# Trigonometry Question

• Dec 1st 2008, 09:14 AM
GregoryT
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!
• Dec 1st 2008, 10:16 AM
running-gag
Hello
$cosx\;=\;\sqrt{3}-1$

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

$cos(2x)\;=\;2(\sqrt{3}-1)^2-1$
$cos(2x)\;=\;2(4-2\sqrt{3})-1$
$cos(2x)\;=\;7-4\sqrt{3}$
• Dec 1st 2008, 10:36 AM
Soroban
Hello, GregoryT!

Quote:

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

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

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

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

Quote:

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. . ??

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

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

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

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

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

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

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

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

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

• Dec 1st 2008, 10:46 AM
GregoryT
Thanks Soroban and running gag, You have been invaluable help. Especially on that last one soroban! My formal Board thanks to both of you!
• Dec 1st 2008, 01:07 PM
GregoryT
Sorry, one more quick question, can you explain what is happening on the line immediately before the answer.