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in the interval of $0< \theta <360$ solve for $\theta$ , the equation $(1)$, I would just say $\tan \theta =-1$ and solve that.

$$\sin\theta=-\cos\theta\tag1$$

but what if i said $\sin θ +\cos θ =0$ then $\cos θ (\tan θ +1)=0$. I'd have solutions for $\cos θ =0$ as well as $\tan θ =-1$ , why is doing this method wrong since arithmetically going through the workings out it seems fine , but the solutions for $\cosθ=0$ are wrong?

I'm trying to get a verdict on if the 2nd method is still right or wrong ? i have a feeling its wrong since the solutions dont work, but the process of me getting to $\cos(θ)⋅[\tanθ+1]=0$ seemed sound that I wouldnt realise it was wrong ?

How do I stop myself from carrying out this sort of working out ,is there a way for me to intuitively prove its wrong even before the equation ends up in the form of $\cos(θ)⋅[\tanθ+1]=0$ because if i carried out a similar working out for this question " Solve for theta ,$\sinθ=3\sinθ\cosθ$" and solving it similarly like so I'd end up with $\sinθ(1-3\cosθ)=0$ ,and it would give the right solutions.

Átila Correia
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j jose
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    In the 1st method, you divided both sides by $\cos\theta$, which is illegitimate if $\cos\theta=0$. – Gerry Myerson Jan 20 '24 at 05:21
  • In the 2nd method, you converted $\sin\theta = \cos\theta\tan\theta$, which is undefined when $\cos\theta = 0$. – peterwhy Jan 20 '24 at 05:25
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    Either method, when you choose to divide by $\cos\theta$, consider the two cases: either 1) $\cos\theta = 0$ then no solutions, or 2) $\cos\theta \ne 0$ then you can safely divide. – peterwhy Jan 20 '24 at 05:40
  • @peterwhy so you are saying whenever i divide by a trig function consider if that trigfunction=0 would there be solutions to it or not , so in the case of cosθ=0 there wouldnt be solutions either way , so either method i did would be right ?also when you say i chose to divde by cosθ in the 2nd scenario you mean when i pulled out cosθ to make a set of factors cos(θ)⋅[tanθ+1]=0 – j jose Jan 20 '24 at 06:33
  • In the 2nd method, to pull out $\cos\theta$ from $\sin\theta$, you either 1) multiplied and divided by $\cos\theta$: $\sin\theta \overset?= \cos\theta \cdot \frac{\sin\theta}{\cos\theta}$, or 2) applied $\sin\theta \overset?= \cos\theta \tan\theta$. For interpretation 2, check for two cases: either $\tan\theta$ is undefined (same as when $\cos\theta=0$, then no solutions), or defined (then you can safely pull, and still reject the $\cos\theta=0$ case). – peterwhy Jan 20 '24 at 17:38
  • @peterwhy so you are saying is that there is no real way to tell whether when you divide by cosθ or any other trig ,if you will lose the solutions for when it =0 ,so its best to divide anyways and just keep a mental note to check if the cosθ=0 work or not

    or are you saying that in the case of sinθ=3sinθcosθ , when you divide by sinθ you are getting rid of sin so as a result you have considered all solutions of sinθ except when sinθ=0.But in sineθ=-cosθ, when you divide by cosθ or pull out cosθ from sinθ+cosθ=0 you have actually gotten rid of cosθ ,its still in the equation .

     – j jose Jan 21 '24 at 10:04
  • @peterwhy continuing from previous comment: and so as a result all solutions of cosθ are still kept in the equation – j jose Jan 21 '24 at 10:05
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    @jjose: There's an important nuance here. The rule that allows us to say $ab=0$ when $a=0$ or $b=0$ pre-supposes that $a$'s zero-ness doesn't cause trouble for $b$ (or vice-versa). Your second strategy is like solving $x-3=0$ by factoring to get $x\left(1-\frac3x\right)=0$, and then claiming $x=0$ is a soln because it makes the first factor vanish; however, $x=0$ makes the second factor undefined, which is problematic. (Moreover, if that strategy worked, then $0$ would be a soln to every eqn!) You're calling attention to the fact that we can't consider each factor in complete isolation. – Blue Jan 21 '24 at 11:51

2 Answers2

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In the first method,

$${\sin\theta=-\cos\theta\\\implies \begin{cases} \tan x=-1,& \text{if } \cos \theta\ne 0\\ \sin\theta=0, & \text{if } \cos \theta= 0\rightarrow\text{No solution.}\text{ As, $\cos \theta$ and $\sin \theta$ both cannot $0$ simultaniously. } \end{cases}\\} $$

Hence, your first method is right.

