7.9.1. Symmetric Relation of Inverse Trigonometric Functions
1. `\sin^{-1}\left(x\right)=\cos ec^{-1}\left(\frac1x\right)` when `-1\leq x\leq1`.
2. `\cos ec^{-1}\left(x\right)=\sin^{-1}\left(\frac1x\right)` when `x\leq-1\;or\;x\geq1`.
3. `\cos^{-1}\left(x\right)=\sec^{-1}\left(\frac1x\right)` when `-1\leq x\leq1`.
4. `\sec^{-1}\left(x\right)=\cos^{-1}\left(\frac1x\right)` when `x\leq-1\;or\;x\geq1`.
5. `\tan^{-1}\left(x\right)=\cot^{-1}\left(\frac1x\right)`.
6. `\cot^{-1}\left(x\right)=\tan^{-1}\left(\frac1x\right)`.
Proof: 1. Let `\sin\theta=x\Rightarrow\theta=\sin^{-1}x\cdots\cdots\cdots(1)`.
Again, `\sin\theta=x\Rightarrow\frac1{\cos ec\theta}=x\Rightarrow\cos ec\theta=\frac1x\Rightarrow\theta=\cos ec^{-1}\left(\frac1x\right)\Rightarrow\sin^{-1}\left(x\right)=\cos ec^{-1}\left(\frac1x\right)`.
Similarly, we can prove others relations from 2. to 6.
7. `\sin\left(\sin^{-1}x\right)=x`.
8. `\cos\left(\cos^{-1}x\right)=x`
9. `\tan\left(\tan^{-1}x\right)=x`.
10. `\cot\left(\cot^{-1}x\right)=x`.
11. `\sec\left(\sec^{-1}x\right)=x`.
12. `\cosec\left(\cosec^{-1}x\right)=x`.
Proof:
7. Let `\sin x=\theta\Rightarrow x=\sin^{-1}\theta\Rightarrow x=\sin^{-1}\left(\sin x\right)\Rightarrow \sin^{-1}\left(\sin x\right)=x`.
Similarly, we can prove the relations from 8 to 12.
13. `\sin^{-1}\left(\sin x\right)=x`
14. `\cos^{-1}\left(\cos x\right)=x`.
15. `\tan^{-1}\left(\tan x\right)=x`.
16. `\cot^{-1}\left(\cot x\right)=x`.
17. `\sec^{-1}\left(\sec x\right)=x`
18. `\cosec^{-1}\left(\cosec x\right)=x`.
Proof:
13. `\sin x=\theta\Rightarrow x=\sin^{-1}\theta\Rightarrow x=\sin^{-1}\left(\sin x\right)\Rightarrow\sin^{-1}\left(\sin x\right)=x`.
Similarly, we can prove the relations from 14 to 118.
Example and Assessment:1. `\sin^{-1}\left(\frac1{\sqrt2}\right)`
`=\sin^{-1}\left(\sin\left(\frac{\pi}4\right)\right)`
`=\frac{\pi}4`.
2. `\sin{\cos^{-1}(\frac{\sqrt3}2)}=\sin{\cos^{-1}(\cos(\frac{\pi}6))}=\sin\frac{\pi}6=\frac1\2`.
3. `\sin^{-1}\left(-\frac1{\sqrt2}\right)=?`
4. `\tan{\cos^{-1}(\frac1\{\sqrt3})}=?`
5. `\sin^{-1}(-1)=\sin^{-1}{-sin(\frac\pi\2)}=\sin^{-1}{sin(-\frac\pi\2)}=-\frac\pi\2`.
6. `\tan[\tan^{-1}{-\frac1{\sqrt3}}]=\tan[\tan^{-1}{-tan(\frac\pi\6)}]`
`\tan^{-1}x+\tan^{-1}y=\tan^{-1}(\frac(x+y)(1-xy))` when `xy<1`.
Example
1. `\tan^{-1}(\frac1\2)+\tan^{-1}(\frac3\2)=?`
Solution:
So, `\tan^{-1}(\frac1\2)+\tan^{-1}(\frac3\2)`
`=tan^{-1}(\frac(\frac1\2+frac3\2)(1-\frac1\2\times\frac3\2))` since `[xy=\frac1\2\times\frac3\2=\frac3\4<1]`
`=tan^{-1}(\frac(2+3)(4-3))`
`=tan^{-1}5`
2. Prove that, `\tan^{-1}(\frac(7)(11))+\tan^{-1}(\frac(1)(7))=\frac(\pi)(4)`.
Solution: Try yourself.
