Rewrite with only sin x and cos x. sin 3x
sin (3x) like this ?
yes i have to rewrite it with only sin(x) and cos(x)
\[\large\rm \sin(\color{red}{3x}) = \sin(\color{red}{2x+x})\]
Simplify the numerator with the sine expression by taking A = B = x and simplify the denominator using A = x and B = 2x. The resulting denominator will have sin(2x) and cos(2x) which you can write in terms of sin(x) and cos(x) using the expressions above. Finally if you use the fact that sin2(x) + cos2(x) = 1 you can write sin(2x)/sin(3x) as a function of cos(x) alone.
you can separate 3x as 2x+x now use the identity \[\large\rm \sin(a+b)= \sin a * \cos b + \cos a *\sin b\]
sin(x+2x)=sinxcos2x+sinsxcosx
hmm first term is correct and 2nd one u forgot something there :=))
Yeah
oh yeah sorry i meant to put a 2 after sin
yes right \[\large\rm \sin(x+2x)= \sin (x)*\cos(2x) + \cos (x) *\sin(2x)\]
what is the next step?
ummm
hmm wait a sec plz let me do it first :3p-_;
alright now we have to use Double angle formula \[\rm \sin(2a)= 2 \sin(a)\cos(a)\]\[\large\rm \cos(2a)= 2\cos^2(a)-1\]
hey
\[\large\rm \sin (x)* \color{Red}{\cos(2x)} + \cos (x) * \color{green}{\sin(2x)}\]
stop making new accounts -.-
\[\rm \sin(2a)=\color{green}{ 2 \sin(a)\cos(a)}\]\[\large\rm \cos(2a)=\color{red}{ 2\cos^2(a)-1}\] substitute cos(2x) and sin(2x) for these identities \[\large\rm \sin (x)* \color{Red}{2cos(x)^2-1} + \cos (x) * \color{green}{2sin(x)cos(x)}\] now simplify
I think i know this
?
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