Question

If sin x=2/9,x, in quadrant i, then find(without findingx): sin(2x), cos(2x), tan(2x)

Answer 1

We have that
sin x=2/9

we know that
sin² x+cos² x=1---------> cos² x=1-sin² x-----> cos² x=1-(2/9)²
cos² x=1-(4/81)------> (81-4)/81-----> 77/81
cos x=√(77/81)

Sin 2x = 2 sin x cos x-------> 2*(2/9)*(√(77/81))----> (4/81)*√77---> 0.43

Sin 2x=0.43
so
cos² 2x=1-sin² 2x---------> 1-[0.43]²-----> 1-[0.1878]----> 0.81
cos² 2x=0.81-----------> cos 2x=0.9

tan 2x=sin 2x/cos 2x--------> tan 2x=0.43/0.9---------> tan 2x=0.48

the answers are
sin 2x=0.43
cos 2x=0.9
tan 2x=0.48