Formula 2:
`\tan^{-1}x+\tan^{-1}y=\frac(\pi)(2)+\tan^{-1}(\frac(x+y)(1-xy))` when `xy>1`
Example
3. `\tan^{-1}3+\tan^{-1}2=?`
Solution:
So, `\tan^{-1}3+\tan^{-1}2`
`=\pi+\tan^{-1}(\frac(3+2)(1-2\times3))` since `xy=3\times2=6>1`
`=\pi+\tan^{-1}(-1)`
`=\pi-\tan^{-1}(1)`
`=\pi-\frac(\pi)4`
`=\frac(3\pi)4`
Formula 3:
`\tan^{-1}x+\tan^{-1}y=\frac\pi2` when `xy=1`
Example
5. `\tan^{-1}(\frac3\2)+\tan^{-1}(\frac2\3)=?`
Solution:
So, `\tan^{-1}(\frac3\2)+\tan^{-1}(\frac2\3)`
`=\frac\pi\2` Since `[xy=(\frac3\2)\times(\frac2\3)=1]`
6. `\tan^{-1}(\frac5\11)+\tan^{-1}(\frac11\5)=?`
Formula 4:
`\tan^{-1}x-\tan^{-1}y=\tan^{-1}(\frac(x-y)(1+xy))`
Example
7. Show that, `\tan^{-1}(\frac5\6)-\tan^{-1}(\frac49\71)=\tan^{-1}(\frac1\11)`
Solution:
So, `\tan^{-1}(\frac5\6)-\tan^{-1}(\frac49\71)`
`=\tan^{-1}{\frac(\frac(5)(6)-\frac49\71)(1+(\frac5\6)\times(\frac49\71))}`
`=\tan^{-1}{\frac(355-294)(426+245)}`
`=\tan^{-1}(\frac(1)(11))`. Proved.
8. Prove that, `\tan^{-1}(\frac(1)(3))-\tan^{-1}(\frac(1)(5))+\tan^{-1}(\frac(1)(7))=\tan^{-1}(\frac(1)(11))`.
Solution: Try yourself.
9. `\tan^{-1}(\frac2\10)-\tan^{-1}(\frac3\5)=?`
Solution: Try yourself.
Formula 5:
`\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}(\frac(x+y+z-xyz)(1-xy-yz-zx))` when `xy+yz+zx<=1`.
Example
12. Show that, `\tan^{-1}(\frac(1)(4))+\tan^{-1}(\frac(1)(5))+\tan^{-1}(\frac(1)(8))=-\tan^{-1}(\frac(91)(2560))`
Solution:
`\tan^{-1}(\frac(1)(4))+\tan^{-1}(\frac(1)(5))+\tan^{-1}(\frac(1)(8))`
`=\tan^{-1}{\frac(\frac1\4+\frac1\5+\frac1\8-(\frac1\4)(\frac1\5)(\frac1\8))(1-(\frac1\4)(\frac1\5)-(\frac1\5)(\frac1\8)-(\frac1\8)(\frac1\4))}`
since `xy+yz+zx=(\frac(1)(4))(\frac(1)(5))+(\frac(1)(5))(\frac(1)(8))+(\frac(1)(8))(\frac(1)(4))=(\frac(1)(20))+(\frac(1)(40))+(\frac(1)(32))=\frac(17)(160)<1`
`=\tan^{-1}{\frac(40+32+20-1)(160-(40)(32)-(32)(20)-(20)(40))}`
`=\tan^{-1}{\frac(91)(160-1280-640-800)}`
`=\tan^{-1}{\frac(91)(-2560)}`
`=-\tan^{-1}(\frac(91)(2560))`
Formula 6:
`\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\frac(\pi)(2)+\tan^{-1}(\frac(x+y+z-xyz)(1-xy-yz-zx))` when `xy+yz+zx>1`
Example
13. `\tan^{-1}5+\tan^{-1}8+\tan^{-1}2=`
Solution:
`\tan^{-1}5+\tan^{-1}8+\tan^{-1}2`
since `xy+yz+zx=(5)(8)+(8)(2)+(2)(5)=66>1`
`=\frac(\pi)(2)+\tan^{-1}{\frac(5+8+2-(5)(8)(2))(1-(5)(8)-(8)(2)-(2)(5))}`
