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20 =--, 29 s in the 14. Given that cos =, 0

Solution

14) We have given cos=sqrt(2)/2,0<<pi/2 and cos=-20/29, is in second quadrant

sin=sqrt(1-cos²)=sqrt(1-2/4)=sqrt((4-2)/4)=sqrt(2/4)=sqrt(2)/2

sin=sqrt(2)/2

sin=sqrt(1-cos²)=sqrt(1-400/841)=sqrt((841-400)/841)=sqrt(441/841)=21/29

sin=21/29

a) cos(+)=coscos-sinsin =(sqrt(2)/2)*(-20/29)-(sqrt(2)/2)*(21/29)=(-20*sqrt(2)-21*sqrt(2))/58=(-41*sqrt(2))/58

cos(+)=(-41sqrt(2))/58

b) sin2=2sincos=2*(sqrt(2)/2)*(sqrt(2)/2)=2/2=1

sin2=1

15) We have given 2sinxcosx-cosx=0,where 0<=x<=2pi

cosx(2sinx-1)=0

cosx=0 or (2sinx-1)=0 implies sinx=1/2

x=pi/2,3pi/2 when cosx=0 or x=pi/6,5pi/6 when sinx=1/2

x=pi/2,3pi/2,pi/6,5pi/6