As thetain"Quadrant"IIIθ∈QuadrantIII and sintheta=-15/17=-0.882353sinθ=−1517=−0.882353
costheta=-sqrt(1-(15/17)^2)=-sqrt(1-225/289)=-sqrt(64/289)=-8/17=-0.470588cosθ=−√1−(1517)2=−√1−225289=−√64289=−817=−0.470588
Then sin3theta=3sinx-4sin^3x=3(-0.882353)-4(-0.882353)^3sin3θ=3sinx−4sin3x=3(−0.882353)−4(−0.882353)3
= -2.647059+2.747813=0.100753−2.647059+2.747813=0.100753
and cos3theta=4cos^3x-3cosx=4(-0.470588)^3-3(-0.470588)cos3θ=4cos3x−3cosx=4(−0.470588)3−3(−0.470588)
= -0.416853+1.411764=0.994911−0.416853+1.411764=0.994911
Note that as sin3thetasin3θ and cos3thetacos3θ are positive, 3theta3θ is in Q1Q1,
Now sin6theta=2sin3thetacos3theta=2xx0.100753xx0.994911sin6θ=2sin3θcos3θ=2×0.100753×0.994911
= 0.2004810.200481
and cos6theta=(0.994911)^2-(0.200481)^2=0.949655cos6θ=(0.994911)2−(0.200481)2=0.949655
and 6theta6θ is in Q2Q2
Continuing this way sina2theta=2sin6thetacos6theta=2xx0.200481xx0.949655sina2θ=2sin6θcos6θ=2×0.200481×0.949655
= 0.3807760.380776
and cos12theta=(0.949655)^2-(0.200481)^2=0.861652cos12θ=(0.949655)2−(0.200481)2=0.861652
Hence 12theta12θ is in Q1Q1.