Question

solve

Answer 1

That attachment is blank.

Answer 2

no itss not blank

Answer 3

All is see is an all white sheet

Answer 4

its not blank ...its size is small

Answer 5

WOW - that is small

Answer 6

given that
A + B + C = PI

(sin 2A + sin 2B + sin 2C) / (sin A + sin B + sin C) = 8 * ( sin A/2 sin B/2 sin C/2)

I figured I'd type it out but I think I'll just leave it at that.

Answer 7

sur was doing all that typing and now he's gone?
Well I figured I'd make a new attachment for you (because I have almost no idea how to solve this).

Answer 8

\[A+B+C=\pi ,A+B=\pi -C,\sin \left( A+B \right)=\sin \left( \pi-C \right)=\sin C\]
\[Again A+B+C=\pi ,C=\pi-\left( A+B \right)\]
\[\cos C=\cos \left\{ \pi-\left( A+B \right) \right\}=-\cos \left( A+B \right)\]
\[\sin 2A+\sin 2B+\sin 2C=\2sin \frac{ 2A+2B }{ 2}\cos \frac{ 2A-2B }{ 2 }+\sin C \cos C\]
\[=2\sin \left( A+B \right)\cos \left( A-B \right)+2\sin C \cos C\]
\[=2\sin C \left\{ \cos \left( A-B \right) -\cos \left( A+B \right)\right\}\]
\[=2\sin C \sin A \sin B ....(1)\]
\[=2*2\sin \frac{ C }{ 2 }\cos \frac{ C }{ 2 }*2\sin \frac{ A }{2 }\cos \frac{ A }{ 2 }*2\sin \frac{ B }{ 2 }\cos \frac{ B }{ 2 }\]
\[\sin A+\sin B+\sin C=2\sin \frac{ A+B }{ 2}\cos \frac{ A-B }{2 }+\sin C ...(2)\]
\[A+B+C=\pi,A+B=\pi -C\]
\[\sin \frac{ A+B }{ 2 }=\sin \left( \frac{ \pi }{ 2 }-\frac{ C }{ 2 } \right)=\cos \frac{ C }{ 2 }\]

Answer 9

You can solve further

Answer 10

can u plse explain last 4th step

Answer 11

\[\sin C+\sin D=2\sin \frac{ C+D }{ 2 }\cos \frac{ C-D }{ 2}\]

Answer 12

this step
=2∗2sinC2cosC2∗2sinA2cosA2∗2sinB2cosB2

Answer 13

\[\sin 2A=2\sin A \cos A\]

Answer 14

okay plse see the attachment
that step , i need to know

Answer 15

\[\sin A=2\sin \frac{ A }{ 2 }\cos \frac{ A }{ 2 }\]
Similarly sinB and sin C

Answer 16

okay

Answer 17

I have some problem with my computer,posted three times but failed.

Answer 18

its okay
i m solving it
and post it also, u plse just check it

Answer 19

still if you wish i will complete the question.
o.k i will check

Answer 20

but one pblm , how to solve (ii)

Answer 21

o.k i will solve but be patient.

Answer 22

k

Answer 23

from (2)
\[=2\cos \frac{ C }{2 }\cos \frac{ A-B }{ 2 }+2\sin \frac{ C }{ 2 }\cos \frac{ C }{ 2 }\]
\[=2\cos \frac{ C }{ 2}\left\{ \cos \frac{ A-B }{ 2 } +\sin \frac{ C }{ 2 }\right\}\]

\[[\sin \frac{ C }{2}=\sin \left( \frac{ \pi }{ 2 }-\frac{ A+B }{2} \right)=\cos \frac{ A+B }{ 2 }]\]

Answer 24

\[=2\cos \frac{ C }{ 2 }\left\{ \cos \frac{ A-B }{ 2 } +\cos \frac{ A+B }{ 2 }\right\}\]

Answer 25

\[=2\cos \frac{ C }{ 2 }*2\cos \frac{ \frac{ A-B }{ 2}+\frac{ A+B }{2 } }{ 2 }\cos \frac{ \frac{ A-B }{ 2 }-\frac{ A+B }{2} }{ 2 } \]

Answer 26

\[=2\cos \frac{ C }{2 }\cos \frac{ A }{ 2 }\cos \frac{ -B }{ 2 }\]

Answer 27

\[=2\cos \frac{ C }{ 2 }\cos \frac{ A }{ 2}\cos \frac{ -B }{ 2 }\]

Answer 28

\[\cos \left( -A \right)=\cos A\]
Now you can complete

Answer 29

yup i got the answer