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I have to solve the following equation:

$$\sin x + \cos x = \dfrac{1}{3} $$

I use the following substitution:

$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$

And by operating, I obtain: $$ \sqrt{(1-\cos^2 x)} = \dfrac{1}{3}-\cos x$$

$$ 1 - \cos^2 x = \dfrac{1}{9} + \cos^2 x - \dfrac{2}{3}\cos x$$

$$ -2\cos^2 x + 2/3\cos x +\dfrac{8}{9}=0$$

$$ \boxed{\cos^2 x -\dfrac{1}{3}\cos x -\dfrac{4}{9} = 0}$$

Can I just substitute $\cos x$ by $z$ and solve as if it was a simple second degree equation and then obtain $x$ by taking the inverse cosine? I have tried to do this but I cannot get the right result. If I do this, I obtain the following results:

$$ z_1 = -0.520517 \longrightarrow x_1 = 121.4º\\ z_2= 0.8538509 \longrightarrow x_2 = 31.37º$$

I obtain $x$ from $z$ by taking the inverse cosine.

The correct result should be around 329º which corresponds to 4.165 rad. My question is if what I am doing is wrong because I have tried multiple times and I obtain the same result (or in the worst case, I have done the same mistake multiple times).

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    $\begingroup$ It would be easier to use the identity $\sin x+\cos x=\sqrt2 \sin(x+\pi/4)$. $\endgroup$ – Henning Makholm Mar 31 '16 at 17:07
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    $\begingroup$ Yes, I know that's another way to solve it (and much easier), but it involves recalling too many formulas. Can it be done as I did it? Even though the developing is longer? @HenningMakholm $\endgroup$ – Airman01 Mar 31 '16 at 17:09
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    $\begingroup$ x @Airman, I didn't recall that; I derived it. See here how. $\endgroup$ – Henning Makholm Mar 31 '16 at 17:10
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    $\begingroup$ You have to be careful when taking the inverse cosine, because it only gives you angles between 0 and $\pi$. In general, if $\cos x = z$, then $z = \pm \arccos x$. $\endgroup$ – Gabe Cunningham Mar 31 '16 at 17:23
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    $\begingroup$ Your actual error was when you set $\sin x = \sqrt{1 - \cos^2 x}$. This assumes that $\sin x \ge 0$, which is why you didn't get a workable answer out of $\arccos$. It wasn't able to give a workable answer in the first place. Changing the sign as Gabe says works (when 360 is added) because that changes the sign of $\sin x$ to the negative it needs to be. $\endgroup$ – Paul Sinclair Mar 31 '16 at 18:41
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You can approach this problem with the substitution $$ \begin{cases} X=\cos x\\[4px] Y=\sin x \end{cases} $$ that transforms the equation into $$ \begin{cases} X+Y=\dfrac{1}{3} \\[6px] X^2+Y^2=1 \end{cases} $$ Rewriting the second equation as $(X+Y)^2-2XY=1$, we can substitute and get $$ \begin{cases} X+Y=\dfrac{1}{3} \\[6px] XY=-\dfrac{4}{9} \end{cases} $$ that leads to the solving equation $$ z^2-\frac{1}{3}z-\frac{4}{9}=0 $$ (finding two numbers knowing their sum and product). After rewriting it as $9z^2-3z-4=0$, we find the roots $$ \frac{1-\sqrt{17}}{6},\qquad \frac{1+\sqrt{17}}{6} $$ The solutions to the original problem are therefore $$ \begin{cases} \cos x=\dfrac{1-\sqrt{17}}{6},\\[6px] \sin x=\dfrac{1+\sqrt{17}}{6} \end{cases} \qquad \begin{cases} \cos x=\dfrac{1+\sqrt{17}}{6},\\[6px] \sin x=\dfrac{1-\sqrt{17}}{6} \end{cases} $$ We can express the solutions in terms of the arctangent by noting that in the first case the (principal) angle is in the interval $(\pi/2,\pi)$ and its tangent is $$ \frac{1+\sqrt{17}}{1-\sqrt{17}}=-\frac{9+\sqrt{17}}{8} $$ so the corresponding solution is $$ \pi-\arctan\frac{9+\sqrt{17}}{8}+2k\pi $$ In the second case the (principal) angle is in the interval $(-\pi/2,0)$ and its tangent is $$ \frac{1-\sqrt{17}}{1+\sqrt{17}}=\frac{\sqrt{17}-9}{8} $$ so the corresponding solution is $$ \arctan\frac{\sqrt{17}-9}{8}+2k\pi $$


