# Math Help - Trigonometric functions.

1. ## Trigonometric functions.

Could someone clarify these questions step-by-step?
Much appreciated! (sorry for my earlier thread!)

Solve within [0, 2pi):

cos2x + cosx = 0

cot²x - cos²x = cot²xcos²x

2. cot²x - cos²x = cot²xcos²x

cos^2x/sin^2x - cos^2x=cos^2x/sin^2x cos^2x

devide by cos^2x , => 1/sin^2x - 1 = cos^2x/sin^2x

=> 1-sin^2x = cos^2x

= cos^2x+sin^2x = 1 => 1=1 ... i cant understand the point from this simplification !!! check again your question , the first one will be similar

3. AHHHH!
Wait wait wait, I'm so sorry!
I posted incorrectly.
It's supposed to be solved within [0, 2pi).

4. Originally Posted by Silvahere
Could someone clarify these questions step-by-step?
Much appreciated! (sorry for my earlier thread!)

Simplify the following:

cos2x + cosx = 0

cot²x - cos²x = cot²xcos²x
you need to look at the directions for these two equations again.

trig expressions are simplified ... trig equations are either identities to be verified or conditional equations to be solved.

the first equation is a conditional equation.

the second equation is an identity.

5. Originally Posted by Silvahere
Could someone clarify these questions step-by-step?
Much appreciated! (sorry for my earlier thread!)

Simplify the following:

cos2x + cosx = 0

cot²x - cos²x = cot²xcos²x
$cos(2x) = 2cos^2(x)-1$

$2cos^2(x)-cos(x)-1=0$. Solve using quadratic equation

6. Ahhhhh! I'm sooo sorryyyy. x.x
I posted my questions incorrectly, they're supposed to be solved within [0, 2pi). *Face palm*

7. right the first one is easy to solve

cos2x+cosx=0 => 2cos^2 (x) + cosx - 1 = 0

=> (cosx - 1 ) ( cosx +1/2 ) = 0

=> cosx=1 x=0 or cosx=-1/2 , x= pi - pi/3 , x= pi+pi/3 check the answers again

8. the second identity is valid for any X ,

9. I'm absolutely and completely klutzing out right now.
Let me clarify again, because AGAIN, I made a mistake. >___<

Okay.

Solve the trigonometric functions within [0, 2pi).
a. cos3x + cosx = 0

b. cot²x - cos²x = cot²xcos²x

10. Originally Posted by Silvahere
I'm absolutely and completely klutzing out right now.
Let me clarify again, because AGAIN, I made a mistake. >___<

Okay.

Solve the trigonometric functions within [0, 2pi).
a. cos3x + cosx = 0

b. cot²x - cos²x = cot²xcos²x
$cos(3x) = cos(2x+x) = cos(2x)cos(x)-sin(2x)sin(x)$
$
(2cos^2(x)-1)(cos(x))-2sin^2(x)cos(x) = 2cos^3(x)-cos(x)-2cos(x)+2cos^3(x))$

$= 4cos^3(x)-3cos(x)$

$\therefore \: \: 4cos^3(x)-2cos(x)= 2cos(x)(2cos^2(x)-1)=0$

11. Originally Posted by e^(i*pi)
$cos(3x) = cos(2x+x) = cos(2x)cos(x)-sin(2x)sin(x)$
$
(2cos^2(x)-1)(cos(x))-2sin^2(x)cos(x) = 2cos^3(x)-cos(x)-2cos(x)+2cos^3(x))$

$= 4cos^3(x)-3cos(x)$

$\therefore \: \: 4cos^3(x)-2cos(x)= 2cos(x)(2cos^2(x)-1)=0$
... Uh... I understood up to: $cos(3x) = cos(2x+x) = cos(2x)cos(x)...$ x_x

Edit: Nevermind nevermind, I just got the edit. But where did the sines come from?

12. Originally Posted by Silvahere
... Uh... I understood up to: $cos(3x) = cos(2x+x) = cos(2x)cos(x)...$ x_x
That is the derivation. It is usually sufficient to learn that $cos(3x)=4cos^3(x)-3cos(x)$

I just derived it because I don't know the triple angle formula