Trig equations

**Alan306090** · January 17th 2010, 02:44 PM

I accidently posted this in the Pre-Calc section, if possible can someone delete that?

How do I solve these

Cos^2x + 2sinxcosx - sin^2 x = 0 [0, 360)

sin(x +1) = cos x [0, 360)

Cos 4X - Cos 2X = 0 but range is [-180, 180]

And finally

Logan has a garden in the shape of a sector of a circle; outer rim of garden is 25 ft long and central angle along of sector is 50 degrees, she wants to add a 3 ft wide walk to the outer rim, how many sq ft of paving does she need to build it?

**Prove It** · January 17th 2010, 05:21 PM

Originally Posted by Alan306090

I accidently posted this in the Pre-Calc section, if possible can someone delete that?

How do I solve these

Cos^2x + 2sinxcosx - sin^2 x = 0 [0, 360)

sin(x +1) = cos x [0, 360)

Cos 4X - Cos 2X but range is [-180, 180]

And finally

Logan has a garden in the shape of a sector of a circle; outer rim of garden is 25 ft long and central angle along of sector is 50 degrees, she wants to add a 3 ft wide walk to the outer rim, how many sq ft of paving does she need to build it?

1. $\cos^2{x} + 2\sin{x}\cos{x} - \sin^2{x} = 0$

Let $\cos{x} = X$ and $\sin{x} = Y$ , then

$X^2 + 2XY - Y^2 = 0$

$X^2 + 2XY + Y^2 - Y^2 - Y^2 = 0$

$(X + Y)^2 - 2Y^2 = 0$

$(X + Y)^2 = 2Y^2$

$X + Y = \pm \sqrt{2} Y$

$X = (-1 \pm \sqrt{2})Y$ .

Therefore

$\cos{x} = (-1 - \sqrt{2})\sin{x}$ or $\cos{x} = (-1 + \sqrt{2})\sin{x}$ .

Case 1:

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

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

$-\frac{1}{1 + \sqrt{2}} = \tan{x}$

$-\frac{1 - \sqrt{2}}{1 - 2} = \tan{x}$

$1 - \sqrt{2} = \tan{x}$

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

Now, using the half angle identity for tangent, we have

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

$\tan{\frac{45^\circ}{2}} = \frac{1 - \cos{45^\circ}}{\sin{45^\circ}}$

$= \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}$

$= \frac{\frac{\sqrt{2} - 1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}$

$= \sqrt{2} - 1$

So this means that the basic angle is $\frac{45^\circ}{2} = 22.5^\circ$ .

Since tangent is negative in the second and fourth quadrants, this means that $x = 157.5^\circ$ or $x = 337.5^\circ$ .

Now have a go for Case 2...

**Prove It** · January 17th 2010, 05:27 PM

Originally Posted by Alan306090

I accidently posted this in the Pre-Calc section, if possible can someone delete that?

How do I solve these

Cos^2x + 2sinxcosx - sin^2 x = 0 [0, 360)

sin(x +1) = cos x [0, 360)

Cos 4X - Cos 2X but range is [-180, 180]

And finally

Logan has a garden in the shape of a sector of a circle; outer rim of garden is 25 ft long and central angle along of sector is 50 degrees, she wants to add a 3 ft wide walk to the outer rim, how many sq ft of paving does she need to build it?

2. $\sin{(x + 1)} = \cos{x}$ .

Use the angle sum identity for sine:

$\sin{(\alpha + \beta)} = \sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta}$

we have

$\sin{x}\cos{1} + \cos{x}\sin{1} = \cos{x}$

$\cos{x} - \cos{x}\sin{1} = \sin{x}\cos{1}$

$\cos{x}(1 - \sin{1}) = \sin{x}\cos{1}$

$\frac{1 - \sin{1}}{\cos{1}} = \frac{\sin{x}}{\cos{x}}$

$\frac{1 - \sin{1}}{\cos{1}} = \tan{x}$

$x = \arctan{\left(\frac{1 - \sin{1}}{\cos{1}}\right)}$ .

**Prove It** · January 17th 2010, 05:29 PM

Originally Posted by Alan306090

I accidently posted this in the Pre-Calc section, if possible can someone delete that?

How do I solve these

Cos^2x + 2sinxcosx - sin^2 x = 0 [0, 360)

sin(x +1) = cos x [0, 360)

Cos 4X - Cos 2X but range is [-180, 180]

And finally

Logan has a garden in the shape of a sector of a circle; outer rim of garden is 25 ft long and central angle along of sector is 50 degrees, she wants to add a 3 ft wide walk to the outer rim, how many sq ft of paving does she need to build it?

What is $\cos{4x} - \cos{2x}$ equal to? Or are you trying to find another expression for it?

**Prove It** · January 17th 2010, 06:35 PM

Originally Posted by Alan306090

I accidently posted this in the Pre-Calc section, if possible can someone delete that?

