Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (anonymous):

prove the identity 1/4(cos(3(theta)=cos^3(theta)-3/4(cos(theta)

OpenStudy (anonymous):

Sorry, but could you use equation editor to make it easier to read?

OpenStudy (anonymous):

yeh i will try sorry about it

OpenStudy (anonymous):

so easyit is

OpenStudy (anonymous):

just in one step

OpenStudy (anonymous):

simplify the rhs by the identity

OpenStudy (anonymous):

simplify cos 3 theta

OpenStudy (anonymous):

\[\frac{ 1 }{ 4 }\cos 3\theta=\cos^3\theta-\frac{ 3 }{ 4 }\cos \theta\]

OpenStudy (anonymous):

cos3theta=4 cos^3 theta-3cos theta

OpenStudy (anonymous):

if u simplify u will directly get the answer

OpenStudy (anonymous):

could cos(3theta)=cos(2theta)+cos(theta)?

OpenStudy (anonymous):

\[\cos(3x)\ne \cos(2x) +\cos(x)\] You see from what you're trying to prove is 4cos^3 theta - 3 cos(theta)...

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

then what is gonna be cos(3theta)?

OpenStudy (anonymous):

4 cos^3 theta- 3 cos theta

OpenStudy (anonymous):

this is an identity

OpenStudy (anonymous):

ah! i get it now.. but how d oi prove the identity..

OpenStudy (anonymous):

The derivation of this identity involves Euler's formula and complex numbers...I can explain I guess...

OpenStudy (anonymous):

ya

OpenStudy (anonymous):

now all the proofs of the identities are interrelated now u'll b asking for all those a what

OpenStudy (anonymous):

Okay: \[\cos(3x)=\Re (e^{3ix})=\Re ((e^{ix})^3)=\Re ((\cos(x)+i \sin(x))^3)=\] \[=\Re(\cos^3(x)+i (3 \sin(x) \cos^2(x)-\sin^3(x))-3 \sin^2(x) \cos(x)) \] \[=\cos^3(x)-3 \sin^2(x)\cos(x)=\cos^3(x)-3(1-\cos^2(x))\cos(x)=4\cos^3(x)-3\cos(x)\] \[=\cos^3(x)-3(1-\cos^2(x))\cos(x)=4\cos^3(x)-3\cos(x)\]

OpenStudy (anonymous):

If you want the sin(3x) one then take the imaginary part: \[\sin(3x)=3\sin(x)\cos^2(x)-\sin^3(x))=3\sin(x)(1-\sin^2(x))-\sin^3(x)\] \[=3\sin(x)-4\sin^3(x)\]

OpenStudy (anonymous):

\[LHS=\frac{\cos(x+2x)}{4}=\frac{cosxcos2x-sinxsin2x}{4} \] \[=\frac{cosx(\cos^2x-\sin^2x)-sinx(2sinxcosx)}{4} \] \[=\frac{\cos^3x-cosxsin^2x-2\sin^2xcosx}{4} \] \[=\frac{\cos^3x-3\sin^2xcosx}{4} \] \[=\frac{\cos^3-3(1-\cos^2x)cosx}{4} \] \[=\frac{\cos^3x-3cosx+3\cos^3x}{4} \] \[=\frac{4\cos^3x-3cosx}{4} = \cos^3x - \frac{3cosx}{4}\] .: LH=RHS

OpenStudy (anonymous):

i am closing it sam bye i got the answer lol

OpenStudy (anonymous):

wtf my answer good

OpenStudy (anonymous):

i know i got the answer lol did you work it out?

OpenStudy (anonymous):

yeah look at mine

OpenStudy (anonymous):

i did lol you are so pro man haha

OpenStudy (anonymous):

did u get same as mine?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!