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`        by solving the equation sin2A=cos3A. find the value of sin18 .show your ans. in the form (sq root a)+b/c where a,c,b are natural nos.`
8 years ago

147 Points
```										Dear yashika
given
sin2A=cos3A
2 sinA cosA=4cos3A-3cosA
2 sin A=4 cos2A-3
2 sin A =4(1-sin2A)-3
now solve for sinA
sin A=(-1-√5)/4  or (-1+√5)/4
Put A=18
and for 0<A<180 so value of sin will be positive
sin18=(-1+√5)/4

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8 years ago
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