The Student Room Group

Core 2 ONE mark question... Driving me insane :|

Sin pheta + cos pheta = 0
Show that tan pheta = -1

Can someone quickly solve this, so I can move on with my life and look beyond the 30 minutes I've wasted... :s-smilie:

And er, thanks in advance.
remember that sin/cos=tan...
Reply 2
Original post by Contrad!ction.
remember that sin/cos=tan...

this.


Most likely what' going wrong is that you've forgotten that cancelling out by dividing by something leaves a 1 not a 0. I'm guessing you're arriving at tan(theta) = 0 ?

it's a mistake that I admittedly made quite a few silly times at the start of my maths A-level :frown:
(edited 11 years ago)
Original post by MSI_10
Sin pheta + cos pheta = 0
Show that tan pheta = -1

Can someone quickly solve this, so I can move on with my life and look beyond the 30 minutes I've wasted... :s-smilie:

And er, thanks in advance.


remember that sin/cos = tan

so use that fact to move the cos pheta across to give

sin = - cos

divide through by -cos

-tan = 1

therefore multiply by -1 to get tan as positive. simples..
isn't it theta
sin + cos = 0
sin = -cos
sin/cos = -1
tan = -1
Reply 6
Hmm thanks all :smile:

Is it also possible to multiply the starting equation -1, then end up with tan theta=-1?
sin theta + cos theta = 0

-sin theta - cos theta = 0
-sin theta=1+cos theta
-sin theta/ cos theta = 1
-tan theta=1

tan theta=-1

After posting, I did that and got it right.
Unless it was a fluke :s-smilie:
Reply 7
Original post by MSI_10
Hmm thanks all :smile:

Is it also possible to multiply the starting equation -1, then end up with tan theta=-1?
sin theta + cos theta = 0

-sin theta - cos theta = 0
-sin theta=1+cos theta
-sin theta/ cos theta = 1
-tan theta=1

tan theta=-1

After posting, I did that and got it right.
Unless it was a fluke :s-smilie:


Where did the 1 come from in the second line? It shouldn't be there. You should just have -sin theta=cos theta
Reply 8
Original post by Gemini92
Where did the 1 come from in the second line? It shouldn't be there. You should just have -sin theta=cos theta


Oh okay thanks.

So if I didn't make that mistake, it would still work right?

So to summarize, my mistake was that when doing sin theta / cos theta, the right hand side gets a +1 NOT a 0?
Reply 9
Original post by MSI_10
Oh okay thanks.

So if I didn't make that mistake, it would still work right?

So to summarize, my mistake was that when doing sin theta / cos theta, the right hand side gets a +1 NOT a 0?


Yes.

If you divide cos theta by cos theta you get one :tongue:

Same as if you divide anything most things by themselves :smile:
(edited 11 years ago)
Original post by wibletg
Yes.

If you divide cos theta by cos theta you get one :tongue:

Same as if you divide anything by itself :smile:


0/0 doesn't equal 1.
Original post by Mr M
0/0 doesn't equal 1.


haha, fair point
Reply 12
Original post by Mr M
0/0 doesn't equal 1.


I was tempted to put most things, infinity divided by infinity isn't one either :tongue:
Reply 13
Original post by IamBeowulf
isn't it theta


Nah, it's a new Greek letter this. :biggrin:

sin( \sin(
Unparseable latex formula:

\cjRL{\varphi\put(0,0){\put(0,0){ \put(-5,0){\theta}}}}

)+cos() + \cos(
Unparseable latex formula:

\cjRL{\varphi\put(0,0){\put(0,0){ \put(-5,0){\theta}}}}

)=0) = 0

Pheta. That's how it's made:   φθ\ \ \varphi \theta \ \rightarrow
Unparseable latex formula:

\ \cjRL{\varphi\put(0,0){\put(0,0){ \put(-1,0){\theta}}}}

 \ \rightarrow
Unparseable latex formula:

\ \cjRL{\varphi\put(0,0){\put(0,0){ \put(-2,0){\theta}}}}

 \ \rightarrow
Unparseable latex formula:

\ \cjRL{\varphi\put(0,0){\put(0,0){ \put(-3,0){\theta}}}}

 \ \rightarrow
Unparseable latex formula:

\ \cjRL{\varphi\put(0,0){\put(0,0){ \put(-4,0){\theta}}}}

 \ \rightarrow
Unparseable latex formula:

\ \cjRL{\varphi\put(0,0){\put(0,0){ \put(-5,0){\theta}}}}

Original post by gff
Nah, it's a new Greek letter this. :biggrin:

sin( \sin(
Unparseable latex formula:

\cjRL{\varphi\put(0,0){\put(0,0){ \put(-5,0){\theta}}}}

)+cos() + \cos(
Unparseable latex formula:

\cjRL{\varphi\put(0,0){\put(0,0){ \put(-5,0){\theta}}}}

)=0) = 0

Pheta. That's how it's made:   φθ\ \ \varphi \theta \ \rightarrow
Unparseable latex formula:

\ \cjRL{\varphi\put(0,0){\put(0,0){ \put(-1,0){\theta}}}}

 \ \rightarrow
Unparseable latex formula:

\ \cjRL{\varphi\put(0,0){\put(0,0){ \put(-2,0){\theta}}}}

 \ \rightarrow
Unparseable latex formula:

\ \cjRL{\varphi\put(0,0){\put(0,0){ \put(-3,0){\theta}}}}

 \ \rightarrow
Unparseable latex formula:

\ \cjRL{\varphi\put(0,0){\put(0,0){ \put(-4,0){\theta}}}}

 \ \rightarrow
Unparseable latex formula:

\ \cjRL{\varphi\put(0,0){\put(0,0){ \put(-5,0){\theta}}}}



That's cool. Unfortunately PRSOM.
Original post by wibletg
I was tempted to put most things, infinity divided by infinity isn't one either :tongue:


0/0 isn't infinity, it's solution set is.
Original post by 122025278
0/0 isn't infinity, it's solution set is.


I don't think he suggested it was.

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