Proving Trig Identity: Tanx = Csc2x - Cot2x | Homework Help

[bubblygum]

Homework Statement

tanx=csc2x-cot2x

Homework Equations

Quotient, Reciprocal, Pythagoreans

The Attempt at a Solution

1/sinx + 1/sinx - cosx/sinx - cosx/sinx
= 2/sinx - 2cosx/sinx
= (2-2cosx)/sinx

STUCK~

 

[AdkinsJr]

Homework Statement

tanx=csc2x-cot2x

Homework Equations

Quotient, Reciprocal, Pythagoreans

The Attempt at a Solution

1/sinx + 1/sinx - cosx/sinx - cosx/sinx
= 2/sinx - 2cosx/sinx
= (2-2cosx)/sinx

STUCK~

$$tan(x)=csc(2x)-cot(2x)$$

You need to use double angle formulae too.

 

[eumyang]

Homework Statement

tanx=csc2x-cot2x

Homework Equations

Quotient, Reciprocal, Pythagoreans

The Attempt at a Solution

1/sinx + 1/sinx - cosx/sinx - cosx/sinx
= 2/sinx - 2cosx/sinx
= (2-2cosx)/sinx

STUCK~

I'm sorry, but this is incorrect. While you used the reciprocal identity:
$$\csc 2x = \frac{1}{\sin 2x}$$
The following is not true for all x:
$$\frac{1}{\sin 2x} \ne \frac{1}{\sin x} + \frac{1}{\sin x}$$
(I see students write things like this often. Why is that?)

As AdkinsJr suggested, use the double-angle identities. Have you learned them yet?

 

[bubblygum]

Thank you! Yes, I have learned the double angle ones, i'll give it a shot now.

 

[bubblygum]

I'm still stuck with :

tanx = csc2x - cot2x
RS
1/sin2x - cos2x/sin2x

1/2sinxcosx - cos^2x-sin^2x/2sinxcosx

1-cos^2x-sin^2x/2sinxcosx

1-(cosx+sinx)(cosx-sinx)/sinxcosx+sinxcosx

 

[eumyang]

I'm still stuck with :

tanx = csc2x - cot2x
RS
1/sin2x - cos2x/sin2x

1/2sinxcosx - cos^2x-sin^2x/2sinxcosx

1-cos^2x-sin^2x/2sinxcosx

1-(cosx+sinx)(cosx-sinx)/sinxcosx+sinxcosx

I'm having a difficult time reading this. (I suggest you learn LaTex.) It looks like you wrote the following:

$$\begin{aligned} \csc 2x - \cot 2x &= \frac{1}{\sin 2x} - \frac{\cos 2x}{\sin 2x} \\ &= \frac{1}{2\sin x \cos x} - \frac{\cos^2 x - \sin^2 x}{2\sin x \cos x} \\ &= \frac{1 - \cos^2 x - \sin^2 x}{2\sin x \cos x} \\ \end{aligned}$$
This is wrong. Watch your signs -- it should be a "+" in front of the sin²x.

 

[Mentallic]

The following is not true for all x:
$$\frac{1}{\sin 2x} \ne \frac{1}{\sin x} + \frac{1}{\sin x}$$
(I see students write things like this often. Why is that?)

It's a misunderstanding of both algebra and trigonometry.

For one, $$sin(2x)\neq 2sin(x) = sin(x)+sin(x)$$

and even more importantly (as if this first one wasn't important enough already)

$$\frac{1}{2x}\neq \frac{1}{x}+\frac{1}{x} =\frac{2}{x}$$