# Thread: Trig Identity Help (and solving)

1. ## Trig Identity Help (and solving)

Prove the following identities

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

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

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

Solve

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

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

2. Hello, Hasan1!

Here are the last two . . .

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

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

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

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

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

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

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

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

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

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

3. that second solving question has a nice solution, thanks!

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

thanks again

4. Here's another solution for this one.

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

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

Solution:

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

Of course $\displaystyle \cos4x\ne0.$

The rest follows from Soroban's conclusions.

5. Originally Posted by Hasan1
2. $\displaystyle \frac{\sin (4x) - \sin (2x)}{\sin (2x)} = \frac{\cos (3x)}{\cos (x)}$
Just tackle the LHS.

Apply the following identity

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

The conclusion follows quickly.

6. Originally Posted by Hasan1
3. $\displaystyle \cos (x) + \cos (2x) + \cos (3x) = \cos (2x) (1+ 2 \cos (x))$
The same trick as identity #2, apply

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

--

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

And we're done

7. Is there a way to do those identities without the sum to product formula, but the double angle formula instead?

8. I think those identities were made to apply the formulae I gave you. It's a quickly way.