`=\frac(\pi)(2)+\tan^{-1}{\frac(65)(-65)}`
`=\frac(\pi)(2)+\tan^{-1}(-1)`
`=\frac(\pi)(2)-\tan^{-1}(1)`
`=\frac(\pi)(2)-\frac(\pi)(4)`
`=\frac(\pi)(4)`
Formula 7:
`\2tan^{-1}x=\sin^{-1}\frac(2x)\(1+x^2)=\cos^{-1}\frac(1-x^2)\(1+x^2)=\tan^{-1}(\frac(2x)(1-x^2))`
Example
14. `\tan^{-1}(\frac(1)(2))=\sin^{-1}(?)=\coss^{-1}(?)=\tan^{-1}(?)`
Solution:
`\tan^{-1}(\frac(1)(2))`
`=\frac(1)(2)\times 2\tan^{-1}(\frac(1)(2))`
`=\frac(1)(2)\times\sin^{-1}\frac(2(\frac(1)(2)))\(1+(\frac(1)(2))^2)`
`=\frac(1)(2)\times\sin^{-1}\frac(1)(1+(\frac(1)(4))`
`=\frac(1)(2)\times\sin^{-1}\frac(1)(\frac(5)(4))`
`=\frac(1)(2)\sin^{-1}\(\frac(4)(5))`
`\tan^{-1}(\frac(1)(2))`
`=\frac(1)(2)\times 2\tan^{-1}(\frac(1)(2))`
`=\frac(1)(2)\times\cos^{-1}\frac(1-(\frac(1)(2))^2)\(1+(\frac(1)(2))^2)`
`=\frac(1)(2)\times\cos^{-1}\frac(1-\frac(1)(4))(1+\frac(1)(4))`
`=\frac(1)(2)\times\cos^{-1}(\frac(4-1)(4+1))`
`=\frac(1)(2)\cos^{-1}(\frac(3)(5))`
`\tan^{-1}(\frac(1)(2))`
`=\frac(1)(2)\times 2\tan^{-1}(\frac(1)(2))`
`=\frac(1)(2)\times\tan^{-1}\frac(2(\frac(1)(2)))\(1-(\frac(1)(2))^2)`
`=\frac(1)(2)\times\tan^{-1}\frac(1)(1-(\frac(1)(4))`
`=\frac(1)(2)\times\tan^{-1}\frac(1)(\frac(3)(4))`
`=\frac(1)(2)\tan^{-1}\(\frac(4)(3))`
So, `\tan^{-1}(\frac(1)(2))=\frac(1)(2)\sin^{-1}\(\frac(4)(5))=\frac(1)(2)\cos^{-1}(\frac(3)(5))=\frac(1)(2)\tan^{-1}\(\frac(4)(3))`.
15. Show that, `\4tan^{-1}\frac(1)(5)-\tan^{-1}\frac(1)\(239)=\frac(\pi)(4)`
Solution:
So, `\4tan^{-1}\frac(1)(5)-\tan^{-1}\frac(1)\(239)`
`=\2\times2tan^{-1}\frac(1)(5)-\tan^{-1}\frac(1)\(239)`
`=\2\timestan^{-1}\frac(2\times\frac(1)(5))(1-(\frac(1)(5))^2)-\tan^{-1}\frac(1)\(239)`
`=\2\timestan^{-1}\frac(10)(25-1)-\tan^{-1}\frac(1)\(239)`
`=2\tan^{-1}\frac(5)(12)-\tan^{-1}\frac(1)\(239)`
`=\tan^{-1}\frac(2\times\frac(5)(12))(1-(\frac(5)(12))^2)-\tan^{-1}\frac(1)\(239)`
`=\tan^{-1}\frac(\frac(5)(6))(1-\frac(25)(144))-\tan^{-1}\frac(1)\(239)`
`=\tan^{-1}\frac(120)(144-25)-\tan^{-1}\frac(1)\(239)`
`=\tan^{-1}\frac(120)(119)-\tan^{-1}\frac(1)\(239)`
`=\tan^{-1}\frac(\frac(120)(119)-\frac(1)\(239))(1+(\frac(120)(119))(\frac(1)\(239))`
`=\tan^{-1}(\frac(\frac(2856)(28441))(\frac(2856)(28441)))`
`=\tan^{-1}(1)`
`=\frac(\pi)(4)`
Formula 8:
`\2tan^{-1}x=\pi+\tan^{-1}(\frac(2x)(1-x^2))` when `x>1`
Example
16. `2\tan^{-1}(\frac(3)(2))=?`
