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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.
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 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Askiitians Experts Badiuddin
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
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Askiitians Experts Badiuddin
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