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

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2 years ago

```							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.  RegardsArun
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2 years ago
```							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)^23/4
```
2 years ago
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