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Grade: 12th pass
        
solve the following
  1. cos2105/2-sin245/2
  2. sin24+cos6
  3. cos2108+ cos2144
2 years ago

Answers : (2)

Arun
24740 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
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