I have a trigonometric equation that is defined as:

$\sin(\alpha) - \cos(\alpha) = \frac{1}{2}$

Solving this equation by mathematica will yield $\alpha = 65.70 $ and $\alpha = -155.705 $

But As I solve it analytically, I will obtain different results:

First I exponentiate both sides to the power of two:

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

Now I expand the expressions:

$\sin(\alpha)^2 - 2 \sin(\alpha) \cos(\alpha) + \cos(\alpha)^2 = \frac{1}{4}$

As $\sin(\alpha)^2 + \cos(\alpha)^2 = 1$, I will have:

$1 - 2 \sin(\alpha) \cos(\alpha) = \frac{1}{4}$

Again if I put $2 \sin(\alpha) \cos(\alpha) = \sin(2\alpha)$ I will be left with:

$\sin(2 \alpha) = \frac{3}{4}$

Which will readily give $\alpha = 24.3$

So, why I am having different results? I don't understand.

  • 2
    $\begingroup$ You have $\forall a \; (\sin a -\cos a=1/2 \implies \sin 2a=3/4)$ but the sentence $\forall a \; (\sin 2a=3/4\implies \sin a -\cos a=1/2)$ is false. Similarly $\forall x\; (x-1=1\implies (x-1)^2 =1)$ but it is not true that $\forall x\; (\;(x-1)^2=1\implies x-1=1).$ $\endgroup$ Nov 21, 2016 at 21:04
  • 1
    $\begingroup$ The full solution to $\sin(2\alpha)=\frac34$ is: $$2\alpha \approx 48.59^\circ + n360^\circ \quad\mathrm{ or }\quad 2\alpha \approx 180^\circ - 48.59^\circ + n360^\circ \quad\mathrm{ for }\quad n\in\mathbb{Z}$$ Since you did the squaring, some of the solutions may be false, so you must go back and check each of them with the original equation. $\endgroup$ Nov 22, 2016 at 6:40

6 Answers 6


If you square, you also get the solutions of $$ \sin\alpha-\cos\alpha=-\frac{1}{2} $$

In general, if you have an equation of the form $f(x)=g(x)$ and square both sides, you get, after rearranging, $$ f(x)^2-g(x)^2=0 \tag{*} $$ that can be rewritten $$ (f(x)-g(x))(f(x)+g(x))=0 $$ so the solutions of (*) are the solutions of $f(x)-g(x)=0$ (the original equation) together with the solutions of $f(x)+g(x)=0$.

A safer way to solve your equation is to set $t=\tan(\alpha/2)$, so the equation becomes $$ \frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2}=\frac{1}{2} $$ that reduces to $$ t^2+4t-3=0 $$ with solutions $$ -2+\sqrt{7}\qquad\text{or}\qquad -2-\sqrt{7} $$ The first solutions corresponds to $$ \alpha=2\arctan(\sqrt{7}-2)\approx 65.7^\circ $$ and the second one to $$ \alpha=-2\arctan(\sqrt{7}+2)\approx-155.7^\circ $$ or $\approx204.3^\circ$ if you want a value between $0$ and $360$.

  • 2
    $\begingroup$ You are exactly right. But why at the end I am left with just one solution which is not the solution of the original equation. $\endgroup$
    – KratosMath
    Nov 21, 2016 at 16:20
  • 2
    $\begingroup$ @MsenRezaee You are not left with just one solution; there are four solutions to $\sin(2\alpha)=\frac12$. See Aretino's post. $\endgroup$
    – George Law
    Nov 21, 2016 at 16:23
  • $\begingroup$ @egreg thanks a lot $\endgroup$
    – KratosMath
    Nov 21, 2016 at 16:27
  • $\begingroup$ @MsenRezaee I added a safer procedure to find a solution. $\endgroup$
    – egreg
    Nov 21, 2016 at 16:28
  • $\begingroup$ The actual set of solutions to this problem is $\infty$. There is not just 2 solutions. $\endgroup$
    – J. Manuel
    Nov 23, 2016 at 10:26

Notice first of all that besides $24.3°$ you also have the solution $90°-24.3°=65.7°$ (supplementary angles have the same sine).

Your solution $24.3°$ is to discard, because it is a solution of the equation $\cos\alpha-\sin\alpha=1$ (which of course is the same as the given one when squared).

In summary: when you solve $\sin2\alpha=3/4$ you must consider all solutions, and discard fake ones:

$\alpha=48.6°/2$ (discard)

$\alpha=(48.6°+360°)/2$ (fine, it is the same as $-155.7°$)

$\alpha=(180°-48.6°)/2$ (fine)

$\alpha=(180°-48.6°+360°)/2$ (discard).

  • 1
    $\begingroup$ how do you choose the ones to discard? $\endgroup$
    – KratosMath
    Nov 21, 2016 at 16:24
  • 5
    $\begingroup$ You plug the solutions into the original equation and see if they fit or not. $\endgroup$ Nov 21, 2016 at 16:25
  • $\begingroup$ exactly I was trying to do. Many thanks $\endgroup$
    – KratosMath
    Nov 21, 2016 at 16:26
  • $\begingroup$ "supplementary angles have the same sine"? That's definitely not right. $\endgroup$ Nov 21, 2016 at 20:02
  • 3
    $\begingroup$ ack. You are right. I saw $90°-24.3°=65.7°$ and that made me think of "complementary", but you really do mean "supplementary". $\endgroup$ Nov 21, 2016 at 23:47

Squaring is often a useful (sometimes unavoidable) way to solve an equation, but it always introduces new solutions which may not fit the original problem. When you square both sides of an equation, you should always check your solutions to make sure they are applicable.

