Thread: Trigonometry Question

I figured I would post this in here and hopefully get somebody who likes trig to help me through this.


Given that cos x = (3^.5) − 1, find the value of cos 2x in the form a +b(3^.5) ,
where a and b are integers.

Given that
2 cos (y + 30)= 3 sin (y − 30),
find the value of tan y in the form k 3 where k is a rational constant.


a bit of background here is that I am looking to take my high school maths certificate and I found out that well... Im rubbish at trig!

Hello
$\displaystyle cosx\;=\;\sqrt{3}-1$

Using the formula
cos(2x) = 2cos²x - 1

$\displaystyle cos(2x)\;=\;2(\sqrt{3}-1)^2-1$
$\displaystyle cos(2x)\;=\;2(4-2\sqrt{3})-1$
$\displaystyle cos(2x)\;=\;7-4\sqrt{3}$

Hello, GregoryT!

1) Given that: .$\displaystyle \cos x \:=\:\sqrt{3}-1$,
find the value of $\displaystyle \cos 2x$ in the form $\displaystyle a +b\sqrt{3}$, where $\displaystyle a\text{ and }b$ are integers.

We are expected to know the identity: .$\displaystyle \cos 2x \:=\:2\cos^2\!x - 1$

Then we have: .$\displaystyle \cos2x \;=\;2(\sqrt{3}-1)^2 - 1$

. . $\displaystyle =\;\;2(3 - 2\sqrt{3} + 1) - 1 \;\;=\;\;6 - 4\sqrt{3} + 2 - 1 \;\;=\;\;\boxed{7 -4\sqrt{3}}$

Given that: .$\displaystyle 2\cos(y + 30)\:=\:3\sin(y - 30)$,

find the value of $\displaystyle \tan y$ . . . in the form k 3 where k is a rational constant. . ??

We have: .$\displaystyle 2\cos(y + 30) \:=\:3\sin(y-30) $

. . $\displaystyle 2\bigg(\cos y \cos30 -\sin y\sin30\bigg) \;=\;3\bigg(\sin y\cos30 - \cos y\sin30\bigg) $

. . $\displaystyle 2\bigg(\frac{\sqrt{3}}{2}\cos y - \frac{1}{2}\sin y\bigg) \;=\;3\bigg(\frac{\sqrt{3}}{2}\sin y - \frac{1}{2}\cos y\bigg) $


Multiply by 2: .$\displaystyle 2\sqrt{3}\cos y - 2\sin y \;=\;3\sqrt{3}\sin y - 3\cos y$

. . $\displaystyle 3\sqrt{3}\sin y + 2\sin y \;=\;2\sqrt{3}\cos y + 3\cos y$

. . $\displaystyle (3\sqrt{3}+2)\sin y \;=\;(2\sqrt{3} + 3)\cos y$

. . $\displaystyle \frac{\sin y}{\cos y} \;=\;\frac{2\sqrt{3}+3}{3\sqrt{3}+2} $

. . $\displaystyle \tan y \;=\;\frac{2\sqrt{3}+3}{3\sqrt{3}+2} \cdot{\color{blue}\frac{3\sqrt{3}-2}{3\sqrt{3}-2} } \;=\;\frac{18 - 4\sqrt{3} + 9\sqrt(3) - 6}{27 - 4}$


Therefore: .$\displaystyle \boxed{\tan y \;=\;\frac{12+5\sqrt{3}}{23}} $

Thanks Soroban and running gag, You have been invaluable help. Especially on that last one soroban! My formal Board thanks to both of you!

Sorry, one more quick question, can you explain what is happening on the line immediately before the answer.