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Suppose that you have to prove the trig identity:

$$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}=\sin\theta$$

I have always been told that I should manipulate the left and right sides of the equation separately, until I have transformed them each into something identical. So I would do:

$$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}$$ $$=\frac{\sin\theta(1 - \sin^2\theta)}{\cos^2\theta}$$ $$=\frac{\sin\theta(\cos^2\theta)}{\cos^2\theta}$$ $$=\sin\theta$$

And then, since the left side equals the right side, I have proved the identity. My problem is: why can't I manipulate the entire equation? In this situation it probably won't make things any easier, but for certain identities, I can see ways to "prove" the identity by manipulating the entire equation, but cannot prove it by keeping both sides isolated.

I understand, of course, that I can't simply assume the identity is true. If I assume a false statement, and then derive from it a true statement, I still haven't proved the original statement. However, why can't I do this:

$$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}\not=\sin\theta$$ $$\sin\theta - \sin^3\theta\not=(\sin\theta)(\cos^2\theta)$$ $$\sin\theta(1 - \sin^2\theta)\not=(\sin\theta)(\cos^2\theta)$$ $$(\sin\theta)(\cos^2\theta)\not=(\sin\theta)(\cos^2\theta)$$

Since the last statement is obviously false, is this not a proof by contradiction that the first statement is false, and thus the identity is true?

Or, why can't I take the identity equation, manipulate it, arrive at $(\sin\theta)(\cos^2\theta)=(\sin\theta)(\cos^2\theta)$, and then work backwards to arrive at the trig identity. Now, I start with a statement which is obviously true, and derive another statement (the identity) which must also be true - isn't that correct?

Another argument that I have heard for keeping the two sides isolated is that manipulating an equation allows you to do things that are not always valid in every case. But the same is true when manipulating just one side of the equation. In my first proof, the step

$$\frac{\sin\theta(\cos^2\theta)}{\cos^2\theta}$$ $$=\sin\theta$$

is not valid when theta is $\pi/2$, for example, because then it constitutes division by zero.

Isaac
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Ord
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    The main reason could be just pedagogical: It is more involved and more educational to work out with the two sides independently. – Jon Jan 21 '12 at 18:12
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    One other reason is that one must be careful, when working both sides. If you arrive at a false statement, it's not a problem (you've done a proof by contradiction); but if you arrive at a true statement, you need to make sure that all steps are reversible. This is something that a lot of people don't realize and/or forget. Rather than invite error... – Arturo Magidin Jan 21 '12 at 22:31
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    You may lose control of the logic of the situation. Even if you do not, and write a perfetly correct argument, there is a serious risk that it will (wrongly) be considered incorrect by a grader. – André Nicolas Jun 06 '12 at 06:13

6 Answers6

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Why can't I manipulate the entire equation?

You can. The analytical method for proving an identity consists of starting with the identity you want to prove, in the present case $$ \begin{equation} \frac{\sin \theta -\sin ^{3}\theta }{\cos ^{2}\theta }=\sin \theta,\qquad \cos \theta \neq 0 \tag{1} \end{equation} $$ and establish a sequence of identities so that each one is a consequence of the next one. For the identity $(1)$ to be true is enough that the following holds $$ \begin{equation} \sin \theta -\sin ^{3}\theta =\sin \theta \cos ^{2}\theta \tag{2} \end{equation} $$ or this equivalent one $$ \begin{equation} \sin \theta \left( 1-\sin ^{2}\theta \right) =\sin \theta \cos ^{2}\theta \tag{3} \end{equation} $$ or finally this last one $$ \begin{equation} \sin \theta \cos ^{2}\theta =\sin \theta \cos ^{2}\theta \tag{4} \end{equation} $$

Since $(4)$ is true so is $(1)$.

The book indicated below illustrates this method with the following identity $$ \frac{1+\sin a}{\cos a}=\frac{\cos a}{1-\sin a}\qquad a\neq (2k+1)\frac{\pi }{2} $$

It is enough that the following holds $$ (1+\sin a)(1-\sin a)=\cos a\cos a $$

or $$ 1-\sin ^{2}a=\cos ^{2}a, $$

which is true if $$ 1=\cos ^{2}a+\sin ^{2}a $$

is true. Since this was proven to be true, all the previous indentities hold, and so does the first identity.

enter image description here enter image description here

Reference: J. Calado, Compêndio de Trigonometria, Empresa Literária Fluminense, Lisbon, pp. 90-91, 1967.