If you can avoid squaring, you might try doing so so as not to be confused with extra solutions. For this particular problem, you can solve it without squaring as follows:

$\dfrac12\ =\ \sin\alpha-\cos\alpha\ =\ \sqrt2\,(\cos45^\circ\sin\alpha - \sin45^\circ\cos\alpha)\ =\ \sqrt2\sin(\alpha-45^\circ)$

which gives $\sin(\alpha-45^\circ)=\frac1{2\sqrt2}$ which you can solve to give you the values obtained from Mathematica.

  • 1
    $\begingroup$ Squaring usually but not always adds new solutions. For example the equation $\sqrt{x+1}=0$ doesn't obtain additional solutions if squared. $\endgroup$
    – celtschk
    Nov 21, 2016 at 22:35

Every solution to $\sin(\alpha) - \cos(\alpha) = \frac{1}{2}$ is, indeed, a solution to $\sin(2 \alpha) = \frac{3}{4}$. That work you have done correctly.

Your error is either in misunderstanding what you have done, or what to do with that information.

Generally, the next step is:

  • Find the set $S$ of all solutions to $\sin(2 \alpha) = \frac{3}{4}$

Note that it really is important to find the entire set of solutions. Don't be sloppy and just take one solution.

Once you've determined the set $S$, you know it contains every solution to the original equation $\sin(\alpha) - \cos(\alpha) = \frac{1}{2}$; however, it can (and does!) contain other things too! The next step is

  • Identify which of the solutions in $S$ are or are not solutions to $\sin(\alpha) - \cos(\alpha) = \frac{1}{2}$

Once you've picked out which of the solutions in $S$ also satisfy $\sin(\alpha) - \cos(\alpha) = \frac{1}{2}$, you now have its complete solution set.


By squaring one unwittingly asks for two solutions of

$$\sin(\alpha) - cos(\alpha) = \frac{1}{2} ,\quad\sin(\alpha) - cos(\alpha) = \frac{-1}{2}$$

At the end we should retain only the solutions of interest.


You actually nailed it, you just didn’t know it.

Algebra’s fundamental theorem states: Any $\rm n^{th}$ order polynomial has $\rm n$ roots.

Then, if one increase the order of a polynomial by $\rm m$ (let’s say $\rm 1^{st}$ to $\rm 2^{nd}$) the number of roots will increase from $\rm n$ to $\rm n+m$, Therefore raising the power makes things different.

But raising to the same power both sides of an equation makes the solution set of the $\rm (n+m)^{th}$ order polynomial to contain the solutions set of the $\rm n^{th}$ polynomial. Meaning, it shouldn’t be a problem in your case, your solutions set should be in the square problem.

Start to get the solutions of $\sin(2α)=\frac{1}{2}$. If $0 ≤α ≤360º$, then your solution set will be $\{24.3º, 65.70º, 204.295º (-155.705º), 245.365º\}$ (check the graph below).

Graph of $\sin(2x)=\frac{3}{4}$

Values in the middle are the ones you are looking for (see graph below).

Graph of $sinx-cosx=\frac{1}{2}$

Has you can see by this graph, the actual set of solutions is much bigger. Let’s find the full solution set to your problem. To do so, select the solutions of the square problem which are also a solution to your initial problem in the range $\rm 0≤α≤360$ i.e., $\rm α_1=65.7º, α_2=204.295º$. Now, let’s see how many times the $\sin$ of $2×65.7º$ or $\sin (131.4º)$ repeats.

enter image description here

If we imagine a circle it becomes clear that after every full turn one gets back to same ($\rm 65.7º$) point, i.e., the first set of solutions satisfy $$\rm \Bbb α_{1n}=2nπ+α_1 \approx n×360º+65.7º, \forall n \in \Bbb N$$ But, we can check that every $180º-65.7º$ has the same $\sin$ of $65.7º$. Appling previous reasoning, the second set satisfy $$\rm \Bbb α’_{1n}=(2n+1)π-α_1 \approx n×360º+180º-65.7º=n×360º+114.3º, \forall n \in \Bbb N$$ We do the same with $\rm α_2$. The result is $$\rm \Bbb α_{2n}=2nπ+α_2 \approx n×360º+204.295º, \forall n \in \Bbb N$$ $$\rm \Bbb α’_{2n}=(2n+1)π-α_2 \approx n×360º+180º-204.295º=n×360º-24,295º, \forall n \in \Bbb N$$

The total set of solutions is: $$\rm \Bbb S= α_{1n} \cup α’_{1n} \cup α_{2n} \cup α’_{2n}, \forall n \in \Bbb N$$

  • $\begingroup$ very wise and complete explanation. thanks a lot $\endgroup$
    – KratosMath
    Nov 22, 2016 at 23:12

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