Trigonometry equation?

SodaCup · #1 $Old$ November 13th, 2009, 05:09 AM

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° = ???

Please help. Thanks

Soroban · #2 $Old$ November 13th, 2009, 05:56 AM

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

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

SodaCup · #3 $Old$ November 13th, 2009, 07:01 AM

Quote:

Originally Posted by Soroban $View Post$

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 $View Post$

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

stapel · #4 $Old$ November 13th, 2009, 08:33 AM

Quote:

Originally Posted by SodaCup $View Post$

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?

SodaCup · #5 $Old$ November 13th, 2009, 09:07 AM

Quote:

Originally Posted by stapel $View Post$

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?

1/2 = cos 60°
the limit is -2..? right?
I still don't get it, how do you find √3 ??