Please help :/ Express 2sin3theta sin5theta as an algebraic sum.
Use: \[\large \color{green}{2\sin(\frac{C+D}{2})\sin(\frac{D-C}{2}) = \cos(C) - \cos(D)}\]
Here: \[\frac{C+D}{2} = 3\] \[\frac{D-C}{2} = 5\] Solve them to find C and D here..
Ok, just a sec
\[C + D = 6\] \[-C + D =1 0\] Add them now you will get D..
16
Left hand side it will be: 2D = 16 Now find D from here..
8?
Yes.. Now you have D so find C now by plugging in D value in any one equation...
its looks like that double angle formula can be used.
\[C + D = 16 \implies C = 6 - D \implies C = 6 - 8 \implies C = -2\]
@muhammad9t5 which double angle will you use here ??
And do you know what question wants?/ It wants Algebraic Sum.. Meaning Plus or Minus separated terms..
So you will have now C = -2 and D = 8 \[\large \cos(-2) - \cos(8) \implies \color{green}{\cos(2) - \cos(8)}\] \[\cos(- \theta) = \cos(\theta)\]
This will help. Here are the options: a. -cos8theta+cos2theta b. cos8theta-cos2theta c. -cos2theta-cos2theta d. cos8theta+cos2theta
How about first one ??
Yes, that's what I was thinking. @muhammad9t5 can you confirm?
\[\large \cos(-2\theta) - \cos(8 \theta) \implies \color{green}{\cos(2\theta) - \cos(8\theta)}\]
I confirm !
Yes, @waterineyes, perfect. hanks @Neemo. Thanks to everyyone who helped!
Welcome dear.. Interact when you are asked something @IloveCharlie
yw !
@waterineyes sorry it was sum to product identity..
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