Question

Given that `tan2A = 3/4` and that the angle `A` is acute, calculate, without using tables, the values of: (i) `cos2A` (ii) `sinA` (iii) `tanA` (iv) `tan3A` ?

Answer 1

`cos2A=4/5`; `sinA=1/sqrt10`; `tanA=1/3` and `tan3A=13/9`

Explanation:

As `tan2A=3/4`, we have

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

or `8tanA=3-3tan^2A`

or `3tan^2A+8tanA-3=0`

or `(3tanA-1)(tanA+3)=0`

i.e. `tanA=1/3` or `tanA=-3`

but as `A` is acute, we cannot have `tanA=-3`

Hence `tanA=1/3`

and `tan3A=tan(A+2A)=(tanA+tan2A)/(1-tanAtan2A)`

= `(1/3+3/4)/(1-1/3xx3/4)=((4+9)/12)/(1-1/4)=13/12xx4/3=13/9`

`cos2A=(cos^2A-sin^2A)/(cos^2A+sin^2A)=(1-tan^2A)/(1+tan^2A)`

= `(1-1/9)/(1+1/9)=(8/9)/(10/9)=8/10=4/5`

As `2cos^2A-1=cos2A` i.e. `cos^2A=(1+4/5)/2=9/10`

`cosA=3/sqrt10`

and `sinA=cosAxxtanA=3/sqrt10xx1/3=1/sqrt10`