solve the following

1. cos²105/2-sin²45/2
2. sin24+cos6
3. cos²108+ cos²144

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.

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

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