Question

SinA=1/2 ho to tan3A=?

Answer 1

`tan 3A = tan 90^circ` which is undefined.

Explanation:

I now get ill when I see `sin A=1/2 .` Can't question writers come up with another triangle?

I do know it means `A=30^circ` or `A=150^circ`, not to mention their coterminal brethren.

So `tan 3A = tan 3(30^circ) or tan(3(150^circ))`

`tan 3A = tan 90^circ or tan 450^circ =tan90^circ`

So either way, `tan 3A = tan 90^circ` which sadly is undefined.

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There's another way to solve these. Let's do it in general.

Given `s = \sin A` find all possible values of `tan(3A).`

The sine is shared by supplementary angles, and there's no reason their triples will have the same slope. So we expect two values.

Those supplementary angles have opposite cosines, indicated by the `pm`:

`c = cos A = \pm \sqrt{1 - sin^2 A} = \pm \sqrt{1-s^2}`

We can use the usual triple angle formula for sine directly, but let's generate a customized one that mixes cosine and sine to use here for the cosine:

`cos(3x) = cos(2x + x) =cos(2x) cos x - sin (2x) sin x`

`= cos x( 1 - 2 sin ^2 x ) - 2 sin ^2 x cos x`

`cos 3x = cos x(1 - 4 sin ^2 x)`

We don't see that form every day, but it's useful here:

`tan 3x = {sin 3x}/{cos 3x} = {3 sin x - 4 sin^3 x} / {cos x( 1 - 4 sin^2 x) } = {sin x (3 - 4 sin^2 x)} / {cos x( 1 - 4 sin^2 x) }`

# tan 3A = {s(3 - 4 s^2)} / {c( 1 - 4 s^2) } = \pm {s(3 - 4 s^2)} / { ( 1 - 4 s^2)sqrt{1-s^2} } #

We see `s=1/2` as asked gives `tan 3A` undefined.

Answer 2

`tan3A` is undefined

Explanation:

For simplicity,we take `0^circ <= A <= 90^circ`

`:.sinA=1/2=>A=30^circ=>3A=90^circ`

We know that,

`tan3A=tan90^circ is` undefined

Also we note that,

`SinA=1/2=>cosA=sqrt3/2,`where, `0^circ <= A <= 90^circ`

`:.tan3A=(sin3A)/(cos3A)`

`=(3sinA-4sin^3A)/(4cos^3A-3cosA)`

`=(3(1/2)-4(1/2)^3)/(4(sqrt3/2)^3-3(sqrt3/2))`

`=(3/2-1/2)/((3sqrt3)/2-(3sqrt3)/2)`

`=>tan3A=1/0.=>tan3A` is undefined