Gabzinc
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Yes, Gabzinc is asking for help once again. Sorry.
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Having some difficulty with an entrance exam question which evidently has something to do with trigonometry. Can anybody offer clues as to the first step in working?

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Thanks!
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the bear
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divide each term by cosx....
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Mr Moon Man
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Divide by cos(x) then work out tan(x)= -1
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Prince Philip
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(Original post by Mr Moon Man)
How is that difficult?
It's simple for A Level but this is an entrance exam for a GCSE student.

If you're a GCSE student and are not familiar with these type of questions then this is definitely not easy.
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Mr Moon Man
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(Original post by notnek)
It's simple for A Level but this is an entrance exam for a GCSE student.

If you're not familiar with this type of question and haven't seen the identity before then this is definitely not easy.
Oh, I didn't read that part, my bad.
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dididid
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(Original post by Mr Moon Man)
Oh, I didn't read that part, my bad.
don't think the op mentioned it anywhere on this thread anyway lol.
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Gabzinc
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Thanks for the help so far guys, but I'm still a little confused :3

so dividing each side by cos x gives me:

tan x/cos x = sin x/cos^2 x

then using tan x = -1 gives me x= -45 (why do I do this?)

so both sides are equal to -root 2....

and then I'm lost. Sorry if I didn't make it clear, yes I'm a GCSE student. I don't understand this even with the new GCSE syllabus which has been made quite a bit harder :/

any more tips guys?
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Ayman!
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(Original post by Gabzinc)
Thanks for the help so far guys, but I'm still a little confused :3

so dividing each side by cos x gives me:

tan x/cos x = sin x/cos^2 x

then using tan x = -1 gives me x= -45 (why do I do this?)

so both sides are equal to -root 2....

and then I'm lost. Sorry if I didn't make it clear, yes I'm a GCSE student. I don't understand this even with the new GCSE syllabus which has been made quite a bit harder :/

any more tips guys?
They meant \dfrac{\sin x + \cos x}{\cos x} = 0, not the first equation
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Prince Philip
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(Original post by Gabzinc)
Thanks for the help so far guys, but I'm still a little confused :3

so dividing each side by cos x gives me:

tan x/cos x = sin x/cos^2 x

then using tan x = -1 gives me x= -45 (why do I do this?)

so both sides are equal to -root 2....

and then I'm lost. Sorry if I didn't make it clear, yes I'm a GCSE student. I don't understand this even with the new GCSE syllabus which has been made quite a bit harder :/

any more tips guys?
You divided the wrong equation by \cos x. Try dividing this equation by \cos x:

\sin x + \cos x = 0.

Try that and see what happens. Post you working if you get stuck.
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Gabzinc
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(Original post by Ayman!)
They meant \dfrac{\sin x + \cos x}{\cos x} = 0, not the first equation
(Original post by notnek)
You divided the wrong equation by \cos x. Try dividing this equation by \cos x:

\sin x + \cos x = 0.

Try that and see what happens. Post you working if you get stuck.
Ohhh, right, will do. Doing the problem now:
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NotNotBatman
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(Original post by Gabzinc)
Thanks for the help so far guys, but I'm still a little confused :3

so dividing each side by cos x gives me:

tan x/cos x = sin x/cos^2 x

then using tan x = -1 gives me x= -45 (why do I do this?)

so both sides are equal to -root 2....

and then I'm lost. Sorry if I didn't make it clear, yes I'm a GCSE student. I don't understand this even with the new GCSE syllabus which has been made quite a bit harder :/

any more tips guys?
No  tanx = \frac{sinx}{cosx} is what your given not what you need to solve.

you need to solve  sinx + cosx =0
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Gabzinc
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Riiiiiigghhht... making progress I guess?

(sin x + cos x)/cos x = 0, so sin x /cos x + cos x/ cos x = 0

if tan x = -1, x must be -45

sin -45 / cos -45 = -1

cos -45/ cos -45 must be equal to one as anything divided by itself is 1.

-1 + 1 = 0!

so one the values are -45,
then +180 to make it fit the range, right?

so 45 and 135 are the answers, right? i hope
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Prince Philip
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(Original post by Gabzinc)
Riiiiiigghhht... making progress I guess?

(sin x + cos x)/cos x = 0, so sin x /cos x + cos x/ cos x = 0

if tan x = -1, x must be -45

sin -45 / cos -45 = -1

cos -45/ cos -45 must be equal to one as anything divided by itself is 1.

-1 + 1 = 0!

so one the values are -45,
then +180 to make it fit the range, right?

so 45 and 135 are the answers, right? i hope
tan x = -1 is correct but I feel you've got that from another post and not derived it yourself. Let me know if I'm wrong.

Can you see how to get from here:

sin x /cos x + cos x/ cos x = 0

to here:

tan x = -1

?
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Gabzinc
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(Original post by notnek)
tan x = -1 is correct but I feel you've got that from another post and not derived it yourself. Let me know if I'm wrong.

