#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# What is the value of Cos(2Cos-1x+Sin-1x) at x=1/5? 10 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

one year 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 Kushagra Madhukar
one year ago
Dear student,

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