan1(x)=tan1x,xRtan^{-1}(-x) = -tan^{-1}x, x \in R

(iii) cos1(x)=πcos1x,1x1cos^{-1}(-x)= \pi - cos^{-1}x, -1 \leq x \leq 1

(iv) cot1(x)=πcot1x,xRcot^{-1}(-x) = \pi - cot^{-1}x, x \in R

(V) sec1(x)=πsec1x,x1 or x1sec^{-1}(-x) = \pi - sec^{-1}x, x \leq -1 \ or \ x \geq 1

(vi) cosec1(x)=cosec1x,x1 or x1cosec^{-1}(-x) = -cosec^{-1}x, x \leq -1 \ or \ x \geq 1


(e)

(i) sin1x+cos1x=π2,1x1sin^{-1}x + cos^{-1}x = \frac{\pi}{2}, -1 \leq x \leq 1

(ii) tan1x+cot1xπ2,xRtan^{-1}x + cot^{-1}x - \frac{\pi}{2}, x \in R

(iii) cosec1x+sec1x=π2,x1cosec^{-1}x + sec^{-1}x = \frac{\pi}{2}, | x | \geq 1


(f) (i) tan1x+tan1y={tan1(x+y1xy),where x>0,y>0 & xy<1π+tan1(x+y1xy),where x>0,y>0 & xy>1π2,where x>0,y>0 & xy=1tan^{-1}x + tan^{-1}y = \begin{cases} tan^{-1} \bigg(\dfrac{x+y}{1-xy}\bigg), where \ x > 0, y > 0 \ \& \ xy < 1 \\ \pi +tan^{-1}\bigg(\dfrac{x+y}{1-xy}\bigg), where \ x > 0, y > 0 \ \& \ xy > 1 \\ \frac{\pi}{2}, where \ x > 0, y > 0 \ \& \ xy = 1 &\text\\ \end{cases}

(ii) tan1xtan1y=tan1(xy1+xy),where x>0,y>0tan^{-1}x - tan^{-1}y = tan^{-1} \bigg(\dfrac{x-y}{1+xy}\bigg), where \space x > 0, y > 0

(iii) sin1x+sin1y=sin1[x1y2+y1x2],      where x>0, y>0 & (x2+y2)<1sin^{-1}x + sin^{-1}y = sin^{-1}[x\sqrt{{1-y}^{2}} + y \sqrt{{1-x}^{2}}],\\ \space \space \space \space \space \space where \space x > 0, \space y > 0 \ \& \ ({x}^{2} + {y}^{2}) < 1

      \space \space \space \space \space \space Note that : x2+y2<10<sin1x+sin1y<π2{x}^{2} + {y}^{2} < 1 \rArr 0 < sin^{-1}x + sin^{-1}y < \frac{\pi}{2}

(iv) sin1x+sin1y=πsin1[x1y2+y1x2],      where x>0, y>0 & x2+y2>1sin^{-1}x + sin^{-1}y = \pi - sin^{-1} [x \sqrt{{1-y}^{2}} + y\sqrt{{1-x}^{2}}],\\ \space \space \space \space \space \space where \space x > 0, \space y > 0 \space \& \space {x}^{2} + {y}^{2} > 1

      \space \space \space \space \space \space Note that : x2+y2>1π2<sin1x+sin1y<π{x}^{2} + {y}^{2} > 1 \rArr \frac{\pi}{2} < sin^{-1}x + sin^{-1}y < \pi

(v) sin1xsin1y=sin1[x1y2y1x2] where x>0,y>0sin^{-1}x - sin^{-1}y = sin^{-1}[x \sqrt{{1-y}^{2}} - y \sqrt{{1-x}^{2}}] \ where \space x > 0, y > 0

(vi) cos1x+cos1y=cos1[xy1x21y2], where x>0,y>0cos^{-1}x + cos^{-1}y = cos^{-1}[xy - \sqrt{{1-x}^{2}} \sqrt{{1-y}^{2}}], \ where \ x > 0, y > 0

(vii) cos1xcos1y={cos1(xy+1x21y2);x<y,x,y>0cos1(xy+1x21y2);x>y,x,y>0cos^{-1}x - cos^{-1}y = \begin{cases} cos^{-1}(xy + \sqrt{{1-x}^{2}} \sqrt{{1-y}^{2}}); x < y, x, y > 0 \\ -cos^{-1}(xy + \sqrt{{1-x}^{2}} \sqrt{{1-y}^{2}}); x > y, x, y >0 &\text\\ \end{cases}

(viii) tan1x+tan1y+tan1z=tan1[x+y+zxyz1xyyzzx]       if x>0,y>0,z>0 & xy+yz+zx<1tan^{-1}x + tan^{-1}y + tan^{-1} z = tan^{-1} \bigg[\dfrac{x+y+z-xyz}{1-xy-yz-zx}\bigg] \\ \space \space \space \space \space \space \space if \space x > 0, y > 0, z > 0 \space \& \space xy + yz +zx < 1

Note : In the above results x & yx \space \& \space y are taken positive. In case if these are given as negative, we first apply P-4 and then use above results.


