Question

sin9cos9/2sin48sin12=?

Answer 1

\[\large \frac{\sin(9)\cos(9)}{2\sin(48)\sin(12)}\]

Answer 2

Yes it is..

Answer 3

Numerator can be made as:

\[\frac{\sin(18)}{2}\]

Answer 4

that is right, for the top, @waterineyes, its using the identity sin(2u) = 2sinucosu.

Answer 5

In numerator should we use:

\[2sinAsinB = \cos(A-B) - \cos(A+B)\]

??

Answer 6

Sorry denominator..

Answer 7

that is right, @waterineyes. :))

Answer 8

Need not to tag me violy I am here only..

Answer 9

my apologies.

Answer 10

I am just informing you and not demanding sorry from you..

Chill..

Now we will help jainik together..

Answer 11

yes, just plug-in the corresponding value for A and B on the identity waterineyes gave you :))

Answer 12

Since I am not seeing any term cancelling or simplifying in better way so you have to use this direct results there:

\[\large \color{green} {\sin{18^{\circ}} = \frac{\sqrt{5} - 1}{4}}\]

\[\large \color{green} {\cos{36^{\circ}} = \frac{\sqrt{5} + 1}{4}}\]

Answer 13

You can solve the denominator as:

\[2\sin(48)\sin(12) = \cos(48-12) - \cos(48 + 12) \implies \cos(36) - \cos(60)\]

\[\implies \cos(36) - \frac{\sqrt{3}}{2}\]

Answer 14

Sorry :

\[\implies \cos(36) - \frac{1}{2}\]

Answer 15

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..