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how do i solve sin(theta)cos(theta)-sin(theta)+cos(theta) = 1?
Reply 1
Avatar for Zacken
Zacken
22
Original post by alesha98
how do i solve sin(theta)cos(theta)-sin(theta)+cos(theta) = 1?


If you write sin⁡θ=x\sin \theta = x and cos⁡θ=y\cos \theta = y it becomes xy−x+y−1=0xy - x + y - 1= 0 which factorises neatly as (x+1)(y−1)=(sin⁡θ+1)(cos⁡θ−1)=0(x+1)(y-1) = (\sin \theta +1)(\cos \theta - 1) = 0 i.e:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\sin \theta \cos \theta - \sin \theta + \cos \theta - 1 \equiv (\sin \theta +1)(\cos \theta - 1)\end{equation*}



So that you get two family of solutions:

sin⁡θ+1=0\sin \theta + 1 = 0 or cos⁡θ−1=0\cos \theta - 1= 0.
(edited 8 years ago)
Reply 2
Avatar for razzor
razzor
10
Original post by alesha98
how do i solve sin(theta)cos(theta)-sin(theta)+cos(theta) = 1?


Bring the 1 over to the other side and see if you can put the LHS into two brackets.

Spoiler

Express sin(theta) as a function of cos(theta) then you should be able to obtain a constant term from the function of cos(theta). This gives you the sin(theta) term but I didn't notice what Zacken had put. Don't listen to me.
(edited 8 years ago)
Reply 4
Avatar for Zacken
Zacken
22
Original post by alesha98
how do i solve sin(theta)cos(theta)-sin(theta)+cos(theta) = 1?


If you're not entirely sure why it factorises the way it does in my post, then consider this:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\sin \theta \cos \theta - \sin \theta + \cos \theta - 1 \equiv \sin \theta (\cos \theta -1) + (\cos \theta - 1) \equiv (\cos \theta - 1)(\sin \theta +1)\end{equation*}



By pulling out the common factor out (cos⁡θ−1)(\cos \theta - 1).
Reply 5
Avatar for alesha98
alesha98
OP
4
Original post by Zacken
If you're not entirely sure why it factorises the way it does in my post, then consider this:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\sin \theta \cos \theta - \sin \theta + \cos \theta - 1 \equiv \sin \theta (\cos \theta -1) + (\cos \theta - 1) \equiv (\cos \theta - 1)(\sin \theta +1)\end{equation*}



By pulling out the common factor out (cos⁡θ−1)(\cos \theta - 1).


Thankyou so much
Reply 6
Avatar for Zacken
Zacken
22
Original post by alesha98
Thankyou so much


You're very welcome!
Original post by Zacken
If you write sin⁡θ=x\sin \theta = x and cos⁡θ=y\cos \theta = y it becomes xy−x+y−1=0xy - x + y - 1= 0 which factorises neatly as (x+1)(y−1)=(sin⁡θ+1)(cos⁡θ−1)=0(x+1)(y-1) = (\sin \theta +1)(\cos \theta - 1) = 0 i.e:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\sin \theta \cos \theta - \sin \theta + \cos \theta - 1 \equiv (\sin \theta +1)(\cos \theta - 1)\end{equation*}



So that you get two family of solutions:

sin⁡θ+1=0\sin \theta + 1 = 0 or cos⁡θ−1=0\cos \theta - 1= 0.


so what if i take out sin⁡θ \sin \theta as a common factor? do i lose some answers?
Original post by thefatone
so what if i take out sin⁡θ \sin \theta as a common factor? do i lose some answers?


If you divide an equation by something which could be zero then you will lose solutions, for example x^2 - 2x = 0; dividing by x gives x - 2 = 0 -> x = 2, which misses the solution x = 0.
Reply 9
Avatar for Zacken
Zacken
22
Original post by thefatone
so what if i take out sin⁡θ \sin \theta as a common factor? do i lose some answers?


Well, depends on what you mean. But I don't see how it would be useful.
Original post by Zacken
Well, depends on what you mean. But I don't see how it would be useful.


i did this

sin⁡θ(cos⁡θ−1+cos⁡θ)=1\sin \theta \left(\cos \theta -1+ \cos \theta \right)=1
which gave me
sin⁡θ(2cos⁡θ−1)=1\sin \theta \left(2\cos \theta -1 \right)=1
then i would solve from there but that doesn't seem right...
Reply 11
Avatar for Zacken
Zacken
22
Original post by thefatone
i did this

sin⁡θ(cos⁡θ−1+cos⁡θ)=1\sin \theta \left(\cos \theta -1+ \cos \theta \right)=1
which gave me
sin⁡θ(2cos⁡θ−1)=1\sin \theta \left(2\cos \theta -1 \right)=1
then i would solve from there but that doesn't seem right...


It is right, it's just massively useless. How would you proceed from here?
Original post by Zacken
It is right, it's just massively useless. How would you proceed from here?


lol i know TeeEm would kill me for this

but i'd continue to solve for values of theta

so

sin⁡θ=1\sin\theta=1
and
2cos⁡θ−1=12\cos\theta-1=1
Reply 13
Avatar for Zacken
Zacken
22
Original post by thefatone
...


That's not how it works. You can only do equate each factor to zero if your equation is of the form f(x)=0f(x) = 0.

Otherwise, do you think we can do this: (x−1)2=x2−2x+1=0⇒x(x−2)=1(x-1)^2 = x^2 - 2x + 1 = 0 \Rightarrow x(x-2) = 1 so x=1x=1 and x−2=1⇒x=3x-2=1 \Rightarrow x = 3? Is this valid.

Nopes, it is not the case that ab=cab = c then a=ca=c or b=0b=0. This is only true if c=0.
Original post by Zacken
That's not how it works. You can only do equate each factor to zero if your equation is of the form f(x)=0f(x) = 0.

Otherwise, do you think we can do this: (x−1)2=x2−2x+1=0⇒x(x−2)=1(x-1)^2 = x^2 - 2x + 1 = 0 \Rightarrow x(x-2) = 1 so x=1x=1 and x−2=1⇒x=3x-2=1 \Rightarrow x = 3? Is this valid.

Nopes, it is not the case that ab=cab = c then a=ca=c or b=0b=0. This is only true if c=0.


there we go that's what i was looking for
thanks :smile:
Reply 15
Avatar for Zacken
Zacken
22
Original post by thefatone
there we go that's what i was looking for
thanks :smile:


You're welcome.

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