Circle Theorem and Trig Equation Problem!

**acid_rainbow** · May 30th 2008, 03:10 AM

Hey i have two problems that i am stuck on

1) Solve the following quation for 0 < theta < 180

(i will use x for theta, i know how to solve using the ASTC graph however i cannot rearrange this)

Cos^2(x) / Sinx + Sin^2(x)= 2

We have been told earlier that this is equal to 1-sinx/sinx

**acid_rainbow** · May 30th 2008, 03:11 AM

Circle Problem

Show that the centre of the circle with equation:

x^2 + y^2 = 6x + 8y is (3,4) and find the radius

Thanks =]

**mr fantastic** · May 30th 2008, 03:28 AM

Originally Posted by acid_rainbow

Hey i have two problems that i am stuck on

1) Solve the following quation for 0 < theta < 180

(i will use x for theta, i know how to solve using the ASTC graph however i cannot rearrange this)

Cos^2(x) / Sinx + Sin^2(x)= 2

We have been told earlier that this is equal to 1-sinx/sinx

You should make sure you know why $\frac{\cos^2 x}{\sin x + \sin^2 x} = \frac{1 - \sin x}{\sin x}$ . Then you will understand why the solutions to the equation $\frac{\cos^2 x}{\sin x + \sin^2 x} = 2$ are got by solving the following two equations:

1. $\frac{1 - \sin x}{\sin x} = 2 \Rightarrow 1 - \sin x = 2 \sin x \Rightarrow \sin x = \frac{1}{3}$ .

2. $1 + \sin x = 0 \Rightarrow \sin x = -1$ .

**Simplicity** · May 30th 2008, 03:29 AM

Originally Posted by acid_rainbow

Hey i have two problems that i am stuck on

1) Solve the following quation for 0 < theta < 180

(i will use x for theta, i know how to solve using the ASTC graph however i cannot rearrange this)

Cos^2(x) / Sinx + Sin^2(x)= 2

We have been told earlier that this is equal to 1-sinx/sinx

$\frac{\cos ^2 \theta}{\sin \theta + \sin ^2 \theta} = 2$

$\frac{1 - \sin ^2 \theta}{\sin \theta + \sin ^2 \theta} = 2$

$\frac{1 - \sin ^2 \theta}{\sin \theta ( 1 + \sin \theta)} = 2$

$\frac{(1 - \sin \theta)(1 + \sin \theta)}{\sin \theta ( 1 + \sin \theta)} = 2$

$\frac{1 - \sin \theta}{\sin \theta } = 2$

$1 - \sin \theta = 2 \sin \theta$

$1 = 3 \sin \theta$

$\sin \theta = \frac 13$

$\theta = \sin ^{-1} \left( \frac 13 \right)$

$\theta = 19.5^\circ, 161^\circ \ \ \text{to 3 sf}$

**acid_rainbow** · May 30th 2008, 03:31 AM

thanks air, you made the first problem clear

**mr fantastic** · May 30th 2008, 03:31 AM

Originally Posted by acid_rainbow

Circle Problem

Show that the centre of the circle with equation:

x^2 + y^2 = 6x + 8y is (3,4) and find the radius

Thanks =]

$x^2 + y^2 = 6x + 8y \Rightarrow (x^2 - 6x) + (y^2 - 8y) = 0 \Rightarrow (x - 3)^2 - 9 + (y - 4)^2 - 16 = 0$ .......

**mr fantastic** · May 30th 2008, 03:34 AM

Originally Posted by acid_rainbow

i still dont understand.

i tried simplifying it and ended up geting -sinx = 2 which is impossible

What don't you understand? What is the it that you tried simplifying and ended up getting -sinx = 2? Have you read my earlier post?

Note: $\frac{\cos^2 x}{\sin x + \sin^2 x} = \frac{1 - \sin^2 x}{\sin x (1 + \sin x} = \frac{(1 - \sin x)(1 + \sin x)}{\sin x (1 + \sin x)}$ ......

**acid_rainbow** · May 30th 2008, 03:37 AM

ahhh i get it all now.

thanks both of you

i just missed the embarassingly simple maths =[

thanks for the circle help too

**Simplicity** · May 30th 2008, 03:38 AM

Originally Posted by acid_rainbow

Circle Problem

Show that the centre of the circle with equation:

x^2 + y^2 = 6x + 8y is (3,4) and find the radius

Thanks =]

$x^2 + y^2 = 6x + 8y$

$x^2 - 6x + y^2 - 8y = 0$

A general form is $x^2 + y^2 + 2gx + 2fy + c = 0$ where center of circle: $(-g, -f)$ and the radius of circle: $\sqrt{g^2 + f^2 - c}$ .

$\therefore$ Center of circle: $\left(-\left(\frac{-6}{2}\right), -\left(\frac{-8}{2}\right)\right) = (3, 4)$ .

$\therefore$ Radius of circle: $\sqrt{3^2 + 4^2 - 0} = 5$ .

**Soroban** · May 30th 2008, 03:50 AM

Hello, acid_rainbow!

1) Solve the following equation for $0^o \leq \theta \leq 180^o$

Cos^2(x) / Sinx + Sin^2(x) = 2 . . . . What is in the denominator?

We have been told earlier that this is equal to $\frac{1-\sin x}{\sin x}$ . . . . Ah, this helps!

We have: . $\frac{\cos^2\!x}{\sin x + \sin^2\!x} \:=\:2 \quad\Rightarrow\quad 1 - \sin^2\!x \;=\;2(\sin x + \sin^2\!x)$

which simplifies to: . $3\sin^2\!x + 2\sin x - 1 \:=\:0\quad\Rightarrow\quad (\sin x + 1)(3\sin x - 1) \:=\:0$

Then: . $\sin x + 1 \:=\:0\quad\Rightarrow\quad\sin x \:=\:-1\quad\Rightarrow\quad x \:=\:270^o \quad\hdots$ not in the interval

And: . $3\sin x -1\:=\:0\quad\Rightarrow\quad\sin x \:=\:\frac{1}{3}\quad\Rightarrow\quad\boxed{ x \;\approx\;19.47^o,\;160.53^o}$

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