The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Maths C3 - Trigonometry... Help??

Scroll to see replies

Original post by ValerieKR
sec(theta)=-2.5
cos(theta)=1/sec(theta)=1/(-2.5)=-0.4

Thank you so much!!
:smile:
Ok so I think my Maths knowledge is lacking... :/

Rewrite the following as a power of sec, cosec, or cot
1−sin⁡2θsin⁡2θ\frac{1-\sin^2\theta}{\sin^2 \theta}

I realise now that I'm suppose to factorise but what from this equation indicates that I am meant to start off by doing that? :frown:I feel like I don't always know what I'm suppose to start off doing when solving equations. :frown:
I know the whole BODMAS thingy but I don't know how to always apply it.
Original post by RDKGames
Eating lunch while doing maths has been scientifically been proven to lead to confusion. Don't do it. :smile:
source?
Original post by Philip-flop
Ok so I think my Maths knowledge is lacking... :/

Rewrite the following as a power of sec, cosec, or cot
1−sin⁡2θsin⁡2θ\frac{1-\sin^2\theta}{\sin^2 \theta}

I realise now that I'm suppose to factorise but what from this equation indicates that I am meant to start off by doing that? :frown:I feel like I don't always know what I'm suppose to start off doing when solving equations. :frown:
I know the whole BODMAS thingy but I don't know how to always apply it.


1-sin^2(x)=cos^2(x)


You should know those identity and you should be able to carry on with that
Original post by metrize
1-sin^2(x)=cos^2(x)


You should know those identity and you should be able to carry on with that


Thank you!! I forgot about that identity!! Managed to work out that the answer is cot⁡2θ\cot^2 \theta
Original post by Philip-flop
Thank you!! I forgot about that identity!! Managed to work out that the answer is cot⁡2θ\cot^2 \theta


No offense but that is not an identity you 'forget'. Many others are derived straight from sin⁡2x+cos⁡2x≡1\sin^2x+\cos^2x \equiv 1. Try not to forget it next time.
(edited 7 years ago)
Original post by RDKGames
No offense but that is not an identity you 'forget'. Many others are derived straight from sin⁡2x+cos⁡2x≡1\sin^2x+\cos^2x \equiv 1. Try not to forget it next time.


You're right! It's not an identity I forgot, it's an identity that I didn't spot to use for this example. Obviously my AS-level Maths knowledge is a bit rusty after the summer (although my knowledge was never amazing :P) I will do better to try and think more strategically instead of panicking and thinking that I can't solve something :smile:
Ok, so yet again I find myself struggling with Trig Identities :frown: I've just come across this question which is...

Q) Write down the value of cotx ....
5sinx=4cosx5sinx=4cosx

I don't seem to know what to start with first :frown:
Reply 28
Avatar for B_9710
B_9710
18
Original post by Philip-flop
Ok, so yet again I find myself struggling with Trig Identities :frown: I've just come across this question which is...

Q) Write down the value of cotx ....
5sinx=4cosx5sinx=4cosx

I don't seem to know what to start with first :frown:


Find sin⁡x/cos⁡x \sin x / \cos x and then take reciprocal.
Original post by B_9710
Find sin⁡x/cos⁡x \sin x / \cos x and then take reciprocal.


I don't quite understand :frown:

So far I've done...

5sinx=4cosx5sinx=4cosx

5sinx4cosx=0 \frac{5sinx}{4cosx}=0

Which then becomes this...?

54×1tanx\frac{5}{4}\times\frac{1}{tanx}

54cot⁡x\frac{5}{4} \cot x

Have I done this right so far? If so, what do I do next?
Reply 30
Avatar for Zacken
Zacken
22
Original post by Philip-flop
I don't quite understand :frown:

So far I've done...

5sinx=4cosx5sinx=4cosx

5sinx4cosx=0 \frac{5sinx}{4cosx}=0

Which then becomes this...?

54×1tanx\frac{5}{4}\times\frac{1}{tanx}

54cot⁡x\frac{5}{4} \cot x

Have I done this right so far? If so, what do I do next?


That's not true! 4cos⁡x4cos⁡x=1\frac{4\cos x}{4\cos x} = 1 not 0. Like 11=0\frac{1}{1} = 0.

So with 5sin⁡x=4cos⁡x5 \sin x = 4\cos x divide both sides by cos⁡x\cos x: this gives you 5sin⁡xcos⁡x=4cos⁡xcos⁡x\frac{5\sin x}{\cos x} = \frac{4\cos x}{\cos x}.