3. Simplified Inverse Trignometric Functions: {toc#3}

     \space \space \space \space \space (a) y=f(x)=sin12x1+x2              ={2tan1xif  x1π2tan1xif   x>1(π+2tan1xif x<1y = f(x) = sin^{-1} \bigg \lgroup \dfrac{2x}{1+x^{2}} \bigg \rgroup \\ \space \space \space \space \space \space \space \space \space \space \space \space \space \space = \begin{cases} 2tan^{-1}x &\text{if } \space |x|\leq 1 \\ \pi - 2 tan^{-1}x &\text{if } \space \space x > 1 \\ -(\pi + 2 tan^{-1}x &\text{if } x < -1 \end{cases}

     \space \space \space \space \space (b) y=f(x)=cos11x21+x2              ={2tan1xif  x02tan1xif   x<0y = f(x) = cos^{-1} \bigg \lgroup \dfrac{1-x^{2}}{1+x^{2}} \bigg \rgroup \\ \space \space \space \space \space \space \space \space \space \space \space \space \space \space = \begin{cases} 2tan^{-1}x &\text{if } \space |x| \geq 0 \\ -2 tan^{-1}x &\text{if } \space \space x < 0 \\ \end{cases}

     \space \space \space \space \space (c) y=f(x)=tan12x1x2           ={2tan1xif  x<1π+2tan1xif   x<1(π2tan1xif   x>1y = f(x) = tan^{-1} \dfrac{2x}{1-x^{2}} \\ \space \space \space \space \space \space \space \space \space \space \space = \begin{cases} 2tan^{-1}x &\text{if } \space |x| < 1 \\ \pi + 2 tan^{-1}x &\text{if } \space \space x < - 1 \\ -(\pi - 2 tan^{-1}x &\text{if } \space \space x > 1 \end{cases}

     \space \space \space \space \space (d) y=f(x)=sin1(3x4x3)          ={(π+3sin1x)if  1x12       3sin1xif   12x12  π3sin1xif       12x1y = f(x) = sin^{-1} (3x -4x^{3}) \\ \space \space \space \space \space \space \space \space \space \space = \begin{cases} -(\pi + 3sin^{-1}x) &\text{if } \space - 1 \leq x \leq - \frac{1}{2} \\ \space \space \space \space \space \space \space 3sin^{-1}x &\text{if } \space \space - \frac{1}{2} \leq x \leq \frac{1}{2} \\ \space \space \pi - 3sin^{-1}x &\text{if } \space \space \space \space \space \space \frac{1}{2} \leq x \leq1 \end{cases}

     \space \space \space \space \space (e) y=f(x)=cos1(4x33x)           ={3cos1x2πif  1x122π3cos1xif   12x12  3cos1xif       12x1y = f(x) = cos^{-1} (4x^{3} -3x) \\ \space \space \space \space \space \space \space \space \space \space \space = \begin{cases} 3cos^{-1} x -2\pi &\text{if } \space - 1 \leq x \leq - \frac{1}{2} \\ 2 \pi - 3cos^{-1}x &\text{if } \space \space - \frac{1}{2} \leq x \leq \frac{1}{2} \\ \space \space 3cos^{-1}x &\text{if } \space \space \space \space \space \space \frac{1}{2} \leq x \leq1 \end{cases}

     \space \space \space \space \space (f) sin1(2x1x2)          ={(π+2sin1x)1x122sin1x              12x12π2sin1x         12x1sin^{-1} \big(2x \sqrt {1-x^{2}}\big) \\ \space \space \space \space \space \space \space \space \space \space = \begin{cases} -(\pi + 2 sin^{-1}x) - 1 \leq x \leq - \frac{1}{\sqrt2} \\ 2sin^{-1}x \space \space \space \space \space \space \space \space \space \space \space \space \space \space - \frac{1}{\sqrt2} \leq x \leq \frac{1}{\sqrt2} \\ \pi - 2sin^{-1}x \space \space \space \space \space \space \space \space \space \frac{1}{\sqrt2} \leq x \leq1 \end{cases}

     \space \space \space \space \space (g) cos1(2x21)              ={2cos1x                0x12π2cos1x      1x0cos^{-1} (2x^{2}-1) \\ \space \space \space \space \space \space \space \space \space \space \space \space \space \space = \begin{cases} 2 cos^{-1}x \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space 0 \leq x \leq 1 \\ 2\pi - 2cos^{-1}x \space \space \space \space \space \space -1 \leq x \leq 0 \end{cases}

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