# Thread: URGENT HELP!!! lol (Trig Problems)

1. ## URGENT HELP!!! lol (Trig Problems)

1. (sinx/1+sinx) - (sinx/1-sinx)

2. tanx(cotx-cosx) = ?

3. cos4x

4. (tanx+1)(2sinx-sqrt3) = 0

5. sin2x + sinx = 0

6. sin^2(2)x=1

If yall could help me that would be great, i really just kinda suck at all of this, im more of a visual learner if you get what im saying.

2. Originally Posted by math/retarded
1. (sin/1+sin) - (sin/1-sin)

2. tan(cot-cos) = ?

3. cos4

4. (tanx+1)(2sinx-sqrt3) = 0

5. sin2x + sinx = 0

6. sin^2(2)x=1

If yall could help me that would be great, i really just kinda suck at all of this, im more of a visual learner if you get what im saying.
A lot of what you've posted is nonsense, because $\displaystyle \sin, \cos, \tan, \cot, \csc, \sec$ have no meaning unless you have a number or variable attached.

Also, you haven't specified what you wanted to do with these equations.

I'm assuming for 1 and 2, you're using identites to simplify, and for 4, 5, 6, you're solving the equations.

1. $\displaystyle \frac{\sin{x}}{1 + \sin{x}} - \frac{\sin{x}}{1 - \sin{x}}$.

Here multiply the top and bottom of each term by a cleverly disguised 1.

$\displaystyle = \frac{\sin{x}}{1 + \sin{x}}\times \frac{1 - \sin{x}}{1 - \sin{x}} - \frac{\sin{x}}{1 - \sin{x}}\times \frac{1 + \sin{x}}{1 + \sin{x}}$

$\displaystyle = \frac{\sin{x}(1 - \sin{x})}{(1 + \sin{x})(1 - \sin{x})} - \frac{\sin{x}(1 + \sin{x})}{(1 - \sin{x})(1 + \sin{x})}$

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

$\displaystyle = \frac{\sin{x} - \sin^2{x} - \sin{x} - \sin^2{x}}{1 - \sin^2{x}}$

$\displaystyle = -\frac{2\sin^2{x}}{\cos^2{x}}$

$\displaystyle = -2\tan^2{x}$.

3. Originally Posted by math/retarded
1. (sin/1+sin) - (sin/1-sin)

2. tan(cot-cos) = ?

3. cos4

4. (tanx+1)(2sinx-sqrt3) = 0

5. sin2x + sinx = 0

6. sin^2(2)x=1

If yall could help me that would be great, i really just kinda suck at all of this, im more of a visual learner if you get what im saying.
2. $\displaystyle \tan{x}(\cot{x} - \cos{x}) = \frac{\sin{x}}{\cos{x}}\left(\frac{\cos{x}}{\sin{x }} - \cos{x}\right)$

$\displaystyle = 1 - \sin{x}$.

3. Just put it into the calculator.

4. $\displaystyle (\tan{x} + 1)(2\sin{x} - \sqrt{3}) = 0$

By the null factor law $\displaystyle \tan{x} + 1 = 0$ or $\displaystyle 2\sin{x} - \sqrt{3} = 0$.

Case 1: $\displaystyle \tan{x} + 1 = 0$

$\displaystyle \tan{x} = -1$

$\displaystyle x = \left\{\pi - \frac{\pi}{4}, 2\pi - \frac{\pi}{4}\right\} + 2\pi n, n \in \mathbf{Z}$

$\displaystyle x = \left\{\frac{3\pi}{4}, \frac{7\pi}{4}\right\} + 2\pi n, n\in \mathbf{Z}$.

Case 2: $\displaystyle 2\sin{x} - \sqrt{3} = 0$

$\displaystyle 2\sin{x} = \sqrt{3}$

$\displaystyle \sin{x} = \frac{\sqrt{3}}{2}$

$\displaystyle x = \left\{\frac{\pi}{3}, \pi - \frac{\pi}{3}\right\} + 2\pi n, n \in \mathbf{Z}$

$\displaystyle x = \left\{\frac{\pi}{3}, \frac{2\pi}{3}\right\} + 2\pi n, n \in \mathbf{Z}$.

So putting them together

$\displaystyle x = \left\{ \frac{\pi}{3}, \frac{3\pi}{4}, \frac{2\pi}{3}, \frac{7\pi}{4}\right\} + 2\pi n, n \in \mathbf{Z}$.

4. Originally Posted by math/retarded
1. (sinx/1+sinx) - (sinx/1-sinx)

2. tanx(cotx-cosx) = ?

3. cos4x

4. (tanx+1)(2sinx-sqrt3) = 0

5. sin2x + sinx = 0

6. sin^2(2)x=1

If yall could help me that would be great, i really just kinda suck at all of this, im more of a visual learner if you get what im saying.
5. $\displaystyle \sin{(2x)} + \sin{x} = 0$

$\displaystyle 2\sin{x}\cos{x} + \sin{x} = 0$

$\displaystyle \sin{x}(2\cos{x} + 1) = 0$

Can you go from here?

6. $\displaystyle \sin^2{(2x)} = 1$

$\displaystyle \sin{(2x)} = \pm 1$.

Can you go from here?