Let thetaθ be an angle where: "1)" theta in "Quadrant III"1)θQuadrant III and "2)" sin( theta ) = - 15/172)sin(θ)=1517. What Quadrant does 12theta12θ belong to ? No Calculators !!

1 Answer
Feb 16, 2018

12theta12θ is in Q1Q1.

Explanation:

As thetain"Quadrant"IIIθQuadrantIII 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=1225289=64289=817=0.470588

Then sin3theta=3sinx-4sin^3x=3(-0.882353)-4(-0.882353)^3sin3θ=3sinx4sin3x=3(0.882353)4(0.882353)3

= -2.647059+2.747813=0.1007532.647059+2.747813=0.100753

and cos3theta=4cos^3x-3cosx=4(-0.470588)^3-3(-0.470588)cos3θ=4cos3x3cosx=4(0.470588)33(0.470588)

= -0.416853+1.411764=0.9949110.416853+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.