Question

If `tan A = (A + sqrt(1-A))/(A + sqrt(1+A))`, find the value of `sin 4A`?

Answer 1

Please see below.

Explanation:

`sin2A=2sinAcosA=(2sinAcosA)/(sin^2A+cos^2A)`

= `((2sinAcosA)/cos^2A)/(sin^2A/cos^2A+1)=(2tanA)/(1+tan^2A)`

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

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

and hence `sin4A=2sin2Acos2A=(4tanA(1-tan^2A))/(1+tan^2A)^2`

I am treating as `tanA=(a+sqrt(1-a))/(a+sqrt(1+a))`, to distinguish it from angle `A`.

then `tan^2A=(a+sqrt(1-a))^2/(a+sqrt(1+a))^2`

or `(a+1-a+2asqrt(1-a))/(a+1+a+2asqrt(1+a))`

i.e. `tan^2A=(1+2asqrt(1-a))/(1+2a+2asqrt(1+a))`

and `1+tan^2A=(2+2a+2a(sqrt(1-a)+sqrt(1+a)))/(1+2a+2asqrt(1+a))`

and using componendo dividendo

`(1-tan^2A)/(1+tan^2A)=((1+2a+2asqrt(1+a))-(1+2asqrt(1-a)))/((1+2a+2asqrt(1+a))+(1+2asqrt(1-a)))`

= `(2a+2a(sqrt(1+a)-sqrt(1-a)))/(2+2a+2a(sqrt(1+a)+sqrt(1-a)))`

and #(2tanA)/(1+tan^2A)=2(a+sqrt(1-a))/(a+sqrt(1+a))xx (1+2a+2asqrt(1+a))/(2+2a+2a(sqrt(1-a)+sqrt(1+a)))#

Hence #sin4A=2(a+sqrt(1-a))/(a+sqrt(1+a))xx (1+2a+2asqrt(1+a))/(2+2a+2a(sqrt(1-a)+sqrt(1+a)))xx(2a+2a(sqrt(1+a)-sqrt(1-a)))/(2+2a+2a(sqrt(1+a)+sqrt(1-a)))#