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What is the value of Cos(2Cos -1 x+Sin -1 x) at x=1/5?

```
9 years ago

```							cos-1x + sin-1x = pi/2   ............1                   (identity)

cos(2cos-1x+sin-1x) = cos[(cos-1x+sin-1x) + cos-1x]
=cos[pi/2 + cos-1x]                             (by using eq 1)

(    we know , cos[pi/2+A] = -sinA   )

=-sin[cos-1x]
=-sin[sin-1(1-x2)1/2]                              (cos-1x = sin-1(1-x2)1/2  )
=-(1-x2)1/2
at x = 1/5
=-(2root6)/5

```
9 years ago
```							It’s just a play of some equations :  cos-1x + sin-1x = pi/2   ............1                   (identity) cos(2cos-1x+sin-1x) = cos[(cos-1x+sin-1x) + cos-1x]=cos[pi/2 + cos-1x]                             (by using eq 1) (    we know , cos[pi/2+A] = -sinA   ) =-sin[cos-1x]=-sin[sin-1(1-x2)1/2]                              (cos-1x = sin-1(1-x2)1/2  )=-(1-x2)1/2 at x = 1/5=-(2root6)/5
```
4 months ago
```							Feel free to ask querry We will happy to help you.Good Luck Cheers ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
```
3 months ago 605 Points
```							Dear student,Please find the attached answer to your problem.  cos[2cos-1x + sin-1x] = cos[(cos-1x + sin-1x) + cos-1x] = cos[π/2 + cos-1x]= – sin[cos-1x] = – (1 – x2)½           (Identity) = – (1 – [1/5]2)½   = 2√6/5  Hope it helps.Thanks and regards,Kushagra
```
3 months ago
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