# Math Help - trigonometric equations

1. ## trigonometric equations

1. Prove that tan x sinx + cosx = 1/cosx

2. Prove that Tanx = sinx + sin²x/ cos x (1+ sinx)

3.prove that sinx - cosx = sinx-cosx

4. solve for x, where O◦ ≤ x ≤ 360◦

a) cosx (1-2sinx) = 0 (two answers)

b) 4sin²x-1 = 0 (two answers)

c) 2cosx + cosx -1 =0 (two answers)

d) 3sinx + 4sin x =4 (two answers)

I'm so lost with this stuff right now. I'm sorry its so much.

2. Originally Posted by musicismylife2007
1. Prove that tan x sinx + cosx = 1/cosx
$\tan x \sin x + \cos x = \frac {1}{ \cos x}$

Consider LHS

$\tan x \sin x + \cos x = \frac {\sin x}{ \cos x } \sin x + \cos x$

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

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

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

$= RHS$

Thus we have $LHS \equiv RHS$

Questions like these are not as hard as you may think. here are some tips:

- Start on the most complicated side, it gives you more options to change stuff
- It is usually a good idea to change everything to sines and cosines. these are the functions students are most familiar with and so it is easy to see connections
- Always work one side to get the other, if you can't, make one side as simple as possible, stop, and then try to work the other side to get the simplified expression.

Now, try the other proving questions

3. Originally Posted by musicismylife2007

3.prove that sinx - cosx = sinx-cosx

4. solve for x, where O◦ ≤ 360◦

a) cosx (1-2sinx) = 0 (two answers)

b) 4sin²x-1 = 0 (two answers)

c) 2cosx + cosx -1 =0 (two answers)

d) 3sinx + 4sin x =4 (two answers)
I see ' over some of the functions, should there be squares there? and some of the questions don't make sense unless they should be squares. double check to make sure you don't have typos, otherwise, it wastes everyone's time

4. Originally Posted by musicismylife2007
4. solve for x, where O◦ ≤ 360◦
you mean, where $0 \leq x \leq 360$

a) cosx (1-2sinx) = 0 (two answers)
if two numbers being multiplied give zero, it means one or the other has to be zero

$\cos x (1 - 2 \sin x ) = 0$

$\Rightarrow \cos x = 0$ or $1 - 2 \sin x = 0$

$\Rightarrow \cos x = 0$ or $\sin x = \frac {1}{2}$

If $\cos x = 0$, we have $x = \frac { \pi}{2}$ or $\frac {3 \pi}{2}$ = $90^{\circ}$ or $270^{\circ}$

If $\sin x = \frac {1}{2}$, we have $x = \frac { \pi}{6}$ or $\frac {5 \pi}{6}$ = $30^{\circ}$ or $150^{\circ}$

so there are your two sets of x-values

5. Sorry made a mistake. Umm this is the way the questions are writen I hope it makes more sence now.

6. Originally Posted by musicismylife2007
Sorry made a mistake. Umm this is the way the questions are writen I hope it makes more sence now.
so those apostrophes i see over the s's and n's mean nothing?

if so, why would they write $2 \cos x + \cos x - 1 = 0$?

wouldn't it be easier to write $3 \cos x - 1 = 0$ ?

are you sure it should not be $2 \cos^2 x + \cos x - 1 = 0$ ??

7. Thats what the question is. I wrote it out like that. Maybe i used a type of character you cant see???

8. $4sin^2x -1 = 0$
$3sin^2x + 4sinx = 4$

9. Originally Posted by musicismylife2007

4. solve for x, where O◦ ≤ 360◦

a) cosx (1-2sinx) = 0 (two answers)

b) 4sin²x-1 = 0 (two answers)

c) 2cosx + cosx -1 =0 (two answers)

d) 3sinx + 4sin x =4 (two answers)
a) $\cos x(1-2\sin x)=0$
$\cos x=0\Rightarrow x\in\left\{\frac{\pi}{2},\frac{3\pi}{2}\right\}$
$1-2\sin x=0\Rightarrow \sin x=\frac{1}{2}\Rightarrow x\in\left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}$

b) $4\sin^2x-1=0\Rightarrow\sin^2x=\frac{1}{4}\Rightarrow\sin x=\pm\frac{1}{2}$
If $\sin x=\frac{1}{2}\Rightarrow x\in\left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}$
If $\sin x=-\frac{1}{2}\Rightarrow x\in\left\{\frac{7\pi}{6},\frac{11\pi}{6}\right\}$

c) $2\cos^2x+\cos x-1=0$
Let $\cos x=t\Rightarrow 2t^2+t-1=0\Rightarrow t_{1,2}=\frac{-1\pm\sqrt{1+8}}{4}=\frac{-1\pm 3}{4}$
$t=-1\Rightarrow\cos x=-1\Rightarrow x=\pi$
$t=\frac{1}{2}\Rightarrow\cos x=\frac{1}{2}\Rightarrow x\in\left\{\frac{\pi}{3},\frac{5\pi}{3}\right\}$

d) $3\sin^2x+4\sin x=4$
Let $\sin x=t\Rightarrow 3t^2+4t-4=0\Rightarrow t_1=-2,t_2=\frac{2}{3}$
If $\sin x=-2\Rightarrow x\in\emptyset$
if $\sin x=\frac{2}{3}\Rightarrow x\in\left\{\sin^{-1}\left(\frac{2}{3}\right),\pi-\sin^{-1}\left(\frac{2}{3}\right)\right\}$