# Trigonometry equation?

• Nov 13th 2009, 04:09 AM
SodaCup
Trigonometry equation?
Given the equation cos2x +7cosx-3 = 0 find the tan value of x

I've tried using 2cos^2x-1 +7cosx-3=0 but I don't know what to do next??

and one more: sin 105° - sin 15° = ???

• Nov 13th 2009, 04:56 AM
Soroban
Hello, SodaCup!

Quote:

Given the equation: . $\cos2x +7\cos x-3 \:=\: 0$, find $\tan x$

I've tried using: . $2\cos^2\!x-1 +7\cos x-3\:=\:0$
but I don't know what to do next.

How about combining the -1 and -3?
We have: . $2\cos^2\!x + 7\cos x - 3 \:=\:0$

Factor: . $(\cos x + 4)(2\cos x - 1) \:=\:0$

And we have two equations to solve:

. . $\cos x + 4 \;=\:0 \quad\Rightarrow\quad \cos x \:=\:-4\quad\hdots\;\text{ no real roots}$

. . $2\cos x - 1 \:=\:0 \quad\Rightarrow\quad \cos x \:=\:\tfrac{1}{2} \quad\Rightarrow\quad x \:=\:\pm\tfrac{\pi}{3}$

Therefore: . $\tan\left(\pm\tfrac{\pi}{3}\right) \:=\:\pm\sqrt{3}$

Quote:

Evaluate: . $\sin105^o - \sin15^o$
We can use two identities: . $\begin{array}{ccc}\sin\theta &=& \sqrt{\dfrac{1-\cos2\theta}{2}} \\ \\[-3mm] \cos\theta &=& \sqrt{\dfrac{1+\cos2\theta}{2}} \end{array}$

So that: . $\begin{array}{ccccccccc}
\sin105^o &=& \sqrt{\dfrac{1 - \cos210^o}{2}} &=& \sqrt{\dfrac{1-\left(-\frac{\sqrt{3}}{2}\right)}{2}} &=& \sqrt{\dfrac{2+\sqrt{3}}{4}} &=& \dfrac{\sqrt{2+\sqrt{3}}}{2} \\ \\[-3mm]

\sin15^o &=& \sqrt{\dfrac{1-\cos30^o}{2}} &=&
\sqrt{\dfrac{1 - \frac{\sqrt{3}}{2}}{4}} &=&
\sqrt{\dfrac{2-\sqrt{3}}{4}} &=& \dfrac{\sqrt{2-\sqrt{3}}}{2}
\end{array}$

Then: . $\sin105^o - \sin15^o \;=\; \frac{\sqrt{2+\sqrt{3}}}{2} - \frac{\sqrt{2-\sqrt{3}}}{2}$

Note that: . $\begin{array}{ccc}2+\sqrt{3} &=& \dfrac{(\sqrt{3}+1)^2}{2} \\ \\[-3mm]
2 - \sqrt{3} &=& \dfrac{(\sqrt{3}-1)^2}{2} \end{array}$

$\text{Therefore: }\;\sin105^o - \sin15^o \;=\;\dfrac{\dfrac{\sqrt{3}+1}{\sqrt{2}} - \dfrac{\sqrt{3}-1}{\sqrt{2}}}{2} \;=\;\frac{1}{\sqrt{2}}$

• Nov 13th 2009, 06:01 AM
SodaCup
Quote:

Originally Posted by Soroban
Hello, SodaCup!

And we have two equations to solve:

. . $\cos x + 4 \;=\:0 \quad\Rightarrow\quad \cos x \:=\:-4\quad\hdots\;\text{ no real roots}$

. . $2\cos x - 1 \:=\:0 \quad\Rightarrow\quad \cos x \:=\:\tfrac{1}{2} \quad\Rightarrow\quad x \:=\:\pm\tfrac{\pi}{3}$

Therefore: . $\tan\left(\pm\tfrac{\pi}{3}\right) \:=\:\pm\sqrt{3}$

Um sorry I still don't get it, how do you find √3 from cosx = -4 and cosx = 1/2?

Quote:

Originally Posted by Soroban
We can use two identities: . $\begin{array}{ccc}\sin\theta &=& \sqrt{\dfrac{1-\cos2\theta}{2}} \\ \\[-3mm] \cos\theta &=& \sqrt{\dfrac{1+\cos2\theta}{2}} \end{array}$

So that: . $\begin{array}{ccccccccc}
\sin105^o &=& \sqrt{\dfrac{1 - \cos210^o}{2}} &=& \sqrt{\dfrac{1-\left(-\frac{\sqrt{3}}{2}\right)}{2}} &=& \sqrt{\dfrac{2+\sqrt{3}}{4}} &=& \dfrac{\sqrt{2+\sqrt{3}}}{2} \\ \\[-3mm]

\sin15^o &=& \sqrt{\dfrac{1-\cos30^o}{2}} &=&
\sqrt{\dfrac{1 - \frac{\sqrt{3}}{2}}{4}} &=&
\sqrt{\dfrac{2-\sqrt{3}}{4}} &=& \dfrac{\sqrt{2-\sqrt{3}}}{2}
\end{array}$

Then: . $\sin105^o - \sin15^o \;=\; \frac{\sqrt{2+\sqrt{3}}}{2} - \frac{\sqrt{2-\sqrt{3}}}{2}$

Note that: . $\begin{array}{ccc}2+\sqrt{3} &=& \dfrac{(\sqrt{3}+1)^2}{2} \\ \\[-3mm]
2 - \sqrt{3} &=& \dfrac{(\sqrt{3}-1)^2}{2} \end{array}$

$\text{Therefore: }\;\sin105^o - \sin15^o \;=\;\dfrac{\dfrac{\sqrt{3}+1}{\sqrt{2}} - \dfrac{\sqrt{3}-1}{\sqrt{2}}}{2} \;=\;\frac{1}{\sqrt{2}}$

Very clear explanation. Thanks
• Nov 13th 2009, 07:33 AM
stapel
Quote:

Originally Posted by SodaCup
Um sorry I still don't get it, how do you find √3 from cosx = -4 and cosx = 1/2?

Check the basic reference angles for the cosine ratio. For what angle value(s) is the cosine equal to 1/2?

Also, check the properties of the cosine wave. What are the limits on the values of the cosine? Does -4 fall within this interval of values?

(Wink)
• Nov 13th 2009, 08:07 AM
SodaCup
Quote:

Originally Posted by stapel
Check the basic reference angles for the cosine ratio. For what angle value(s) is the cosine equal to 1/2?

Also, check the properties of the cosine wave. What are the limits on the values of the cosine? Does -4 fall within this interval of values?

(Wink)

1/2 = cos 60°
the limit is -2..? right?
I still don't get it, how do you find √3 ?? (Wondering)
• Nov 13th 2009, 08:11 AM
stapel
Quote:

Originally Posted by SodaCup
1/2 = cos 60°

And another value, as provided to you earlier. Then take the tangent of these angles to find the required values. (Wink)

Quote:

Originally Posted by SodaCup
the limit is -2..? right?

Where are you taking limits...? (Wondering)
• Nov 13th 2009, 09:08 AM
SodaCup
Quote:

Originally Posted by stapel
And another value, as provided to you earlier. Then take the tangent of these angles to find the required values. (Wink)

Where are you taking limits...? (Wondering)

I think I get it now..

cos 60° = 1/2
sin 60° = 1/2 √3

sin/cos = √3 (Rofl)

anyway Thanks to all who helped.