Solution:
`2\tan^{-1}(\frac(3)(2))`
`=\pi+\tan^{-1}(\frac(2\times\frac(3)(2))(1-(\frac(3)(2))^2))`
`=\pi+\tan^{-1}(\frac(3)(1-(\frac(9)(4))))`
`=\pi+\tan^{-1}(\frac(3)(4-9)`
`=\pi-\tan^{-1}(\frac(3)(5))`
17. Prove that, `2\tan^{-1}(\frac(3)(2))+tan^{-1}(\frac(3)(5))=\pi`
18. Prove that, `\sin^{-1}(\frac(3)(5))+\cot^{-1}(\frac(17)(19))=\tan^{-1}(\frac(127)(11))`
Solution:
`\sin^{-1}(\frac(3)(5))+\cot^{-1}(\frac(17)(19))`
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19. Prove that, `\sin^{-1}(\frac(4)(5))+\cos^{-1}(\frac(2)(\sqrt5))=\tan^{-1}(\frac(11)(2))`
Solution: Try yourself.
20. Prove that, `4{\sin^{-1}(\frac(1)(\sqrt5))+\cot^{-1}3)}=\pi`
Solution: Try yourself.
21. Prove that, `\tan^{-1}(\frac(1)(4))+\tan^{-1}(\frac(2)(9))=\frac(1)(2)\cos^{-1}(\frac(3)(5))`
Solution: Try yourself.
22. Prove that, `\cos^{-1}(\frac(63)(65))+2\tan^{-1}(\frac(1)(5))=\tan^{-1}(\frac(3)(4))`
Solution: Try yourself.
23. Prove that, `\cos^{-1}(\frac(1)(\sqrt5))-\frac(1)(2)\sin^{-1}(\frac(3)(5))+\tan^{-1}(\frac(1)(3))=\tan^{-1}2`
Solution: Try yourself.
24. Prove that, `\sin^{-1}(\frac(3)(5))+\frac(1)(2)\cos^{-1}(\frac(5)(13))-\cot^{-1}2=\tan^{-1}(\frac(28)(29))`
Solution: Try yourself.
25. Prove that, `\tan^{-1}(\frac(2)(3))=\frac(\pi)(2)-\sec^{-1}(\frac(sqrt13)(2)`.
Solution: Try yourself.
26. Prove that, `\cos^{-1}(\sqrt(\frac(2)(3)))-\cos^{-1}(\frac(\sqrt6+1)(2sqrt3))=\frac(\pi)(6)`
Solution: Try yourself.
27. Prove that, `\cos(2\tan^{-1}(\frac(1)(7)))=\frac(\pi)(6)`
Solution: Try yourself.
28. Prove that, `\tan(2\tan^{-1}x)=2\tan(\tan^{-1}x+\tan^{-1}x^3)`. ***
Solution: Try yourself.
29. Prove that, `\cot^{-1}(\tan2x)+\cot^{-1}(-\tan3x)=x`
Solution: Try yourself.
30. Prove that, `\tan^{-1}(\frac(1)(2)\tan2\theta)+\tan^{-1}(\cot\theta)+\tan^{-1}(\cot^3\theta)=0`.*
Solution: Try yourself.
31. Prove that, `\tan^{-1}{(\sqrt2+1)\tan\theta}-tan^{-1}{(\sqrt2-1)tan\theta}=\tan^{-1}(\sin2\theta)`.*
Solution: Try yourself.
32. If , `\sin^{-1}(\frac(2p)(1+p^2))+\cos^{-1}(\frac(1-q^2)(1+q^2))=2\tan^{-1}x` then prove that, `x=\frac(p-q)(1+pq)`.*
Solution: Try yourself.
33. Prove that, `\tan^{-1}(\sqrtx)=\frac(1)(2)cos^{-1}(\frac(1-x)(1+x))`.*
Solution: Try yourself.
34. Prove that, `\tan^{-1}x=\frac(1)(2)cosec^{-1}(\frac(1+x^2)(2x))`.*
Solution: Try yourself.