In the second method,

When we take out common factor $\cos\theta$, we are dividing $\sin\theta$ by $\cos\theta$ so we have assumed that $\cos\theta\ne0$, $$\sin\theta+\cos\theta=0{\implies\cos \theta (\tan \theta +1)=0\quad \text{When $\cos\theta\ne0$}\\\implies \tan\theta=-1\quad(\because \cos\theta\ne0)}$$

Edit: We will prove that $\sin\theta+\cos\theta=\cos \theta (\tan \theta +1)$ is invalid when $\cos\theta=0$:

LHS$=\pm1+0=\pm1$ but, RHS$=0\times(\text{undefined}+1)=\text{undefined}$

But, in the case $\sinθ-3\sinθ\cosθ=\sinθ(1−3\cosθ)$ it is valid for $\sin\theta=0$:

LHS$=0-3\times0\times(\pm1)=0$ and, RHS$=0\times(1-3\times(\pm1))=0$

Alternative aaproach,

$$\sin\theta+\cos\theta=0\\ {\implies\sqrt2\sin\left(\theta+\frac{\pi}{4}\right)=0\\ \implies\theta+\frac{\pi}{4}=n\pi \quad\text{Where, } n\in\Bbb Z}$$

O M
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  • you mentioned how when i factor out cosθ i assume its not equal to 0 , which in many questions will be seen as a mistake as you will miss out on solutions , so how am i supposed to know when to divide by a trig function and if i will still have all the solutions even when i divde through by a trig . Like in this case sinθ=3sinθcosθ goes to sinθ(1−3cosθ)=0 , if i were to divide by sin rather than factoring it out i would have lost solutions as when you divided by sinθ you assumed its not=0 – j jose Jan 20 '24 at 20:14
  • @jjose No. You diving $\sin\theta$ by $\cos\theta$ to get $\tan \theta$ so, . If something like $\sin θ-3\sin θ\cos θ=0$ then you can factor out $\sin θ(1−3\cos θ)=0$ here you not dividing any thing. I edited my answer and add your query – O M Jan 21 '24 at 09:23
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As noticed with the first method the equivalence

$$\sin\theta=-\cos\theta \iff \cos \theta (\tan \theta +1)=0$$

is true only when $\cos \theta \neq 0$ and since $\cos \theta=0$ is not a solution to the original equation, indeed

$$\cos \theta =0 \implies \sin\theta=\pm 1$$

therefore we can solve for $$\tan \theta +1=0$$

As another alternative we have

$$\sin\theta=-\cos\theta \iff \cos \left(\frac \pi 2 -\theta\right)=\cos(\pi\pm\theta)$$

then use

$$\cos x = \cos y \iff x=y+2k\pi \;\lor\; x=-y+2k\pi$$

which lead to

  • $\frac \pi 2 -\theta=\pi\pm\theta+2k\pi \implies \theta=-\frac \pi 4+k\pi$
  • $\frac \pi 2 -\theta=-\pi\mp\theta+2k\pi \implies \theta=-\frac 3 4\pi+k\pi$

and the solutions are $x=135°$ and $x=315°$.

Another way by squaring

$$\sin\theta=-\cos\theta \iff \sin\theta+\cos\theta=0 \implies \sin 2\theta =-1$$

which leads to

$$2\theta =\frac 3 2 \pi +2k\pi \implies \theta = \frac 3 4 \pi +k\pi$$

The case $\sin\theta=3\sin\theta\cos\theta$ is slightly different, here we can factor out as follows (this is always true, we have no limitations on $\theta$)

$$\sin\theta=3\sin\theta\cos\theta \iff \sin\theta(1-3\cos\theta)=0$$

and the latter is equivalent to

$$\sin \theta =0 \;\lor \; 1-3\cos\theta=0$$

user
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  • @jjose Thanks I’ve fixed it! Hope now it is clear. – user Jan 21 '24 at 11:48
  • :505767 how does cosθ=0 imply that sinθ=+ or - 1 and therfore mean that tanθ+1?

    sorry im not great with maths speak

    are you saying that when you divide cosθ when you have sinθ=-cosθ or when you factors out cosθ from sinθ+cosθ=0 , you wont lose solutions as you would if you were to divide sinθ=3sinθcosθ by sinθ because in this latter case you are canceling sinθ from the equation ,but in the former cases you are not

     – j jose Jan 21 '24 at 11:55
  • im trying to see how the case of sinθ=3sinθcosθ⟺sinθ(1−3cosθ)=0 instead of dividing by sinθ from the get go is different to

    sinθ=-cosθ >>> dividing cosθ from the get go to solve which is allowed here but not allowed in the former case.

     – j jose Jan 21 '24 at 11:57
  • There are many ways to manipulate and solve the equation, each one requires to pay attention. – user Jan 21 '24 at 12:00
  • For $\sin x=-\cos x$ we can note that $\cos x=0$ is not a solution, then we can assume $\cos x\neq 0$ and divide the original equation by $\cos x$ to obtain $\tan x=-1$ which is equivalent to the original problem. – user Jan 21 '24 at 12:02
  • For $\sin x=3\sin x\cos x$ we can note that $\sin x=0$ is a solution. Then for the remaining cases with $\sin x \neq 0$ we can divide the original equation by $\sin x$ to obtain $1=3\cos x$ which solution completes the problem. – user Jan 21 '24 at 12:05