# Math Help - Trig equations

1. ## Trig equations

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?

2. 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...

3. 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

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

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

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

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

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

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

4. 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?

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

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

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

6. 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

7. 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

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

8. 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?