Question

If sin A = `3/5`, find the values of cos 3A and tan 3A.

Answer 1

Given sin A = `3/5`

cos A = `"Adjacent side"/"Hypotenuse" = 4/5`

and tan A = `Opposite side/"Adjacent side" = 3/4`

We know that cos 3A = 4 cos³ A – 3 cos A

`= 4(4/5)^3 - 3(4/5)`

`= (4/5)[4 xx (4/5)^2 - 3] = 4/5[4 xx 16/25 - 3]`

`= 4/5[(64 - 3 xx 25)/(25)]`

`= 4/5((64 - 75)/25)`

`= 4/5 xx (-11)/25 = (-44)/125`

tan 3A = `(3 tan "A" - 4tan^3 "A")/(1 - 3 tan^2"A")`

`= (3(3/4) - (3/4)^3)/(1 - 3(3/4)^2)`

`= (3/4[3 - (3/4)^2])/(1 - 3 xx 9/16)`

`= (3/4 [(48 - 9)/16])/((16 - 27)/16)`

`= 3/4[39/16 xx 16/(-11)]`

`= (- 117)/44`