1. ## Trigonometry

1. Verify the identity (sinx/1-cosx) -cotx=cscx

2. Show that the equation sin(x+y)=sin x + sin y is not an identity.

3. Write the product 11sin (x/2) cos (x/4) as a sum.

2. I will give you some hints, try them and see what you come up with Originally Posted by johntuan 1. Verify the identity (sinx/1-cosx) -cotx=cscx
- start with the left hand side
- put everything in terms of sine and cosine
- combine all terms into a single term (simplify)

2. Show that the equation sin(x+y)=sin x + sin y is not an identity.
weird question. start on the left hand side, expand it using the addition formula for sine. show that you can't get the right hand side from that

3. Write the product 11sin (x/2) cos (x/4) as a sum.
the product to sum formula for sine-cosine is:

$\displaystyle \sin A \cos B = \frac {1}{2} \big[ \sin (A + B) + \sin (A - B) \big]$

3. ok thanks I will try that. Can you tell me what each part does in this function f(x)=2tan(x+(pi/2))

4. That's a different question, give it its own thread.

5. for 1 I am stuck on this L.H.S. sinx/1-cosx - 1/tanx what can I do from here.

For 2 what would I do from here

sinxcosy+cosxsiny=sinx+siny

6. Originally Posted by johntuan for 1 I am stuck on this L.H.S. sinx/1-cosx - 1/tanx what can I do from here.
i said change everything in terms of sines and cosines did i not? what is tan x doing there. and you still have two terms, combine them into one

7. Originally Posted by johntuan sinxcosy+cosxsiny=sinx+siny
that's pretty much it, i think

in general, those two sides are not equal, only for particular values of x and y, which means they can't be identities

8. Originally Posted by Jhevon i said change everything in terms of sines and cosines did i not? what is tan x doing there. and you still have two terms, combine them into one
What else can cotx equal then if not 1/tanx?

9. $\displaystyle \displaystyle\frac{\sin x}{1-\cos x}-\frac{\cos x}{\sin x}=\frac{\sin^2x-\cos x+\cos^2x}{\sin x(1-\cos x)}=$
$\displaystyle \displaystyle =\frac{1-\cos x}{\sin x(1-\cos x)}=\frac{1}{\sin x}=\csc x$

10. Originally Posted by johntuan What else can cotx equal then if not 1/tanx?
$\displaystyle cot x = \frac{cos x}{sin x}$

AND

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

11. Originally Posted by johntuan What else can cotx equal then if not 1/tanx?
as red_dog pointed out:

remember that we have a formula for tanx itself in terms of sine and cosine

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

$\displaystyle \Rightarrow \cot x = \frac {1}{\tan x} = \frac {1}{ \frac {\sin x }{\cos x}} = \frac {\cos x }{\sin x}$

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