A few HW questions I dont get

**slaying_ice** · December 3rd 2006, 02:25 PM

Using the conditions below
sinA=3/4, A lies in quadrant 2
CosB=12/13, B lies in quadrant 1

Find the following:

Cos (A+B) Tan(A-B)

Cos B/2

Solve the equation in th interval [0,2pie)

2cos^2x-3cosx+1

Use Half angle forumlas to find the exact value of the expression

cos 22.5

cos^2 75 - sin^2 75

**topsquark** · December 3rd 2006, 04:06 PM

Originally Posted by slaying_ice

Using the conditions below
sinA=3/4, A lies in quadrant 2
CosB=12/13, B lies in quadrant 1

Find the following:

Cos (A+B) Tan(A-B)

Cos B/2

First:
$cos(A + B) = cos(A) cos(B) - sin(A) sin(B)$

Second:
$tan(A - B) = \frac{sin(A - B)}{cos(A - B)}$

$tan(A - B) = \frac{sin(A)cos(B) - sin(B)cos(A)}{cos(A)cos(B) + sin(A)sin(B)}$

Now:
$sin(A) = \frac{3}{4}$ and is in the 2nd quadrant.

Thus: $cos(A) = -\sqrt{1 - sin^2(A)} = -\frac{1}{2}$

Again:
$cos(B) = \frac{12}{13}$ and is in the first quadrant.

Thus: $sin(B) = +\sqrt{1 - cos^2(B)} = \frac{5}{13}$

To the problems:
$cos(A + B) = cos(A) cos(B) - sin(A) sin(B)$

$cos(A + B) = -\frac{1}{2} \cdot \frac{12}{13} - \frac{3}{4} \cdot \frac{5}{13} = -\frac{3}{4}$

$tan(A - B) = \frac{sin(A)cos(B) - sin(B)cos(A)}{cos(A)cos(B) + sin(A)sin(B)}$

$tan(A - B) = \frac{\frac{3}{4} \cdot \frac{12}{13} - \frac{5}{13} \cdot -\frac{1}{2}}{-\frac{1}{2} \cdot \frac{12}{13} + \frac{3}{4} \cdot \frac{5}{13}} = - \frac{46}{9}$

So
$cos(A + B) \cdot tan(A - B) = -\frac{3}{4} \cdot -\frac{46}{9} = \frac{23}{6}$
================================================== =======
$cos \left ( \frac{B}{2} \right ) = \pm \sqrt{\frac{1 + cos(B)}{2}}$

$cos \left ( \frac{B}{2} \right ) = \pm \sqrt{\frac{1 + \frac{12}{13}}{2}} = \pm \sqrt{\frac{25}{26}} = \pm \frac{5}{\sqrt{26}}$

Since B is in the 1st quadrant, so is B/2. Thus
$cos \left ( \frac{B}{2} \right ) = \frac{5 \sqrt{26}}{26}$

-Dan

**topsquark** · December 3rd 2006, 04:11 PM

Originally Posted by slaying_ice

Solve the equation in th interval [0,2pie)

2cos^2x-3cosx+1

By the way, the Greek letter $\pi$ is spelled "pi" not "pie."

There's no equation to solve here!

I'll assume you meant:
$2cos^2(x) - 3cos(x) + 1 = 0$

Let's simplify a bit: Let $y = cos(x)$ .
The we have:
$2y^2 - 3y + 1 = 0$

This factors:
$(2y - 1)(y - 1) = 0$

So $y = \frac{1}{2}$ or $y = 1$ .

Thus
$cos(x) = \frac{1}{2}$ ==> $x = \frac{\pi}{3}, \frac{5 \pi}{3}$ rad

or

$cos(x) = 1$ ==> $x = 0$ .

So the full solution set on $[0, 2 \pi)$ is $x = 0, \frac{\pi}{3}, \frac{5 \pi}{3}$ .

-Dan

**slaying_ice** · December 3rd 2006, 04:18 PM

On the second problem you assumed correct on what the equation was.

