Trigonometry equation?

**SodaCup** · November 13th 2009, 04:09 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** · November 13th 2009, 04:56 AM

Hello, SodaCup!

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

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** · November 13th 2009, 06:01 AM

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?

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

**stapel** · November 13th 2009, 07:33 AM

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?

**SodaCup** · November 13th 2009, 08:07 AM

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?

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

**stapel** · November 13th 2009, 08:11 AM

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.

Originally Posted by SodaCup

the limit is -2..? right?

Where are you taking limits...?

**SodaCup** · November 13th 2009, 09:08 AM

Originally Posted by stapel

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

Where are you taking limits...?

I think I get it now..

cos 60° = 1/2
sin 60° = 1/2 √3

sin/cos = √3

anyway Thanks to all who helped.

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