Question

Express in surd form.

Answer 1

\(tan(-15)\)

Answer 2

just to double-check my working.

Answer 3

\[\large\rm \tan(-15)=\tan\left(\frac{-30}{2}\right)\]And then I guess we have some type of Half-Angle Formula, yes?
I don't remember the tangent one off the top of my head, I'll have to look it up.

Answer 4

Ok let's use this one maybe,\[\large\rm \tan\frac{x}{2}=\frac{1-\cos x}{\sin x}\]Single term in the denominator makes it easier to simplify.

Answer 5

\(=tan(45-60)\)
\(\frac{1-\sqrt{3}}{1+\sqrt{3}}\times\frac{1-\sqrt{3}}{1-\sqrt{3}}\)
\(=\frac{(1-sqrt{3})^2}{(1+\sqrt{3})(1-sqrt{3})}\)
\(=\frac{1-2\sqrt{3}+2}{1-3}\)
\(=\frac{4-2\sqrt{3}}{-2}\)
\(=\frac{2(2-\sqrt{3}}{-2}\)
\(=\sqrt{3}-2\)

Answer 6

Oh you took that path instead :) Interesting.

Answer 7

Is the working from start can be accepted?

Answer 8

So you used your angle difference formula for tangent, ok,\[\large\rm \tan(x-y)=\frac{\tan x-\tan y}{1+\tan x \tan y}\]

Answer 9

yep

Answer 10

\[\large\rm \tan(45-60)=\frac{\tan 45-\tan 60}{1+\tan 45 \tan 60}\]

Answer 11

\[\large\rm \tan(-15)=\frac{1-\sqrt3}{1+\sqrt3}\]And you did umm some conjugate business.

Answer 12

yep.

Answer 13

You wrote 1+2sqrt3+2 in the numerator,
I think you meant 1+2sqrt3+3 cause it looks like you fixed it on the next line.

Answer 14

Hmm ya seems right to me! :)

Answer 15

yes,it's a typo.. x'D

Answer 16

1-2sqrt3+3
*

Answer 17

Thank youuu Zepdrix! :D :D :D