ALEKS Question Algebra 2

Question

Find sin2x, cos2x, and tan2x if tanx=1/3 and x terminates in quadrant 3?

Answer 1

sin2x=2sinxcosx=2(-1/sqr10)(-3/sqr10)=6/10=3/5=0.6

cos2x=cos^2(x)- sin^2(x)=9/10- 1/10=8/10=4/5 =0.8

tan2x=sin2x/cos2x= .6/.8= 3/4=0.75

tanx=1/3

sinx=-1/sqr10

cosx=-3/sqr10

tanx=1/3

x=tan^-1(1/3)+180=about 18.435+180=about 198.87 degrees

2x=about 397.74

sin2x=sin397.74=sin37.74=about 0.6

x is in quadrant 3, but 2x is in quadrant 1, in case you wonder why sinx and cosx <0 but sin2x and cos2x >0