For the equation sin 3θ + cos 3θ = 1 - sin2θ then? Please note that the question has multiple answers, i.e. one or more options may be correct.

A: tanθ = 1 is possible
B: cosθ = 0 is possible
C: tanθ/2 = -1 is possible
D: cosθ/2 = 0 is possible

1 Answer
P dilip_k
Aug 8, 2017

Given equation

#sin3theta+cos3theta=1-sin2theta#

#=>3sintheta-4sin^3theta+4cos^3theta-3costheta-(1-sin2theta)=0#

#=>3(sintheta-costheta)-4(sin^3theta-cos^3theta)-(sin^2theta+cos^2theta-2sinthetacostheta)=0#

#=>3(sintheta-costheta)-4(sintheta-costheta)(sin^2theta+cos^2theta+sinthetacostheta)-(sin^2theta+cos^2theta-2sinthetacostheta)=0#

#=>3(sintheta-costheta)-4(sintheta-costheta)(1+sinthetacostheta)-(sintheta-costheta)^2=0#

#=>(sintheta-costheta)(3-4-4sinthetacostheta-sintheta+costheta)=0#

#=>(sintheta-costheta)(costheta-1-4sinthetacostheta-sintheta)=0#

So

#sintheta-costheta=0#

#=>sintheta=costheta#

#=>tantheta=1->"option A possible"#

Checking opotion B, #costheta=0# or #theta =90# for 2nd equation

#costheta-1-4sinthetacostheta-sintheta#

#=cos90-1-4*sin90*cos90-sin90#

#=0-1-0-1=-2#

So option B is not possible

Checking opotion C #tan(theta/2)=-1# or #theta/2 =-45# or #theta=-90# for 2nd equation

#costheta-1-4sinthetacostheta-sintheta#

#=cos(-90)-1-4*sin(-90)*cos(-90)-sin(-90)#

#=0-1-0+1=0#

So option C is possible

Checking opotion D, #cos(theta/2)=0# or #theta =180^@# for 2nd equation

#costheta-1-4sinthetacostheta-sintheta#

#=cos180-1-4*sin180*cos180-sin180#

#=-1-1-0-0=-2#

So option D is not possible

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