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How does cos alpha divided by cos theta = cos alpha?

This is from an MEI 2006 C4 paper, in the mark scheme they have divided through by cos theta, but I can't understand how this leaves you with just tan theta, cos alpha, not tan theta cos alpha/cos theta?

thanks!
Original post by mathshelplol
This is from an MEI 2006 C4 paper, in the mark scheme they have divided through by cos theta, but I can't understand how this leaves you with just tan theta, cos alpha, not tan theta cos alpha/cos theta?

thanks!


What you are saying is the equivalent of saying xyy=xyyy\frac{xy}{y}=\frac{x}{y}\cdot \frac{y}{y}

Clearly it is wrong.
(edited 6 years ago)
Original post by RDKGames
What you are saying is the equivalent of saying xyy=xyyy\frac{xy}{y}=\frac{x}{y}\cdot \frac{y}{y}

Clearly it is wrong.

If you were to take an odd and even number as being two separate values and one plus one then you know this wouldn't work here because it is arithmetic but cosine and sine are the sum of a perfect geometric series so it's possible they take the values x and y and you're proof might be correct as long as the values are in phase and y gives the value 1 all times.
Original post by Anfanny
If you were to take an odd and even number as being two separate values and one plus one then you know this wouldn't work here because it is arithmetic but cosine and sine are the sum of a perfect geometric series so it's possible they take the values x and y and you're proof might be correct as long as the values are in phase and y gives the value 1 all times.


What...?
Original post by Anfanny
If you were to take an odd and even number as being two separate values and one plus one then you know this wouldn't work here because it is arithmetic but cosine and sine are the sum of a perfect geometric series so it's possible they take the values x and y and you're proof might be correct as long as the values are in phase and y gives the value 1 all times.


:s-smilie:
Original post by mathshelplol
This is from an MEI 2006 C4 paper, in the mark scheme they have divided through by cos theta, but I can't understand how this leaves you with just tan theta, cos alpha, not tan theta cos alpha/cos theta?

thanks!

You were only asked to get to the end sometimes, and I checked with a guy using a program like Matlab and sometimes the questions aren't really solvable. But you just have to get to the end there you don't have to find the value.

It could be taken from another set of questions from a university book just know how to get a solution from the left hand side of a trig identity when it's given to you.
Original post by RDKGames
What...?


Original post by the bear
:s-smilie:


I have explained above.
Original post by Anfanny
I have explained above.


I'm not sure what you've explained, or tried to...
Original post by Anfanny
You were only asked to get to the end sometimes, and I checked with a guy using a program like Matlab and sometimes the questions aren't really solvable. But you just have to get to the end there you don't have to find the value.

It could be taken from another set of questions from a university book just know how to get a solution from the left hand side of a trig identity when it's given to you.


:hmmmm2:
Reply 9
Original post by Anfanny
I have explained above.


Your posts are complete rubbish to be frank!

Please don't troll other people's threads if you don't know what you're talking about - you will only confuse the OP.
Answering the part you don't understand, the only cos theta there was, was divided by itself so just became 1. The other functions with theta were sin and so divided by cos became tan. When you have two functions together and you divide by another you only divide it once so if one can be simplified by the division you show that.
(edited 6 years ago)
Original post by RDKGames
What you are saying is the equivalent of saying xyy=xyyy\frac{xy}{y}=\frac{x}{y}\cdot \frac{y}{y}

Clearly it is wrong.


patronising arse :smile:))
Original post by mathshelplol
patronising arse :smile:))


Well that is what you said you put it as.
Original post by davros
Your posts are complete rubbish to be frank!

Please don't troll other people's threads if you don't know what you're talking about - you will only confuse the OP.


You're not listening. I should have said,
Just learn how to get the right hand side from starting at the left hand sideIf you don't just follow the rules you'll confuse yourself.
Original post by Vikingninja
Well that is what you said you put it as.


I didn't say I put anything, I was asking for an explanation to how that method works as I don't understand it...
Reply 15
Original post by mathshelplol
I didn't say I put anything, I was asking for an explanation to how that method works as I don't understand it...



I'll use T for theta and A for alpha, so I don't have to worry about the Latex!

Are you confused where they go from:

sinTcosA + cosTsinA = 2sinT

to

tanTcosA + sinA = 2tanT ?

All they've done is divide both sides by cos T.

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