How do I solve these

Cos^2x + 2sinxcosx - sin^2 x = 0 [0, 360)

sin(x +1) = cos x [0, 360)

Cos 4X - Cos 2X but range is [-180, 180]

And finally

Logan has a garden in the shape of a sector of a circle; outer rim of garden is 25 ft long and central angle along of sector is 50 degrees, she wants to add a 3 ft wide walk to the outer rim, how many sq ft of paving does she need to build it?

Worded problem:

First, you need to find the radius of this sector.

To do this, remember that $C = 2\pi r$ .

In your case, since your angle is $50^\circ$ , this means you have $\frac{50}{360} = \frac{5}{36}$ of the circle.

So you have

$\frac{5C}{36} = \frac{5\cdot 2\pi r}{36}$

$\frac{5C}{36} = \frac{5\pi r}{18}$ .

You know that this part of the circumference is $25\,\textrm{ft}$ .

So $25\,\textrm{ft} = \frac{5\pi r}{18}$

$\frac{25\cdot 18}{5\pi}\,\textrm{ft} = r$

$r = \frac{90}{\pi}\,\textrm{ft}$ .

Now, you are planning to make a path around the outside of this sector.

If you draw it, you'll notice that you have a sector inside a sector.

The first sector has radius $\frac{90}{\pi}\,\textrm{ft}$ , and since the pavers are $3\,\textrm{ft}$ wide, the second sector has radius $\frac{90}{\pi} + 3\,\textrm{ft}$ .

$A_{\textrm{path}} = A_{\textrm{sector 2}} - A_{\textrm{sector 1}}$ .

$A_{\textrm{sector 1}} = \pi r^2$

$= \pi\left(\frac{90}{\pi}\,\textrm{ft}\right)^2$

$= \frac{8100}{\pi} \,\textrm{ft}^2$ .

$A_{\textrm{sector 2}} = \pi r^2$

$= \pi \left(\frac{90}{\pi} + 3\,\textrm{ft}\right)^2$

$= \pi \left(\frac{8100}{\pi^2} + \frac{540}{\pi} + 9\right) \,\textrm{ft}^2$

$= \left(\frac{8100}{\pi} + 540 + 9\pi\right)\,\textrm{ft}^2$ .

So $A_{\textrm{path}} = A_{\textrm{sector 2}} - A_{\textrm{sector 1}}$

$= \left(\frac{8100}{\pi} + 540 + 9\pi\right)\,\textrm{ft}^2 - \frac{8100}{\pi} \,\textrm{ft}^2$

$= \left(540 + 9\pi\right)\,\textrm{ft}^2$ .

**Alan306090** · January 17th 2010, 08:33 PM

It should be

Cos 4X - Cos 2X = 0 but range is [-180, 180]

And prove it, your answers don't seem to correlate to the answers in the back of the book which says:

3pi\8, 7pi\8, 11pi\8 and 15pi\8 for the first one
and .285, 3.427 for the second

And answer to the last should be 121 square ft

**Prove It** · January 17th 2010, 08:47 PM

Originally Posted by Alan306090

It should be

Cos 4X - Cos 2X = 0 but range is [-180, 180]

$\cos{4x} - \cos{2x} = 0$

$\cos^2{2x} - \sin^2{2x} - \cos{2x} = 0$

$\cos^2{2x} - (1 - \cos^2{2x}) - \cos{2x} = 0$

$2\cos^2{2x} - \cos{2x} - 1 = 0$ .

Let $\cos{2x} = X$ so that you have

$2X^2 - X - 1 = 0$

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

$2X(X - 1) + 1(X - 1) = 0$

$(X - 1)(2X + 1) = 0$

$X - 1 = 0$ or $2X + 1 = 0$

$X = 1$ or $X = -\frac{1}{2}$ .

Case 1:

$\cos{2x} = 1$

$2x = \left\{0^\circ, 180^\circ \right\} + 360^\circ n$ , where $n$ is an integer representing how many times you have gone around the unit circle.

$x = \left\{ 0^\circ, 90^\circ \right\} + 180^\circ n$

$x = \left\{-180^\circ , -90^\circ , 0^\circ, 90^\circ, 180^\circ \right\}$ .

Case 2:

$\cos{2x} = -\frac{1}{2}$ .

$\cos{X} = \frac{1}{2}$ when $X = 60^\circ$ , and cosine is negative in the second and third quadrants.

So

$2x = \left\{ 120^\circ, 240^\circ \right\} + 360^\circ n$

$x = \left\{ 60^\circ, 120^\circ \right\} + 180^\circ n$

$x = \left \{-120^\circ , -60^\circ , 60^\circ, 120^\circ \right\}$ .

Putting it together you have

$x = \left\{ -180^\circ, -120^\circ, -90^\circ, -60^\circ, 0^\circ, 60^\circ, 90^\circ, 120^\circ, 180^\circ \right \}$ .

**Alan306090** · January 18th 2010, 12:39 PM

Thank you prove it, though for Case 1,

The only time Cos2x = 1 [-180, 180] should be 0, no?

So then 2x = 0 + 360k

x = 0 + 180k

Then it should be, -180, 0, and 180? Does this seem right?

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