Now remember that cos⁡xcos⁡x=1\frac{\cos x}{\cos x} = 1. So you get 5sin⁡xcos⁡x=45 \frac{\sin x}{\cos x} = 4.

But you know that tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}, so you have 5tan⁡x=45 \tan x = 4.

Now divide both sides by 55 to get 5tan⁡x5=45\frac{5\tan x}{5} = \frac{4}{5} but you know that 55=1\frac{5}{5} = 1 so you have

5tan⁡x5=tan⁡x\frac{5\tan x}{5} = \tan x, which means tan⁡x=45\tan x = \frac{4}{5}. Now take the reciprocal of both sides to get 1tan⁡x=145=54\displaystyle \frac{1}{\tan x} = \frac{1}{\frac{4}{5}} = \frac{5}{4}
Original post by Zacken
That's not true! 4cos⁡x4cos⁡x=1\frac{4\cos x}{4\cos x} = 1 not 0. Like 11=0\frac{1}{1} = 0.

So with 5sin⁡x=4cos⁡x5 \sin x = 4\cos x divide both sides by cos⁡x\cos x: this gives you 5sin⁡xcos⁡x=4cos⁡xcos⁡x\frac{5\sin x}{\cos x} = \frac{4\cos x}{\cos x}.

Now remember that cos⁡xcos⁡x=1\frac{\cos x}{\cos x} = 1. So you get 5sin⁡xcos⁡x=45 \frac{\sin x}{\cos x} = 4.

But you know that tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}, so you have 5tan⁡x=45 \tan x = 4.

Now divide both sides by 55 to get 5tan⁡x5=45\frac{5\tan x}{5} = \frac{4}{5} but you know that 55=1\frac{5}{5} = 1 so you have

5tan⁡x5=tan⁡x\frac{5\tan x}{5} = \tan x, which means tan⁡x=45\tan x = \frac{4}{5}. Now take the reciprocal of both sides to get 1tan⁡x=145=54\displaystyle \frac{1}{\tan x} = \frac{1}{\frac{4}{5}} = \frac{5}{4}


OMG I am going to kick myself. When I can't work something out I go into panic mode and start making Maths up!! :frown: I originally had...

5sinx=4cosx5sinx=4cosx

5sinxcosx=4\frac{5sinx}{cosx}=4

5tanx=45tanx=4

tanx=45tanx=\frac{4}{5}

But then I got stuck because I thought I went wrong somewhere as I wasn't sure how to get the equation to become cot(x)
As I knew that cot⁡x⇔1tanx\cot x \Leftrightarrow \frac{1}{tan x} I assumed that I wouldn't be able to get the answer from what I had originally had done. So I started all over again, and from there on my brain decided to make up Maths rules :frown: :frown: :frown:

But seriously, Thanks again Zacken!! I'm sure you must be fed up with my noobie mistakes by now!! :P
How did you identify whether to take the reciprocal of both sides?
(edited 7 years ago)
Reply 32
Avatar for Zacken
Zacken
22
Original post by Philip-flop
[...] How did you identify whether to take the reciprocal of both sides?



Like you said, you know that cot⁡x=1tan⁡x\cot x = \frac{1}{\tan x}, which means that whenever you know what tan x is, you automatically know what cot x it, it's just 1/(whatever tan x is). And vice-versa. Whenever you know cot x, you know tan x. Whenever you know sin x you know cosec x, whenever you know cos x, you know sec x, etc...

So here we know that tan x = 4/5, which means we automatically know that cot x = 1/(4/5) = 5/4.
Original post by Zacken
Like you said, you know that cot⁡x=1tan⁡x\cot x = \frac{1}{\tan x}, which means that whenever you know what tan x is, you automatically know what cot x it, it's just 1/(whatever tan x is). And vice-versa. Whenever you know cot x, you know tan x. Whenever you know sin x you know cosec x, whenever you know cos x, you know sec x, etc...

So here we know that tan x = 4/5, which means we automatically know that cot x = 1/(4/5) = 5/4.


Yeah that's it! I know these things in the back of my mind but have trouble adopting these techniques :frown:
Maths will not defeat me!! :P

Thanks again Zacken!
Ok, I'm having another dumb moment with this question...

