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C3 Trig

So basically I've been trig homework all night because it's so hard haha!
Just wondering if anyone can find a solution to:

Cotx - Cosx / (1 - sinx) = cotx

If you could explain your workings out itd be really helpful.

Thanks
James


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Original post by james0902
So basically I've been trig homework all night because it's so hard haha!
Just wondering if anyone can find a solution to:

Cotx - Cosx / (1 - sinx) = cotx

If you could explain your workings out itd be really helpful.

Thanks
James


Posted from TSR Mobile


Subtract cot⁡(x)\cot(x) from both sides then it's just when cos⁡(x)=0\cos(x)=0
Reply 2
Avatar for B_9710
B_9710
18
Multiply by sinx/sinx.
Is this question basically proving the identity cot⁡x−cos⁡x1−sin⁡x\displaystyle \frac{\cot x -\cos x}{1-\sin x} ?
(edited 7 years ago)
Reply 3
Avatar for james0902
james0902
OP
10
I mean you have to simplify the left to find the right. The two sides cannot interact with one another.


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Reply 4
Avatar for james0902
james0902
OP
10
Original post by B_9710
Multiply by sinx/sinx.
Is this question basically proving the identity cot⁡x−cos⁡x1−sin⁡x\displaystyle \frac{\cot x -\cos x}{1-\sin x} ?


Yes it's proving the identity.


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Reply 5
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B_9710
18
Original post by james0902
Yes it's proving the identity.


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Multiply both numerator and denominator of LHS by sinx.
Reply 6
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james0902
OP
10
Original post by B_9710
Multiply both numerator and denominator of LHS by sinx.


I'm really struggling so what would that give me? Would that make

sinx (cotx - cosx)

Is that right?


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Reply 7
Avatar for B_9710
B_9710
18
Original post by james0902
I'm really struggling so what would that give me? Would that make

sinx (cotx - cosx)

Is that right?


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Think about what cotx is in terms of cosx and sinx. What is sinxcotx?
Original post by james0902
Yes it's proving the identity.


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Personally I'd multiply top and bottom by 1+sin⁡(x)1+\sin(x)

This gets a nice difference of two square on the denominator which is equal to cos⁡2(x)\cos^2(x) then the numerator just takes a bit messing around with to get get into a fraction in terms of sine/cosine, by which point loads of things cancel and you're soon left with cot⁡(x)\cot(x). But I'm not sure how simple you'd find messing around with a fraction as a numerator, should be straight forward enough if you know what's happening.
Reply 9
Avatar for BJA98
BJA98
4
Original post by james0902
So basically I've been trig homework all night because it's so hard haha!
Just wondering if anyone can find a solution to:

Cotx - Cosx / (1 - sinx) = cotx

If you could explain your workings out itd be really helpful.

Thanks
James






Easy solution!!

1. We start off by multiplying both sides of your expression by (1-Sinx)
2. Divide both sides by Cotx.You should get (Cotx-Cosx)/Cotx=1-Sinx
3. Now realise that you can split the fraction on the left so the LHS becomes (Cotx/Cotx)-(Cosx/Cotx)=1-Sinx
4. Hence it simplifies to 1-(Cosx/Cotx).
5. Cotx is 1/tanx therefore it now becomes 1-Cosxtanx
6. Tanx=Sinx/Cosx hence the two Cosx's cancel and you are left with 1-Sinx on LHS which is equal to the RHS.

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Original post by BJA98
Easy solution!!

1. We start off by multiplying both sides of your expression by (1-Sinx)
2. Divide both sides by Cotx.You should get (Cotx-Cosx)/Cotx=1-Sinx
3. Now realise that you can split the fraction on the left so the LHS becomes (Cotx/Cotx)-(Cosx/Cotx)=1-Sinx
4. Hence it simplifies to 1-(Cosx/Cotx).
5. Cotx is 1/tanx therefore it now becomes 1-Cosxtanx
6. Tanx=Sinx/Cosx hence the two Cosx's cancel and you are left with 1-Sinx on LHS which is equal to the RHS.

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wat

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