Can you see how to get from here:

sin x /cos x + cos x/ cos x = 0

to here:

tan x = -1

?
Yeah, someone mentioned tan x = -1 so I just put -1 into inverse tan on my calculator. But admittedly I have no idea how to get tan x = -1 from (sin x+ cos x)/cos x

the worst bit is that there are questions like this all over the paper. If I can't do it while spending an hour on it at home, I won't be able to do it in the exam
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Ayman!
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(Original post by Gabzinc)
Yeah, someone mentioned tan x = -1 so I just put -1 into inverse tan on my calculator. But admittedly I have no idea how to get tan x = -1 from (sin x+ cos x)/cos x

the worst bit is that there are questions like this all over the paper. If I can't do it while spending an hour on it at home, I won't be able to do it in the exam
Do you know how to find least common multiples?

Remember that \dfrac{a}{b} + \dfrac{c}{b} = \dfrac{a+c}{b}.

I can essentially say that for an example where x = 2, x =2 = 1 + 1 \iff \dfrac{x}{2} = \dfrac{1+1}{2} = \dfrac{1}{2} + \dfrac{1}{2}.

Do you get how that works?
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RDKGames
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(Original post by Gabzinc)
Yeah, someone mentioned tan x = -1 so I just put -1 into inverse tan on my calculator. But admittedly I have no idea how to get tan x = -1 from (sin x+ cos x)/cos x

the worst bit is that there are questions like this all over the paper. If I can't do it while spending an hour on it at home, I won't be able to do it in the exam
\sin(x)+\cos(x)=0

Divide each term by \cos(x), assuming \cos(x) \not= 0

\frac{\sin(x)}{\cos(x)}+\frac{ \cos(x) }{\cos(x)}=\frac{0}{\cos(x)}

Then the first term becomes \tan(x) from the identity, and the second becomes 1, and the third is still 0.

So you're left with \tan(x)+1=0 and rearrange for \tan(x)
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Prince Philip
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(Original post by Gabzinc)
Yeah, someone mentioned tan x = -1 so I just put -1 into inverse tan on my calculator. But admittedly I have no idea how to get tan x = -1 from (sin x+ cos x)/cos x

the worst bit is that there are questions like this all over the paper. If I can't do it while spending an hour on it at home, I won't be able to do it in the exam
Try to ignore other people's working and do the question yourself.

So you got to this equation which is correct:

sin x /cos x + cos x/ cos x = 0

There are two terms in this equation and both of them can be changed/simplified.

sinx / cosx is ?

cosx / cosx is ?

Try this and again post your thoughts if you get stuck.
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Zacken
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(Original post by Gabzinc)
If I can't do it while spending an hour on it at home, I won't be able to do it in the exam
You start off by spending an hour on it at home - doesn't mean these sort of things will always take you an hour at home to do; with practice and understanding you'll be able to do it a lot quicker.

Anywho: from \displaystyle \frac{\sin x+ \cos x}{\cos x} = \frac{0}{\cos x} you get \displaystyle \frac{\sin x}{\cos x} + \frac{\cos x}{\cos x} = 0

Can you see why this is true? (rememember that \frac{a+b}{c} = \frac{a}{c} + \frac{b}{c})

Then, can you simplify this? If so, what do you get?
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Gabzinc
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(Original post by notnek)
...
(Original post by RDKGames)
...
omg. It was in my face the whole time and I STILL didn't see this?
HUGE FACEPALM MOMENT

At least now in exams I won't make the same fatal mistake.

so sin x/cos x = tan x (as given in the formula)

and cos x/ cos x = 1

tan x + 1 =0
tan x = -1
x= -45

I'm so upset that I did not recognise this earlier!

Thanks a whole lot guys, but I have just one remaining question: why would you divide (sin x+ cos x) by cos x in the first place? Just in case a similar question comes up in the future.


(Original post by Zacken)
You start off by spending an hour on it at home - doesn't mean these sort of things will always take you an hour at home to do; with practice and understanding you'll be able to do it a lot quicker.
True. I am also planning to practice a little but of A/AS Level trig to help with my understanding a little.

EDIT: Also, I want to give out more rep to everyone but I can't rep you guys multiple times in a row
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RDKGames
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(Original post by Gabzinc)
omg. It was in my face the whole time and I STILL didn't see this?
HUGE FACEPALM MOMENT

At least now in exams I won't make the same fatal mistake.

so sin x/cos x = tan x (as given in the formula)

and cos x/ cos x = 1

tan x + 1 =0
tan x = -1
x= -45

I'm so upset that I did not recognise this earlier!

Thanks a whole lot guys, but I have just one remaining question: why would you divide (sin x+ cos x) by cos x in the first place? Just in case a similar question comes up in the future.
Glad to help You divide by cosine because then you have an identity you can use, it's just a decision made by inspecting the equation for long enough. It helps because then you simplify your equation down to a single variable of \tan(x) for which you can solve.

But, er, while x=-45 is true for the equation to hold, it is not a valid solution for this question because the range for it is 0\leq x \leq 180. Are you able to hack around this one??
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