Question

if sin A is 3/5, find values of cos 3A & tan 3A

Answer 1

sinA=3/5.
we have ,
sin^2 A+cos^2 A=1
=(3/5)^2+cos^2 A=1
=9/25+cos^2 A=1
=cos^2 A=1-9/25
=cos^2 A=16/25
=cosA=4/5.
tan A=sin A/cos A
=3/5÷4/5
=3/4
cos 3A=4cos^3 A- 3cos A
=4×(4/5)^3 - 3(4/5)
=4×64/125 - 12/5
=256/125 - 300/125
=44/125
=0.352
tan 3A=(3 tan A-tan^3 A)/(1-3tan^2 A)
=(3×3/4-(3/4)^3)/(1-3 (3/4)^2)
=(9/4-27/64)/(1-27/16)
=(144-27/64)/(16-27/16)
=117/64×16/-11
=117/4×-1/11
=(-117/44)
=2.65
hope it helps

Answer 2

I think its the right solution