Q) sec⁡2(x)cos5(x)+cot⁡(x)cosec(x)sin4(x)\sec ^2(x) cos^5(x)+\cot (x) \mathrm{cosec}(x) sin^4(x)

I think I've made a mistake somewhere, so far I've got...

cos5x(1cos2x)+sin4x(cosxsinx)(1sinx)cos^5x(\frac{1}{cos^2x})+sin^4x(\frac{cosx}{sinx})(\frac{1}{sinx})

cos5x(1cos2x)+sin4x(cosxsin2x)cos^5x(\frac{1}{cos^2x})+sin^4x(\frac{cosx}{sin^2x})

cos3x+sin2xcosxcos^3x+sin^2xcosx

Edit: Never mind I think i've got it by factorising now...
(edited 7 years ago)
Reply 35
Avatar for Zacken
Zacken
22
Original post by Philip-flop
Ok, I'm having another dumb moment with this question...

Q) sec⁡2(x)cos5(x)+cot⁡(x)cosec(x)sin4(x)\sec ^2(x) cos^5(x)+\cot (x) \mathrm{cosec}(x) sin^4(x)

I think I've made a mistake somewhere, so far I've got...

cos5x(1cos2x)+sin4x(cosxsinx)(1sinx)cos^5x(\frac{1}{cos^2x})+sin^4x(\frac{cosx}{sinx})(\frac{1}{sinx})

cos5x(1cos2x)+sin4x(cosxsin2x)cos^5x(\frac{1}{cos^2x})+sin^4x(\frac{cosx}{sin^2x})

cos3x+sin2xcosxcos^3x+sin^2xcosx


Whwat are you meant to do? What's the question?
Original post by Philip-flop
Ok, I'm having another dumb moment with this question...

Q) sec⁡2(x)cos5(x)+cot⁡(x)cosec(x)sin4(x)\sec ^2(x) cos^5(x)+\cot (x) \mathrm{cosec}(x) sin^4(x)

I think I've made a mistake somewhere, so far I've got...

cos5x(1cos2x)+sin4x(cosxsinx)(1sinx)cos^5x(\frac{1}{cos^2x})+sin^4x(\frac{cosx}{sinx})(\frac{1}{sinx})

cos5x(1cos2x)+sin4x(cosxsin2x)cos^5x(\frac{1}{cos^2x})+sin^4x(\frac{cosx}{sin^2x})

cos3x+sin2xcosxcos^3x+sin^2xcosx


factorise out a cos(x)
Original post by Zacken
Whwat are you meant to do? What's the question?


Never mind I think I've managed to get it by factorising now :smile:
Q) Simplify the following expression...
sec⁡2(x)cos5(x)+cot⁡(x)cosec(x)sin4(x)\sec ^2(x) cos^5(x)+\cot (x) \mathrm{cosec}(x) sin^4(x)

My answer below...
cos5x(1cos2x)+sin4x(cosxsinx)(1sinx)cos^5x(\frac{1}{cos^2x})+sin^4x(\frac{cosx}{sinx})(\frac{1}{sinx})

cos5x(1cos2x)+sin4x(cosxsin2x)cos^5x(\frac{1}{cos^2x})+sin^4x(\frac{cosx}{sin^2x})

cos3x+sin2xcosxcos^3x+sin^2xcosx

Now that I have factorised I've got...
cosx(cos2x+sin2x)cosx(cos^2x+sin^2x)

using the Trig Identity cos^2 x + sin^2 x = 1 .....
cosx(1)cosx(1)

cosxcosx

Original post by ValerieKR
factorise out a cos(x)


Thank you so much. I stupidly didn't think of factorising at first :frown:
(edited 7 years ago)
I'm stuck yet again...

How can I get the L.H.S (Left Hand Side) to equal the R.H.S (Right Hand Side)?? :frown:...

(1−cosx)(1+sec⁡x)=sinxtanx(1-cosx)(1+ \sec x)=sinxtanx

So far I've expanded the bracket to give me...

1+sec⁡x−cosx−cosxsec⁡x1+ \sec x -cosx -cosx\sec x
Original post by Philip-flop
I'm stuck yet again...

How can I get the L.H.S (Left Hand Side) to equal the R.H.S (Right Hand Side)?? :frown:...

(1−cosx)(1+sec⁡x)=sinxtanx(1-cosx)(1+ \sec x)=sinxtanx

So far I've expanded the bracket to give me...

1+sec⁡x−cosx−cosxsec⁡x1+ \sec x -cosx -cosx\sec x


simplify cos(x)sec(x) and put the remaining terms over a cos(x) denominator

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you