Trig Identity Help (and solving)

• December 13th 2007, 03:34 PM
Hasan1
Trig Identity Help (and solving)
Prove the following identities

1. $\sin (\Pi + x) + \cos ( \frac{\Pi}{2} - x) + \tan (\frac{\Pi}{2} + x) = -\cot x$

2. $\frac{\sin (4x) - \sin (2x)}{\sin (2x)} = \frac{\cos (3x)}{\cos (x)}$

3. $\cos (x) + \cos (2x) + \cos (3x) = \cos (2x) (1+ 2 \cos (x))$

Solve

1. $\cos^2 (2x) + 2 \cos (2x) + 1 = 0 [-\Pi , \Pi]$

2. $\tan (4x) - \tan (2x) = 0 [0,\Pi]$

• December 13th 2007, 05:04 PM
Soroban
Hello, Hasan1!

Here are the last two . . .

Quote:

$1)\;\;\cos^2(2x) + 2\cos(2x) + 1 \:= \:0\qquad[-\pi,\:\pi]$
The left side is a square . . .

. . $\bigg[\cos(2x) + 1\bigg]^2 \:=\:0\quad\Rightarrow\quad\cos(2x) + 1 \:=\:0\quad\Rightarrow\quad\cos(2x) \:=\:-1$

Therefore: . $2x \:=\:-\pi,\:\pi\quad\Rightarrow\quad\boxed{ x \:=\:-\frac{\pi}{2},\:\frac{\pi}{2}}$

Quote:

$2)\;\;\tan(4x) - \tan(2x) \:=\: 0\qquad[0,\,\pi]$
This one gets messy . . .

We know the identity: . $\tan2\theta \:=\:\frac{2\tan\theta}{1-\tan^2\!\theta}$
Apply it to $\tan4x$

We have: . $\frac{2\tan2x}{1-\tan^2\!2x} - \tan(2x) \:=\:0$

Multiply by $(1-\tan^2\!2x)\!:\;\;2\tan2x - \tan2x(1 - \tan^2\!2x) \:=\:0$

. . $2\tan2x -\tan2x + \tan^3\!2x \:=\:0\quad\Rightarrow\quad\tan^3\!2x + \tan2x \:=\:0$

Factor: . $\tan2x\left[\tan^2\!2x + 1\right] \:=\:0$

The equation: . $\tan^2\!2x + 1 \:=\:0$ has no real roots.
So we have: . $\tan2x\:=\:0\quad\Rightarrow\quad 2x \:=\:0,\:\pi,\:2\pi\quad\Rightarrow\quad \boxed{x \:=\:0,\:\frac{\pi}{2},\:\pi}$

• December 13th 2007, 05:51 PM
Hasan1
that second solving question has a nice solution, thanks!

I guess I never saw the perfect square trinomial in that first one though :D

thanks again :cool:
• December 13th 2007, 06:31 PM
Krizalid
Here's another solution for this one.

Quote:

Originally Posted by Hasan1
2. $\tan (4x) - \tan (2x) = 0 [0,\Pi]$

Previous knowledge:

$\tan \alpha - \tan \beta = \frac{{\sin \alpha }}
{{\cos \alpha }} - \frac{{\sin \beta }}
{{\cos \beta }} = \frac{{\sin (\alpha - \beta )}}
{{\cos \alpha \cos \beta }}.$

Solution:

$\tan 4x - \tan 2x = 0 \implies \frac{{\sin 2x}}
{{\cos 4x\cos 2x}} = 0\,\therefore \,\tan 2x = 0.$

Of course $\cos4x\ne0.$

The rest follows from Soroban's conclusions.
• December 13th 2007, 07:15 PM
Krizalid
Quote:

Originally Posted by Hasan1
2. $\frac{\sin (4x) - \sin (2x)}{\sin (2x)} = \frac{\cos (3x)}{\cos (x)}$

Just tackle the LHS.

Apply the following identity

$\sin \alpha - \sin \beta = 2\sin \frac{{\alpha - \beta }}
{2} \cdot \cos \frac{{\alpha + \beta }}
{2}.$

The conclusion follows quickly.
• December 13th 2007, 07:22 PM
Krizalid
Quote:

Originally Posted by Hasan1
3. $\cos (x) + \cos (2x) + \cos (3x) = \cos (2x) (1+ 2 \cos (x))$

The same trick as identity #2, apply

$\cos \alpha + \cos \beta = 2\cos \frac{{\alpha + \beta }}
{2} \cdot \cos \frac{{\alpha - \beta }}
{2}.$

--

The last one is easy. Expand $\sin(\pi+x)$ & observe that $\cos \left( {\frac{\pi }
{2} - x} \right) = \sin x.$
(This last identity is a fact that you actually should know.)

And we're done :D:D
• December 14th 2007, 07:56 PM
Hasan1
Is there a way to do those identities without the sum to product formula, but the double angle formula instead?
• December 15th 2007, 12:16 PM
Krizalid
I think those identities were made to apply the formulae I gave you. It's a quickly way.