badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

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

Please check your email for login details.
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
Menu
Grade: 12th pass

                        

solve the following cos 2 105/2-sin 2 45/2 sin24+cos6 cos 2 108+ cos 2 144

2 years ago

Answers : (2)

Arun
25236 Points
							
Dear Dinesh
 
Please ask only one question in one thread.
 
LHS=sin24°+cos6° =sin(90-66)°+cos6°. [We can write sin(90-66)°as cos(66)°.] We get LHS= cos66°+cos6° now just apply formula of cosC+cosD=2cos[(C+D)\2].cos[(C-D)\2] above.. We get cos66°+cos6°= = 2cos[(66+6)\2]×cos[(66-6)\2] = 2cos36°×cos30°= 2×(√5+1)\4×√3\2 = (√5+1)√3\4.
 
 
Regards
Arun
2 years ago
Subham Kumar
19 Points
							
sin24°+cos6°
cos66°+cos6°
2cos30°cos36°
2(√3/2)((√5+1)/4)=√3(√5+1)/4
 
cos^2(105/2)-sin^2(45/2)
sin^2(75/2)--sin^2(45/2)
sin(A+B)sin(A-B)
sin60°sin15°
√3(√3+1)/(4√2)
 
cos^2(108)+cos^2(144)
cos^2(180-72)+cos^2(180-36)
cos^2(72)+cos^2(36)
sin^2(18)+cos^2(36).
((√5-1)/4)^2+((√5+1)/4)^2
3/4
 
 
2 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details