@raheem94 - there are a few errors in my book, and I can follow what you've written, it explains it really well. Trouble is, even if the book didn't have any errors, I'm not sure if I would have got there without your help. What steps did you go through to realsie what you had to do?

Reply 24

raheem94

Original post by gavinlee

First put the LHS over a common denominator and then just simplify it.

It will be hard for you at first, but remember PRACTICE MAKES PERFECT, just practice more and you will be fine with it.

(edited 12 years ago)

Reply 25

roar558

Original post by gavinlee

...but I'm getting confused because it's not

\frac{a}{b} + \frac{c}{d}

it's

\frac{a}{b} + \frac{b}{a}

p.s. I'm more explaining what I'm thinking rather than doubting your help :colondollar:

It doesn't really matter if b=c or a=d, use the same rule ie

\frac{a}{b} + \frac{b}{a}=\frac{b^2 +a^2}{ab}

Original post by gavinlee

The question copied from my book, word for word is:

Prove the identity

\frac {\sin \theta}{1 \cos \theta} + \frac {1 \cos \theta}{\sin \theta} \equiv \frac {2}{\sin \theta}

If that's the question, I think it's a typo, they clearly meant the question to be

\frac {\sin \theta}{1 \cos \theta} + \frac {1 \cos \theta}{\sin \theta} \equiv \frac {2}{\sin 2\theta}

Reply 26

Math12345

Original post by roar558

It doesn't really matter if b=c or a=d, use the same rule ie

\frac{a}{b} + \frac{b}{a}=\frac{b^2 +a^2}{ab}

If that's the question, I think it's a typo, they clearly meant the question to be

\frac {\sin \theta}{1 \cos \theta} + \frac {1 \cos \theta}{\sin \theta} \equiv \frac {2}{\sin 2\theta}

That's the same question? Unless I'm blind. :O

Reply 27

roar558

Original post by Math12345

That's the same question? Unless I'm blind. :O

na in the original question it was equal to

\frac{2}{sin \theta}

rather than

\frac{2}{sin 2\theta}

Reply 28

elldeegee

Original post by gavinlee

Hi, still on the trigonometrical equations…finding them quite difficult to be honest. Got this one here that I’m not sure where to begin with:

\frac {\sin \theta}{1 \cos \theta} + \frac {1 \cos \theta}{\sin \theta} \equiv \frac {2}{\sin \theta}

I’m really not being lazy but I have no idea where to begin with, I didn’t think the

\tan \theta \equiv \frac {\sin \theta}{\cos \theta}

identity would be useful as adding tan into the identity would just cause another variable, and I also didn’t think it would be the

\sin^2 \theta + \cos^2 \theta \equiv 1

identity as there are no squared values, so I’m lost!
:confused:

I'm going to walk through the solution (well, most of it) for you, as quite a few people here are wrong and are confusing matters.

\frac{sin\theta}{cos\theta}

\frac{cos\theta}{sin\theta} = \frac{2}{sin\theta}

this is the original, as stated.
rearrange to get the parts with sin as the denominator on one side:

\frac{sin\theta}{cos\theta} = \frac{2}{sin\theta}-\frac{cos\theta}{sin\theta}

NOW you can subtract the two fractions together because the denominator is the same.

\frac{sin\theta}{cos\theta} = \frac{2-cos\theta}{sin\theta}

Now "cancel out" the cos from the LHS by multiplying both asides by cos:

sin\theta = \frac{(2-cos\theta)cos\theta}{sin\theta}

expand the brackets on the RHS

sin\theta = \frac{2cos\theta - cos^2\theta}{sin\theta}

"cancel out" the sin from the RHS by multiplying by sin:

sin^2\theta=2cos\theta - cos^2\theta

now, rearrange to get:

sin^2\theta + cos^2\theta = 2cos\theta

***

Recall:

sin^2\theta+cos^2\theta = 1

so:
*** =

1 = 2cos\theta

\frac{1}{2}

= cos\theta

now, how do you get

\theta

from this?!

(edited 12 years ago)

Reply 29

elldeegee

Original post by J10

I went ahead to solve for theta and got cos(theta)=0.5
Anybody got to this?

I got this.

Original post by Roshniroxy

I would love to know the answer for this, as I'm confused on the same question

I've managed to "solve for

\theta

Reply 30

elldeegee

It wouldn't make sense for this to be a proof question. (if , in fact the original question was correct and it wasn't suppoed to have + after the 1s.)
for it to be a proof question, you'd expect both the LHS and the RHS to be the same.
but they will not become the same as the two things are not the same.

This can be shown just by putting in any value of

\theta

into the equation.

eg

\theta = 6

on the RHS this gives 9.6195 (to 4 dp)
on the LGS it gives19.1335 (to 4 dp)

which are not the same.

Reply 31

Mr M

Study Forum Helper

Original post by elldeegee

I'm going to walk through the solution (well, most of it) for you, as quite a few people here are wrong and are confusing matters.

Oh the irony!

Reply 32

elldeegee

Original post by raheem94

Is your question,

\displaystyle \frac{sin\theta}{cos\theta}+ \frac{cos\theta}{sin\theta} \equiv \frac2{sin2\theta} \ ?

Original post by roar558

It couldn't be that cause all you would end up with is an equation that says 1=2

no, it wouldnt be 1=2

\frac{sin\theta}{cos\theta}

\frac{cos\theta}{sin\theta} = \frac{2}{sin2\theta}

\frac{sin\theta}{cos\theta} + \frac{cos\theta}{sin\theta} = \frac{2}{2sin\theta cos\theta}

\frac{sin\theta}{cos\theta} + \frac{cos\theta}{sin\theta} = \frac{1}{sin\theta cos\theta}

multiply through by cos and sin gives:

sin^2\theta + cos^2\theta = 1

Which is true, and is therefore proven this would be proven.

(edited 12 years ago)

Reply 33

Mr M

Study Forum Helper

You would probably need to be psychic (or familiar with the OP's other posts) to know this but the actual question is:

Prove this identity:

\frac{\sin\theta}{1-\cos\theta}+\frac{1-\cos\theta}{\sin\theta} = \frac{2}{\sin\theta}

Reply 34

elldeegee

Original post by Mr M

You would probably need to be psychic (or familiar with the OP's other posts) to know this but the actual question is:

Prove this identity:

\frac{\sin\theta}{1-\cos\theta}+\frac{1-\cos\theta}{\sin\theta} = \frac{2}{\sin\theta}

Well, as i am not psychic, and i have obviously not seen his other posts, then i would NOT know this.

And as later on in the thread in the thread he reiterated the question word for word in his book, how would you expect me to know this?

There is no need to be rude about such things.

the original question, whilst having a

\equiv

sign, did not say anything about proving.
So i went with my instinct and solved for theta.

OP, i apologise if i have confused matters.

EDIT:
I have just read back through the posts, and have seen that his "reiteration" indeed mentioned proving, but I was originally going off of the original question, and had only glanced at the reiteration to be able to make sure the "equation" was indeed correct.

(edited 12 years ago)

Reply 35

Mr M

Study Forum Helper

Original post by elldeegee

There is no need to be rude about such things.

I didn't realise I was being rude. You should probably familiarise yourself with the Forum Rules on the provision of full solutions.

By the way, your style is poor. Try to work on the LHS or the RHS but not both at the same time.