# Trigonometry identity proof trouble. (Help much appreciated)

1. Jul 20, 2004

### KnowledgeIsPower

I can do pretty much all the proofs except these two, though i've tried to make a start. Note that in the questions the angle is given as theta, but for the sake of my keyboard i've changed it to angle 'x'.

Q15: Prove:
Tan2xsecx = 2sinxsec2x.
So I need to show the two sides are equal.
First i worked on the left hand side.
Using the double angle conversion formulae it becomes:
=2sinxsec2x
=(2tanx)/(1-tan^2 x)secx
=(2tanx)/(1-tan^2 x)1/cosx
=(2tanx)/(1-tan^2 x)cosx
=(2tanx)/(cosx-tan^2cosx)

I find it difficult to find exactly the right form to convert back to double angles via manipulation so i converted the right hand side to single angles (to see what form i needed it in).
=2sinxsec2x
=2sinx multiplied by 1/(cos2x)
=2sinx multipled by 1/(cos^2x - sin^2x)
=2sinx/(cos^2x - sin^2x)

So far i have:
=(2tanx)/(cosx-tan^2cosx) = 2sinx/(cos^2x - sin^2x)
Is that correct so far and how can i go on to show they are exactly the same?

And the other one:
Q17: Prove that:
1/(cosx-sinx) - 1/(cosx+sinx) = 2sinxsec2x
The LHS consisted entirely of single angles so i left that for the time being and worked on the RHS.
=2sinxsec2x
=2sinx multiplied by 1/(cos2x)
=2sinx multiplied by 1/(cos^2x - sin^2x)
=2sinx/(cos^2x - sin^2x)
=2sinx/(1-sin^2x-sin^2x)
=2sinx/(1-2sin^x)

And i have no idea where to go from there.
Any help on either of the questions would be much appreciated.

2. Jul 20, 2004

### Muzza

First one: work on the LHS, rewrite everything in terms of sine and cosine and apply the double angle formula for sin(2x)...

Second one: again, work on the LHS. Rewrite everything as a single fraction. I'll get you started:

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

$$\frac{(cos(x) + sin(x))}{ (cos(x) - sin(x))(cos(x) + sin(x)) } - \frac{(cos(x) - sin(x))}{(cos(x) + sin(x))(cos(x) - sin(x))}$$

Last edited: Jul 20, 2004
3. Jul 20, 2004

### NateTG

Have you tried using:
$$\tan \theta=\frac{\sin \theta}{\cos \theta}$$
and
$$\sec \theta=\frac{1}{\cos \theta}$$
?

Try simplifying the LHS.

4. Jul 20, 2004

### AKG

$$\tan (2x) \sec x = 2 \sin x \sec (2x)$$

$$\frac{\sin (2x)}{\cos (2x)} \sec x = \frac{2 \sin x}{\cos (2x)}$$

$$\frac{2 \sin x \cos x}{\cos (2x) \cos x} = \frac{2 \sin x}{\cos (2x)}$$

$$\frac{2 \sin x}{\cos (2x)} = \frac{2 \sin x}{\cos (2x)}$$

Q.E.D.
$$\frac{1}{\cos x - \sin x} - \frac{1}{\cos x + \sin x}$$

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

$$= \frac{2\sin x}{\cos (2x)}$$

$$= 2\sin x \sec (2x) =\mbox{ R.H.S.}$$

Q.E.D.

5. Jul 20, 2004

### KnowledgeIsPower

So for the first one, i had previously:
(2tanx)/(cosx-tan^2cosx)
Replace tan with sin/cos.
Move the divide sign, turn everything under the divide upside down and multiply it with the original top side.
Forming:
2sinx/cosx x 1/(cosx-(sin^2x/cos^2x)cosx).
=2sinx/(cos^2x-sin^2x)
=2sinx/cos2x
=2sinxsec2x.
=Done, thanks for the help.

6. Jul 20, 2004

### KnowledgeIsPower

Now for the second one. Did you do simply find the common denominator to transfer 1/(cosx-sinx) - 1/(cosx+sinx) to ((cosx+sinx - (cosx-sinx))/(cos^2x-sin^2x) ?
If so I see how to do that one now.
Thanks to everyone for the input.

I'll have to look up how to write the more understandable posts. Sorry mine look terrible.

