Question

Prove 2SinAsinB =cos(A-B)-cos(A+B) FROM LHS < 13 years ago

Answer 1

what u mean by LHS and RHS ?
u can expand cos (A-B) and cos (a+b ) and sub their common terms . u will find the answer

Answer 2

Prove from Left hand side saying what formula you have used? I know right hand side is addition formula.Prove starting from left side of eqn

Answer 3

Yes we can do that just do opposite steps...

Answer 4

Show what formula you have used and how you manipulate the eqn from LHS?

Answer 5

Can you write \(2sinAsinB = sinAsinB + sinAsinB \) ???

Answer 6

dear
sinAsinB = 1/2(cos ( A-B)-cos(A+B))
just this

Answer 7

LHS = \(sinAsinB + sinAsinB\)

Or:

\[\large = cosAcosB +sinAsinB - (cosAcosB - sinAsinB)\]

\[\large = \cos(A-B) -\cos(A+B) \implies RHS\]

Answer 8

tnx

Answer 9

Can you guys do it step by step stating the formula you have used like "sin addition formula" instead of just showing me the working derived from right side?

Answer 10

I am not getting what you said @vadevalor

Answer 11

Can you state what formula you used to explain what you did for each step?

Answer 12

Firstly I write the LHS as:

\[\large sinAsinB + sinAsinB\]

Got??

Answer 13

Then add or subtract by \(cosAcosB\):

\[\large cosAcosB + sinAsinB - cosAcosB + sinAsinB\]

Group them now:

\[\large (cosAcosB+sinAsinB)−(cosAcosB-sinAsinB)\]

Now use the formula:

\[\large = \cos(A-B) - \cos(A + B)\]

Answer 14

The formula we will be using in this are:

\[\large \color{green}{\cos(A + B) = cosA cosB - sinAsinB}\]

\[\large \color{green}{\cos(A - B) = cosA cosB + sinAsinB}\]

Answer 15

How do you transit from LHS to 1st step?

Answer 16

2x you can write it as x + x or not???

Answer 17

Okay got it thanks...sorry

Answer 18

Welcome dear...