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Grade: 12

                        

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

9 years ago

Answers : (4)

vikas askiitian expert
509 Points
							

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
Krish Gupta
askIITians Faculty
76 Points
							
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
Vikas TU
12284 Points
							
Feel free to ask querry 
We will happy to help you.
Good Luck 
Cheers ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
 
3 months ago
Kushagra Madhukar
askIITians Faculty
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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