35. Prove that, `\tan^{-1}x=\frac(1)(2)cosec^{-1}(\frac(1+x^2)(2x))`.*
Solution: Try yourself.
36. Prove that, `2\tan^{-1}(\sqrt(\frac(x)(y))\tan\frac(\theta)(2))=sin^{-1}{\frac(2\sqrt(xy)\sin\theta)((y+x)+(y-x)\cos\theta)}`.*
Solution:
`2\tan^{-1}(\sqrt(\frac(x)(y))\tan\frac(\theta)(2))`
`=sin^{-1}(\frac(2\sqrt(\frac(x)(y))\tan\frac(\theta)(2))(1+(\sqrt(\frac(x)(y))\tan\frac(\theta)(2))^2))`
`=sin^{-1}(\frac(2\sqrt(\frac(x)(y))\tan\frac(\theta)(2))(1+\frac(x)(y)\tan^2\frac(\theta)(2)))`
`=sin^{-1}(\frac(2\sqrt(xy)\frac(\sin\frac(\theta)(2))(\cos\frac(\theta)(2)))(y+x\frac(\sin^2\frac(\theta)(2))(\cos^2\frac(\theta)(2))))`
`=sin^{-1}(\frac(\sqrt(xy)\times2\sin\frac(\theta)(2)\cos\frac(\theta)(2))(y\cos^2\frac(\theta)(2)+x\sin^2\frac(\theta)(2)))`
`=sin^{-1}{\frac(\sqrt(xy)\times\sin\theta)(\frac(y)(2)\times2\cos^2\frac(\theta)(2)+\frac(x)(2)\times2\sin^2\frac(\theta)(2))}`
`=sin^{-1}{\frac(2\sqrt(xy)\times\sin\theta)(y(1+\cos\theta)+x(1-\cos\theta))}`
`=sin^{-1}{\frac(2\sqrt(xy)\sin\theta)((y+x)+(y-x)\cos\theta)}`.
37. Prove that, `2\tan^{-1}(\sqrt(\frac(x-y)(x+y))\tan\frac(\theta)(2))=cos^{-1}(\frac(y+x\cos\theta)(x+y\cos\theta))`.***
Solution: Try yourself.
38. Prove that, `2\tan^{-1}[\tan\frac(\alpha)(2)\tan(\frac(\pi)(4)-\frac(\beta)(2))]=tan^{-1}(\frac(\sin\alpha\cos\beta)(\sin\beta+\cos\alpha))`.*
Solution: Try yourself.
39. Prove that, `\tan^{-1}(\frac(x\cos\theta)(1-x\sin\theta))-\tan^{-1}(\frac(x-\sin\theta)(cos\theta))=\theta`.*
Solution: Try yourself.
`\sin^{-1}x+\cos^{-1}x=\frac\pi\2` when `-1<=x<=1`
`\tan^{-1}x+\cot^{-1}x=\frac\pi\2` when `x>=0`
`\cosec^{-1}x+\sec^{-1}x=\frac\pi\2` when `x<=-1, x>=1`
3.9.4. Formula for `\sin^{-1}x+-\sin^{-1}y` and `\cos^{-1}x+-\sin^{-1}y`:
`\sin^{-1}x+\sin^{-1}y=\sin^{-1}{x\sqrt(1-y^2)+y\sqrt(1-x^2)}` when `x^2+y^2<=1`
`\sin^{-1}x+\sin^{-1}y=\pi-\sin^{-1}{x\sqrt(1-y^2)+y\sqrt(1-x^2)}` when `x^2+y^2>1`