**topsquark** · December 3rd 2006, 04:20 PM

Originally Posted by slaying_ice

Use Half angle forumlas to find the exact value of the expression

cos 22.5

cos^2 75 - sin^2 75

The first one:
$cos(22.5) = cos \left ( \frac{45}{2} \right ) = \sqrt{ \frac{1 + cos(45)}{2} } = \sqrt{ \frac{1 + \frac{ \sqrt{2} }{2} }{2} } = \frac{ \sqrt{2 + \sqrt{2} }}{2}$

The second one actually isn't a half-angle formula, it's a double angle formula:
$cos(150) = cos(2 \cdot 75) = cos^2(75) - sin^2(75)$

and
$cos(150) = -cos(30) = -\frac{\sqrt{3}}{2}$

-Dan

**slaying_ice** · December 3rd 2006, 04:22 PM

The file you uploaded to solve the double angle problem isnt showing up. Plus I have a couple more problems that need to be solved if you dont mind

Find the exact valuse of Sin2x Cos2x and tan 2x given tan=4/3 in the first quadrant.

and

Find the exact valuse of sin x/2 cos x/2 and tan x/2 when cosx= -5/13 and x is in the third quadrant

Thank you

~Kyle

**topsquark** · December 3rd 2006, 04:34 PM

Originally Posted by slaying_ice

The file you uploaded to solve the double angle problem isnt showing up.

I fixed it. Sorry about that.

-Dan

**slaying_ice** · December 3rd 2006, 04:37 PM

Plus I have a couple more problems that need to be solved if you dont mind

Find the exact valuse of Sin2x Cos2x and tan 2x given tan=4/3 in the first quadrant.

and

Find the exact valuse of sin x/2 cos x/2 and tan x/2 when cosx= -5/13 and x is in the third quadrant

Thank you

~Kyle

**topsquark** · December 3rd 2006, 04:51 PM

Originally Posted by slaying_ice

Plus I have a couple more problems that need to be solved if you dont mind

Find the exact valuse of Sin2x Cos2x and tan 2x given tan=4/3 in the first quadrant.

$tan(x) = \frac{4}{3}$ so

$sin(x) = sin \left ( tan^{-1} \left ( \frac{4}{3} \right ) \right )$

This is a 3 - 4 - 5 triangle, so
$sin(x) = \frac{4}{5}$ .

Similarly:
$cos(x) = \frac{3}{5}$

So:
$sin(2x) = 2 sin(x) cos(x) = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25}$

$cos(2x) = cos^2(x) - sin^2(x) = \left (\frac{4}{5} \right ) ^2 - \left ( \frac{3}{5} \right ) ^2 = \frac{7}{25}$

$tan(2x) = \frac{sin(2x)}{cos(2x)} = \frac{ \frac{24}{25} }{ \frac{7}{25} } = \frac{24}{7}$

-Dan

**topsquark** · December 3rd 2006, 05:00 PM

Originally Posted by slaying_ice

Find the exact valuse of sin x/2 cos x/2 and tan x/2 when cosx= -5/13 and x is in the third quadrant

$cos(x) = -\frac{5}{13}$

Thus
$sin \left ( \frac{x}{2} \right ) = \pm \sqrt{\frac{1 - cos(x)}{2}} = \pm \sqrt{ \frac{1 + \frac{5}{13}}{2} } = \pm \frac{3}{\sqrt{13}}$

Since x is 3rd quadrant, x/2 will be 2nd quadrant.
Thus $sin \left ( \frac{x}{2} \right ) = \frac{3 \sqrt{13}}{13}$

$cos \left ( \frac{x}{2} \right ) = \pm \sqrt{\frac{1 + cos(x)}{2}} = \pm \sqrt{ \frac{1 - \frac{5}{13}}{2} } = \pm \frac{2}{\sqrt{13}}$

Again, since x/2 is 2nd quadrant:
$cos \left ( \frac{x}{2} \right ) = - \frac{2\sqrt{13}}{13}$

$tan \left ( \frac{x}{2} \right ) = \frac{sin \left ( \frac{x}{2} \right )}{cos \left ( \frac{x}{2} \right )} = \frac{ \frac{3 \sqrt{13}}{13} }{ - \frac{2\sqrt{13}}{13} } = - \frac{3}{2}$

-Dan

**slaying_ice** · December 3rd 2006, 05:03 PM

Thank you very much you have saved me from homework doom

**topsquark** · December 4th 2006, 05:07 AM

Originally Posted by slaying_ice

Thank you very much you have saved me from homework doom

But I LIKE Doom! (Doom, DoomII, etc...)

-Dan

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