sin9cos9/2sin48sin12=?
\[\large \frac{\sin(9)\cos(9)}{2\sin(48)\sin(12)}\]
Yes it is..
Numerator can be made as: \[\frac{\sin(18)}{2}\]
that is right, for the top, @waterineyes, its using the identity sin(2u) = 2sinucosu.
In numerator should we use: \[2sinAsinB = \cos(A-B) - \cos(A+B)\] ??
Sorry denominator..
that is right, @waterineyes. :))
Need not to tag me violy I am here only..
my apologies.
I am just informing you and not demanding sorry from you.. Chill.. Now we will help jainik together..
yes, just plug-in the corresponding value for A and B on the identity waterineyes gave you :))
Since I am not seeing any term cancelling or simplifying in better way so you have to use this direct results there: \[\large \color{green} {\sin{18^{\circ}} = \frac{\sqrt{5} - 1}{4}}\] \[\large \color{green} {\cos{36^{\circ}} = \frac{\sqrt{5} + 1}{4}}\]
You can solve the denominator as: \[2\sin(48)\sin(12) = \cos(48-12) - \cos(48 + 12) \implies \cos(36) - \cos(60)\] \[\implies \cos(36) - \frac{\sqrt{3}}{2}\]
Sorry : \[\implies \cos(36) - \frac{1}{2}\]
In numerator you will get: \[\frac{\sqrt{5} - 1}{8}\] In the denominator you will get: \[\frac{\sqrt{5} - 1}{4}\] This is what I can do maximum for you @jainilk .. You have to just divide them and you will have your answer..
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