`\sin^{-1}x-\sin^{-1}y=\sin^{-1}{x\sqrt(1-y^2)-y\sqrt(1-x^2)}`
`\cos^{-1}x+\cos^{-1}y=\cos^{-1}{xy-\sqrt((1-y^2)(1-y^2))}` when `x+y>=0`
`\cos^{-1}x-\cos^{-1}y=\cos^{-1}{xy+\sqrt((1-y^2)(1-y^2))}`
Example and Assessment
1. If `\sin^{-}x+\sin^{-}\frac(\pi)(2)`, then prove that, (i) `x^2+y^2=1` and (ii) `x\sqrt(1-y^2)+y\sqrt(1-x^2)`.
Try yourself
7.9.5. Formula for `2\sin^{-1}x`, `2\cos^{-1}x`, `2\tan^{-1}x`, `3\sin^{-1}x`, `3\cos^{-1}x` and `3\tan^{-1}x`:
Formula
(1) `2\sin^{-1}x=sin^{-1}(2x\sqrt(1-x^2))`
Proof:
(1) Let `\sin^{-1}x=y\Rightarrow x=\siny-----(i)`
We know that,
`\sin2y=2\siny\cosy=2siny\sqrt(1-\sin^2y)=2x\sqrt(1-x^2)`
`\Rightarrow \sin2y=2x\sqrt(1-x^2)`
`\Rightarrow 2y=\sin^{-1}(2x\sqrt(1-x^2))`
`\Rightarrow 2\sin^{-1}=\sin^{-1}(2x\sqrt(1-x^2))`
Formula (2) `2\cos^{-1}x=cos^{-1}(2x^2-1)`
Proof:
Let `\cos^{-1}x=y\Rightarrow x=\cosy-----(i)`
We know that,
`\cos2y=2\cos^2y-1=2x^2-1`
`\Rightarrow \cos2y=2x^2-1`
`\Rightarrow 2y=\cos^{-1}(2x^2-1)`
`\Rightarrow 2\cos^{-1}=\cos^{-1}(2x^2-1)`
Formula (3) `2\tan^{-1}x=tan^{-1}(\frac(2x)\(1-x^2))`
Proof:
Let `\tan^{-1}x=y\Rightarrow x=\tany-----(i)`
We know that,
`\tan2y=\frac(2\tany)\(1-\tany)`
`\Rightarrow 2y=\tan^{-1}(\frac(2\tany)\(1-\tan^2y))`
`\Rightarrow 2\tan^{-1}x=\tan^{-1}(\frac(2x)\(1-x^2))`
Formula (4) `3\sin^{-1}x=sin^{-1}(3x-4x^3)`
Proof:
Let `\sin^{-1}x=y\Rightarrow x=\siny-----(i)`
We know that,
`\sin3y=3\siny-4sin^3y`
`\Rightarrow 2y=\tan^{-1}(\frac(2\tany)\(1-\tan^2y))`
`\Rightarrow 2\tan^{-1}x=\tan^{-1}(\frac(2x)\(1-x^2))`
Example
(1) Prove that, `\sin(3\sin^{-1}x)=3x-4x^3`.
Solution:
`\sin(3\sin^{-1}x)`
`\sin(sin^{-1}(3x-4x^3))`
`=3x-4x^3`. Proved
Formula (5) `3\cos^{-1}x=cos^{-1}(4x^3-3x)`
Proof:
Formula (6) `3\tan^{-1}x=tan^{-1}(\frac(3x-x^3)\(1-3x^2))`
Proof:
Example and Assessment
(1) Prove that, `\cos^{-1}x=2\sin^{-1}\sqrt(\frac(1-x)(2))=2\cos^{-1}\sqrt(\frac(1+x)(2))`.
RUET 2004-05
Solution:
Let `\cos^{-1}x=\theta\Rightarrowx=\cos\theta`.
Now,
`\sin(\frac(\theta)(2))=\sqrt(\frac(2sin^2\frac(\theta)(2))(2))`
`\Rightarrow\sin(\frac(\theta)(2))=\sqrt(\frac(1-\cos\theta)(2))`
`\Rightarrow\sin(\frac(\theta)(2))=\sqrt(\frac(1-x)(2))`
`\Rightarrow\frac(\theta)(2)=\sin^{-1}{\sqrt(\frac(1-\cos\theta)(2))}`
`\Rightarrow\cos^{-1}x=2\sin^{-1}\sqrt(\frac(1-x)(2))`
Again,
`\cos(\frac(\theta)(2))=\sqrt(\frac(2cos^2\frac(\theta)(2))(2))`
`\Rightarrow\cos(\frac(\theta)(2))=\sqrt(\frac(1+\cos\theta)(2))`
`\Rightarrow\cos(\frac(\theta)(2))=\sqrt(\frac(1+x)(2))`
`\Rightarrow\frac(\theta)(2)=\cos^{-1}{\sqrt(\frac(1+\cos\theta)(2))}`
`\Rightarrow\cos^{-1}x=2\cos^{-1}\sqrt(\frac(1+x)(2))`
So, `\cos^{-1}x=2\sin^{-1}\sqrt(\frac(1-x)(2))=2\cos^{-1}\sqrt(\frac(1+x)(2))`.
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