# Need some help with few trig equations...

• Jan 20th 2008, 03:23 PM
toop
Need some help with few trig equations...
Find all solutions of the given equation:

1.) 2 sin x + 1 = 0

2.) sin 2x - cos x = 0

3.) sinē x - sin x = 0

**Not looking for the entire answer, just how to start.

Prove that the given trig identity is true:

cos (a - b) = (cos a) (cos b) + (sin a) (sin b)

**I know that it's true because it states the formula in the book but I just don't know how to prove it, lol.
• Jan 20th 2008, 03:35 PM
Krizalid
1) $\sin x = - \frac{1}
{2}.$

2) $\sin 2x = 2\sin x\cos x$ & factorise.

3) Factorise and proceed as 2).

---

About last formula, it requires a geometric reasoning to prove it, or you can also use complex numbers.
• Jan 21st 2008, 09:14 AM
toop
Could you explain a little further on...all of them?

For #1, -1/2 is what I got on my own, but I don't know how to put the answer in terms of (ex. from another problem: x= pi/3 + 2n(pi) for any int n).

As for #2 and #3, can you give me the first step to factorizing? I'm clueless on where to start.

And #4...I still have no clue.
• Jan 21st 2008, 09:22 AM
janvdl
Quote:

Originally Posted by toop
Could you explain a little further on...all of them?

For #1, -1/2 is what I got on my own, but I don't know how to put the answer in terms of (ex. from another problem: x= pi/3 + 2n(pi) for any int n).

As for #2 and #3, can you give me the first step to factorizing? I'm clueless on where to start.

And #4...I still have no clue.

Krizalid seems to be away for the moment so I'll jump in... :D

1) $sin x = - \frac{1}{2}$

$x = sin^{-1} ( \frac{1}{2} )$ (The value of x must be in quadrants 3 and 4.

2) $2 \ sin x \ cos x - cos x= 0$

$cos x( 2 sin x - 1) = 0$

$Cos x = 0$

OR

$2 Sin x - 1 = 0$

3) $sin^{2} x - sin x = 0$

$sin x(sin x - 1) = 0$

$Sin x = 0$

OR

$sin x - 1 = 0$

4) This is a long process of proving. Look it up on the net, or it is most probably in your textbook.
• Jan 21st 2008, 11:18 AM
Krizalid
Quote:

Originally Posted by toop
cos (a - b) = (cos a) (cos b) + (sin a) (sin b)

First of all, we're gonna find a formula for $\cos (\alpha + \beta ).$

Consider the following sketch:

http://img169.imageshack.us/img169/6003/recqe9br8.png

Construction:

• Consider a rectangle $NOPR.$
• Extend $\overline{ON}.$ We get point $M.$
• Join $M$ & $P.$
• Extend $\overline{NR}.$ We get point $Q.$
• Draw $\overline {QP} \perp\overline {MP} ,$ now join $M$ & $Q.$
• Let $\measuredangle \,OMP = \alpha$ & $\measuredangle \,QMP = \beta .$

Start by showing that $\measuredangle \,OMP = \measuredangle \,PQR = \alpha .$

We have

$\cos (\alpha + \beta ) = \frac{{\overline {MN} }}
{{\overline {QM} }} = \frac{{\overline {OM} - \overline {ON} }}
{{\overline {QM} }} = \frac{{\overline {OM} }}
{{\overline {QM} }} \cdot \frac{{\overline {MP} }}
{{\overline {MP} }} - \frac{{\overline {RP} }}
{{\overline {QM} }} \cdot \frac{{\overline {QP} }}
{{\overline {QP} }}.$

Rearrange

$\cos (\alpha + \beta ) = \frac{{\overline {MP} }}
{{\overline {QM} }} \cdot \frac{{\overline {OM} }}
{{\overline {MP} }} - \frac{{\overline {QP} }}
{{\overline {QM} }} \cdot \frac{{\overline {RP} }}
{{\overline {QP} }} = \cos \beta \cos \alpha - \sin \beta \sin \alpha \quad\blacksquare$

Finally, write $\cos (\alpha - \beta ) = \cos \left[ {\alpha + ( - \beta )} \right]$ and you'll get the desired formula.

---

Now, a proof with complex numbers.

$\cos (\alpha - \beta ) = \text{Re} \Big[ {e^{i(\alpha - \beta )} } \Big] = \text{Re} \Big[ {(\cos \alpha + i\sin \alpha )(\cos \beta - i\sin \beta )} \Big].$

Finally

$\cos (\alpha - \beta ) = \text{Re}\, (\cos \alpha \cos \beta - i\cos \alpha \sin \beta + i\sin \alpha \cos \beta + \sin \alpha \sin \beta ).$

And we happily get $\cos (\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\quad\blacksquare$

As desired.
• Jan 21st 2008, 01:08 PM
ThePerfectHacker
Quote:

Originally Posted by toop

Prove that the given trig identity is true:

cos (a - b) = (cos a) (cos b) + (sin a) (sin b)

The best way is to use $\cos z = = \frac{1}{2}(e^{iz} + e^{-iz})$ --- but that is unfair. :D

Here is another geometric proof that I remember from 10th grade.

1)Construct the circle $x^2+y^2 = 1$.
2)Construct those red lines touching at $P_1,P_2$.
3) $a$ is the bigger angle from the x-axis.
4) $b$ is the smaller angle from the [tex]x-axis.
5)Coordinates of $P_1$ is $(\cos b,\sin b)$.
6)Coordinates of $P_2$ is $(\cos a,\sin a)$
7)The distance from $P_1 \to P_2$ is $\sqrt{(\cos b-\cos a)^2 + (\sin b - \sin a)^2}$.
8)The angle between the two red lines is $a-b$.
9)The length of $P_1P_2$ using law of cosines is $\sqrt{1^2+1^2 - 2 \cos (a-b)}$.
10)Equality of #7 and #9 leads to ...

$\sqrt{(\cos b-\cos a)^2 + (\sin b - \sin a)^2} = \sqrt{2 - 2\cos (a-b)}$
$(\cos b - \cos a)^2 + (\sin b - \sin a)^2 = 2 - 2 \cos (a-b)$
$\cos^2 b - 2\cos b\cos a + \cos^2 a + \sin^2 b - 2\sin a \sin b + \sin^2 a = 2 - 2 \cos (a-b)$
$\cos (a-b) = \cos a \cos b + \sin a \sin b$
• Jan 21st 2008, 02:46 PM
toop
Wow, thanks for the help on #4.

For #1 I got:

$\frac{{7\pi }}
{6} + 2n\pi$

Does that look correct to you guys?

For #2 I got:

$x = \frac{\pi }
{2} + 2n\pi$

For #3:

$x = 2n\pi ;\frac{\pi }
{2} + 2n\pi$

Thanks a lot guys!