(Difficult) Trigonometry

Yes, Gabzinc is asking for help once again. Sorry.

Spoiler

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?

Here it is:

Thanks!

Scroll to see replies

Reply 1

the bear

divide each term by cosx....

Reply 2

Mr Moon Man

Divide by cos(x) then work out tan(x)= -1

(edited 7 years ago)

Reply 3

Notnek

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.

Reply 4

Mr Moon Man

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.

Reply 5

dididid

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.

Reply 6

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?

Reply 7

Ayman!

Original post by Gabzinc

They meant

\dfrac{\sin x + \cos x}{\cos x} = 0

, not the first equation

Reply 8

Notnek

Original post by Gabzinc

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.

Reply 9

NotNotBatman

Original post by Gabzinc

tanx = \frac{sinx}{cosx}

is what your given not what you need to solve.

you need to solve

sinx + cosx =0

Reply 10

Gabzinc

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:

Reply 11

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

Reply 12

Notnek

Original post by Gabzinc

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

?

Reply 13

Gabzinc

Original post by notnek

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 :frown:

Reply 14

Ayman!

Original post by Gabzinc

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?

(edited 7 years ago)

Reply 15

RDKGames

Study Forum Helper

Original post by Gabzinc

\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)

Reply 16

Notnek

Original post by Gabzinc

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.

Reply 17

Zacken

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 :frown:

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?

(edited 7 years ago)

Reply 18

Gabzinc

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 :frown:

(edited 7 years ago)

Reply 19

RDKGames

Study Forum Helper

Original post by Gabzinc

Glad to help :smile:

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