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Mathematics 64 Online
OpenStudy (anonymous):

Prove the identity: sin5x+sin3x/4sinxcos^3(x)-4sin^3(x)cosx = 2cosx

OpenStudy (perl):

you can do a wolfram proof, type it in wolfram and it will say true or false

OpenStudy (anonymous):

@perl It's true, but I need a step-by-step proving. :D

OpenStudy (perl):

wolfram might offer that, dont know

OpenStudy (anonymous):

do you mean: \[\frac{ \sin (5x) + \sin (3x) }{ 4 \sin (x) \cos^3(x) - 4 \sin^3(x)\cos(x) }=2\cos(x)\] or \[\frac{ \sin (5x) + \sin (3x) }{ 4 \sin (x) \cos^3(x)} - 4 \sin^3(x)\cos(x) =2\cos(x)\]

OpenStudy (anonymous):

@DHeart The first one :)

OpenStudy (sidsiddhartha):

\[\frac{ \sin5x+\sin3x }{ 4sinx~\cos^3x- 4\sin^3x~cosx}=~what~ever\] thats the question right?

OpenStudy (anonymous):

Yup :)

OpenStudy (sidsiddhartha):

if thats the question then use these \[4\cos^3x=3cosx+\cos3x\\and\\4\sin^3x=3sinx-\sin3x\]

OpenStudy (sidsiddhartha):

just replace them \[\frac{ \sin5x+\sin3x }{ sinx(\cos3x+3cosx)-cosx(3sinx-\sin3x) }\\=\frac{ \sin5x+\sin3x }{sinxcos3x+cosxsin3x }\\=\frac{ 2\sin \frac{ 5+3 }{ 2 }\cos \frac{ 5-3 }{ 2 }}{ \sin(3+1)x }\\=\frac{ 2\sin4x.cosx }{ \sin4x }=2cosx\]

OpenStudy (sidsiddhartha):

ok buddy? any problem?

OpenStudy (anonymous):

i don't understand how you simplified the denominator.

OpenStudy (anonymous):

Oh wait! Nvm, I get it now :) Thanks!

OpenStudy (sidsiddhartha):

i have used three basic trigonometric formulaes-----\[(1)\cos3x=4\cos^3x-3cosx............(A)\\(2)\sin3x=3sinx-4\sin^3x............(B)\\(3)sinC+sinD=2\sin \frac{ C+D }{ 2 }.\cos \frac{ C-D }{ 2}\] i'm sure u'll get these basic formulas in any text books ok?

OpenStudy (anonymous):

@sidsiddhartha Yeah, I understand it now :) Thank you! :D

OpenStudy (sidsiddhartha):

yW!! happy to help you :)

OpenStudy (anonymous):

I've started to simplify the denominator: \[4\sin(x)(\frac{1}{4}(\cos(3x)+3\cos(x)))-4*\sin(x)(\frac{1}{4}*(3*\sin(x)-\sin(3x)))\] \[\sin(x)\cos(3x)+3\sin(x)\cos(x)-3\sin(x)\cos(x)+\sin(3x)\cos(x)\] \[\sin(x)\cos(3x)+\sin(3x)\cos(x)\] \[\sin(4x)+\frac{1}{2}sen(-2x)+\frac{1}{2}\sin(2x)\] \[\sin(4x)-\frac{1}{2}sen(2x)+\frac{1}{2}\sin(2x)\] \[\sin(4x)\]

OpenStudy (anonymous):

in the end, you equation would look like this: \[\frac{\sin(5x)+\sin(3x)}{\sin(4x)}\]

OpenStudy (anonymous):

I suppose! ;) Anyone, feel free to correct me.

OpenStudy (sidsiddhartha):

nope ure correct and a little more simplification will produce the exact result :)

OpenStudy (anonymous):

thanks, sidsiddharth!

OpenStudy (sidsiddhartha):

yw!!:)

OpenStudy (anonymous):

I'm sorry for double replying, but I didn't saw your reply. I was distracted typing the formulas

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