How do you prove #sin(3theta)/sintheta-cos(3theta)/costheta=2#?

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Answer 1 · 2016-10-07T11:25:13

Starting with the LHS subtract using the rule for subtraction of fractions, and then simplify using Trig identities.

#(sin(3theta)cos(theta)-sin(theta)cos(3theta))/(sin(theta)cos(theta)#

#=(sin(3theta-theta))/(1/2xx2sinthetacostheta#

=#2sin(2theta)/sin(2theta)#

#=2#

with problems like these recognition of the basic Trig. identities is vital.

Answer 2 · 2016-10-07T11:49:22

See below.

According to de Moivre's identity

#sintheta = (e^(itheta)-e^(-itheta))/(2i)# and
#costheta = (e^(itheta)+e^(-itheta))/(2)#

so following

#(a^3-b^3)/(a-b)=a^2+a b + b^2#

# (e^(i3theta)-e^(-i3theta))/ (e^(itheta)-e^(-itheta)) = e^(i2theta)+1+e^(-i2theta)#

and

# (e^(i3theta)+e^(-i3theta))/ (e^(itheta)+e^(-itheta)) = e^(i2theta)-1+e^(-i2theta)#

so

# (e^(i3theta)-e^(-i3theta))/ (e^(itheta)-e^(-itheta)) - (e^(i3theta)+e^(-i3theta))/ (e^(itheta)+e^(-itheta)) =2#