Trig Problems

**Altair_xI** · April 5th 2009, 12:46 PM

I'm having some more difficulties with these two.

This one is I'm supposed to find the solutions in interval $[0,2pi)$
$cos(x)=sin(\frac{x}2)$

This is what I've done so far but I'm unsure how to continue:

$cos(x)=\sqrt{(\frac{1-cos(x)}2)}$

$cos^2(x)=(\frac{1-cos(x)}2)$

---
For the second problem
It is determining if its a identity

$cos(x)csc(x)tan(x)=1$

I began it like this:
$cos(x)\frac{1}{sin(x)}\frac{sin(x)}{cos(x)}=1$

**stapel** · April 5th 2009, 12:54 PM

Originally Posted by Altair_xI

I'm supposed to find the solutions in interval $[0,2pi)$
$cos(x)=sin(\frac{x}{2})$

This is what I've done so far but I'm unsure how to continue:

$cos(x)=\sqrt{(\frac{1-cos(x)}2)}$

$cos^2(x)=(\frac{1-cos(x)}2)$

What you've done is fine!

Now multiply through by "2", move everything over to the left-hand side of the equation, and solve the quadratic for "cos(x)=". Then solve the two trig equations.

Originally Posted by Altair_xI

It is determining if its a identity

$cos(x)csc(x)tan(x)=1$

I began it like this:
$cos(x)\frac{1}{sin(x)}\frac{sin(x)}{cos(x)}=1$

Cancel, and you're done!

**Jhevon** · April 5th 2009, 12:54 PM

Originally Posted by Altair_xI

I'm having some more difficulties with these two.

This one is I'm supposed to find the solutions in interval $[0,2pi)$
$cos(x)=sin(\frac{x}2)$

This is what I've done so far but I'm unsure how to continue:

$cos(x)={\color{red} \pm}~ \sqrt{(\frac{1-cos(x)}2)}$

$cos^2(x)=(\frac{1-cos(x)}2)$

---

good so far: now, multiply by 2 and bring everything to one side. we get:

$2 \cos^2 x + \cos x - 1 = 0$

now notice that we have a quadratic equation in $\cos x$

to see this, let $a = \cos x$ , then we have

$2a^2 + a - 1 = 0$

do you see ow to proceed? solve for $a$ , and then you can solve for $x$

For the second problem
It is determining if its a identity

$cos(x)csc(x)tan(x)=1$

I began it like this:
$cos(x)\frac{1}{sin(x)}\frac{sin(x)}{cos(x)}=1$

yes, though you wrote it out weird. it makes me wonder if you actually know how to prove an identity or you're just rewriting the equation with the left side changed. proceed thusly,

Consider the left hand side:

$\begin{array}{lcl} \cos x \csc x \tan x & = & \cos x \cdot \frac 1{\sin x} \cdot \frac {\sin x}{\cos x} \\ & = & 1 \\ & = & RHS \end{array}$

thus we have shown $LHS \equiv RHS$ , and so we have proven the identity.

EDIT: argh! this is the second time i've been beaten today by less than a minute

**Altair_xI** · April 5th 2009, 01:07 PM

Okay, I understand now. Thanks.
Ah, I wrote it weird cause I'm unfamiliar with the math tag codes. Sorry.

**Jhevon** · April 5th 2009, 01:12 PM

Originally Posted by Altair_xI

Okay, I understand now. Thanks.

good

Ah, I wrote it weird cause I'm unfamiliar with the math tag codes. Sorry.

no apologies necessary. you made a good first effort. what you wrote is understandable, so it is better than most posts made by first time LaTeX users.

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