7. Jul 21, 2004

### KnowledgeIsPower

I did a heap more trigonomic identity proofs last night and worked through them all pretty sucessfully, thanks to the users here.
However, i'm now solving some equations and came across a curious question. I've worked out the correct answer and know how to do it but i thought there was also another mechanism to do it. However, what i thought was an alternative gives an incorrect answer.
I will just list the incorrect path that i was trying at first, any idea why it's wrong?
I'm not sure why:

2sinxcosx=1-2sin^2x
4sin^2xcos^2x=1+4sin^4x
2sin^2x(1-sin^2x)=1+4sin^4x
2sin^2x-2sin^4x=1+4sin^4x
2sin^2x=1+4sin^4x+2sin^4x
2sin^2 x = 1 + 6sin^4 x
2sinx=1+6sin^2 x
6sin^2 x - 2sinx + 1 = 0

b2-4ac < 0
No solution.
What's wrong with this path? I thought you could square and square root everything on both sides of the equals sign.

I know you gain the correct answer by subbing 1-2sin^2 x as cos2x.

8. Jul 21, 2004

### AKG

Did you try to square both sides? Seems like it. Anyways, if that's what you did, remember:

(a + b)² = a² + 2ab + b², not a² + b²

9. Jul 22, 2004

### KnowledgeIsPower

Ahh yes, i forgot it should be bracketed.
Simple yet stupid error, thanks very much for the correction.
I'm just working through the books by myself over the summer as i'm finding it enjoyable/interesting and want to gain a headstart for the next year. Thus, i have no tutors to ask and you people are invaluable.
I can scarcely believe 6 months ago i didn't know the basic sin=o/h etc.
By the end of the summer i hope to be doing integration by parts, chain rule etc.

10. Jul 23, 2004

### KnowledgeIsPower

Sorry for the trouble but i have encountered a small problem with another question.

Q20: Solve 0<x<2pi
20sin(x/2) - 21cos(x/2) = 14.5
I can get the correct answer by multiplying the whole equation by -1, then putting it in the form Rcos(x+a)=-14.5 ((29cos(x/2 +.761) = 14.5) and solving to give 2.67 (2dp)

However, i first tried a different method and i'm not sure why it's wrong.

20sin(x/2) - 21cos(x/2) = 14.5
-21cos(x/2) + 20sin(x/2) = 14.5
a=-21=rcosa
b=20=-rsina
rcosa=-21, rsina=-20
tana = -20/-21, a = .761
R = root(a^2 + b^2) = 29
Which gives: 29cos(x/2 + 0.761) = 14.5 (evidently different from the above, correct version)
This solves to give an incorrect answer.

Why does the equation not work when the coefficient of cos (a) is < 0. Is it a mistake i've done or simply a perequisite of the formulae for a to be > 0?

Just 1 more page of trig to go after 40 pages of it -_-;;.

11. Jul 23, 2004

### arildno

Well, from the looks of it, you must use the 4.quadrant angle in the tangent expression, not the 1.quadrant value 0.761

12. Jul 23, 2004

### arildno

Oops; I meant of course 3.quadrant (where bothe the cosine and sine are negative)!!

13. Jul 23, 2004

### KnowledgeIsPower

If a = 0.761 (and it's a tan angle)
Then in the third quadrant it's still positive so. 0.761+pi = 3.903
So i would have 29cos(x/2 + 3.903) = 14.5
cos(x/2 + 3.903) = 0.5
x/2 + 3.903 = 1.047 or 4.712.
4.712 - 3.903 = 0.809
x = 1.62

It's strange. I don't see why it solves when the coefficient of cos is positive and not when it isn't.
Sorry for being a pain in the backside.

14. Jul 23, 2004

### arildno

But, you DO get the right answer:
We have:
$$29\cos(\frac{x}{2}+0.761+\pi)=14.5$$
But:
$$\cos(w+\pi)=-\cos(w)$$
Or:
$$29\cos(\frac{x}{2}+0.761)=-14.5$$

the reason why it screwed up in the last post, is that you have used an inverse cosine function defined on $$0,\pi$$ but this cannot be used for angles in the 3.(and 4.) quadrants.

EDIT: Corrected a grievous mistake!!

15. Jul 23, 2004

### KnowledgeIsPower

Ahh, i see. Because cos pi = -1.
But how was i supposed to know i had to put it in the third quadrant? Are there any particular rules i should be aware of?
I guess it's just easier to ensure the coefficient of cosx > 0.
Thanks for the help!. That's trigonometry finished (for this book).
Now onto review exercise 1, where i get 98 exam questions waiting for me -_-;;
.

16. Jul 23, 2004

### KnowledgeIsPower

No worries =). Thanks for all the help.
Also, you can delete a post. When you click edit there's an option to delete it.
Have a nice evening.