Américo Tavares
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  • At least to me, "... each one is a consequence of the next one." is hard to read correctly—at first glance, it seemed like $(1)\implies(2)$, when it in fact says $(2)\implies(1)$—so I had to re-read your answer a few times to convince myself that I was misreading it and that what you said was in fact correct. Difficulty with understanding direction of implication (both symbolically and in "A if B"/"A when B"/"A only if B" type forms) is why I'd encourage a student to write $(4)\implies(3)\implies(2)\implies(1)$, even if they got to their proof by working from (1) to (2) to (3) to (4). – Isaac Jan 21 '12 at 23:07
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    (Also, I edited the book page images to lighten the background, darken the text, and reduce the file size—feel free to revert if you don't like my edited images.) – Isaac Jan 21 '12 at 23:17
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    @Isaac: The sentence "each one is a consequence of the next one" is a direct translation of " ... cada uma seja uma consequência da seguinte" in the 2nd. paragraph. As you wrote the correct sequence of implications is $(4)\Rightarrow (3)\Rightarrow (2)\Rightarrow (1)$. Thanks. – Américo Tavares Jan 21 '12 at 23:34
  • Thank you, the example of a real proof was quite helpful! – Ord Jan 23 '12 at 01:26
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You've got a pretty good handle on the situation. It's not so much that you can't manipulate the potential identity as an equation as that, in general, most people shouldn't manipulate the potential identity as an equation. The key part is what you said—use the manipulation to arrive at a true statement (that's your scratch-work), then work backwards to write your proof: starting with a true statement and arriving at the identity.

In your last example, since $\cos\theta$ is in the denominator, $\theta=\frac{\pi}{2}$ would not be in the domain of the identity, so it's okay to simplify to $\sin\theta$.

Isaac
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Prove the trig identity "LHS = RHS" Given: LHS Goal: RHS (or vise versa) The reason that it is not valid to work on both sides at the same time (cross-multiplying, etc) is that you are not given "LHS = RHS", so there is no equation until after you have proven the trig identity. Is it valid to use the equation "LHS = RHS" to prove the trig identity "LHS = RHS"? Steve

Steve Cox
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  • For any $L$ and $R$, the equation $L(\theta) = R(\theta)$ is either true or false for each value of $\theta$. Your goal in proving the identity is to prove that the solution set is all $\theta\in\mathbb{R}$. In this sense the problem is like any algebra problem. You can proceed to solve the equation by applying any transformations which do not add solutions. – Noah Stein Jun 06 '12 at 09:01
  • You can also apply transformations which a priori might add solutions, as long as you check that in this case they do not. For example you can multiply both sides by $\sin(\theta)$ and work on $L(\theta)\sin(\theta) = R(\theta)\sin(\theta)$ as long as you separately verify that $L(\theta) = R(\theta)$ whenever $\theta\in\pi\mathbb{Z}$. – Noah Stein Jun 06 '12 at 09:01
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I agree completely with Noah Stein. I just want to clarify/add the following:

Suppose the identity you are trying to show is $L(\theta)=R(\theta)$ and it is undefined at $\theta\in\left\{k\pi\,:\,k\in\Bbb Z\right\}$ because there is $\sin(\theta)$ in the denominator of $L(\theta)$ or $R(\theta)$ or both. Then you can manipulate it as an equation by multiplying both sides by $\sin(\theta)$. Suppose after several other manipulations/transformation that required no further multiplication by a variable, you arrived at $\cos(\theta)=\cos(\theta)$. Then the identity is proved for all $\theta\ne k\pi$ which is the largest possible set on which $L(\theta)$ and $R(\theta)$ are defined. Hence you proved it is an identity.

Sayel Ali
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According to the definition "identity is a relation which is true for all values of x", so when we manipulate the trigonometric identity just like the trigonometric equations try to find the angles and we ended up a true relation like 0=0 independent of angle means the relation is an identity.

Faisal Hussain
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I think it can be done. We are proving that LHS =RHS, as by assuming so, you arrive at a Universal truth.

Similarly, by assuming it's not true, we arrive at a contradiction , which is called, proof by contradiction.

Revathy
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  • You can't prove "LHS=RHS" by "assuming so". Also, the "universal truth"-thing isn't really suited for this forum, but more for one about philosophy. – Bobson Dugnutt Feb 19 '16 at 15:16