There is another procedure that doesn't introduce extraneous solutions: remember the relations $$ \cos x=\frac{1-t^2}{1+t^2},\sin x=\frac{2t}{1+t^2} $$ where $t=\tan\dfrac{x}{2}$.

This is possible because $x=\pi$ is not a solution of the equation. Then you get $$ \frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}=\frac{1}{3} $$ that simplifies into $$ 2t^2-3t-1=0 $$ so you get $$ \tan\frac{x}{2}=\frac{3+\sqrt{17}}{4} \qquad\text{or}\qquad \tan\frac{x}{2}=\frac{3-\sqrt{17}}{4} $$ and so $$ x=2\arctan\frac{3+\sqrt{17}}{4}+2k\pi \qquad\text{or}\qquad x=2\arctan\frac{3-\sqrt{17}}{4}+2k\pi $$ In degrees, the first solution corresponds to $\approx121.367^\circ$ and the second one to $\approx-58.633^\circ$

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  • $\begingroup$ Thank you. Very well explained and detailed, I appreciate it. $\endgroup$ – Airman01 Mar 31 '16 at 19:01
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Hint: I think the approach of @JanEerland is instructive. Here some thoughts how we could find this kind of substitution.

When looking at \begin{align*} \sin x+\cos x=\frac{1}{3} \end{align*} and we think of the trigonometric addition formulas we know that \begin{align*} \sin(x+a)=\sin x \cos a+\cos x\sin a \end{align*} It would be convenient if $\cos a=\sin a$, so that we can separate them. This is the case if $a=\frac{\pi}{4}$ and we obtain \begin{align*} \sin\left(x+\frac{\pi}{4}\right) &=\sin x\sin \frac{\pi}{4}+\cos x\cos \frac{\pi}{4}\\ &=\frac{\sqrt{2}}{2}\left(\sin x+\cos x\right) \end{align*} It follows \begin{align*} \sin x+\cos x&=\frac{1}{3}\\ \sin\left(x+\frac{\pi}{4}\right)&=\frac{1}{3\sqrt{2}} \end{align*}

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$$\cos(x)+\sin(x)=\frac{1}{3}\Longleftrightarrow$$


Use:

$$\cos(x)+\sin(x)=\sqrt{2}\left[\frac{\cos(x)}{\sqrt{2}}+\frac{\sin(x)}{\sqrt{2}}\right]=$$ $$\sqrt{2}\left(\sin\left(\frac{\pi}{4}\right)\cos(x)+\cos\left(\frac{\pi}{4}\right)\sin(x)\right)=\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)$$


$$\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)=\frac{1}{3}\Longleftrightarrow$$ $$\sin\left(\frac{\pi}{4}+x\right)=\frac{1}{3\sqrt{2}}$$

Now, when we take the inverse sine of both sides, we got two options, with $n_1\space\wedge\space n_2\in\mathbb{Z}$:

  • $$\frac{\pi}{4}+x=\pi-\arcsin\left(\frac{1}{3\sqrt{2}}\right)+2\pi n_1\Longleftrightarrow x=\frac{3\pi}{4}-\arcsin\left(\frac{1}{3\sqrt{2}}\right)+2\pi n_1$$
  • $$\frac{\pi}{4}+x=\arcsin\left(\frac{1}{3\sqrt{2}}\right)+2\pi n_1\Longleftrightarrow x=\arcsin\left(\frac{1}{3\sqrt{2}}\right)+2\pi n_2-\frac{\pi}{4}$$
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  • $\begingroup$ Thank you, nice approach to solve it and clear explanation. $\endgroup$ – Airman01 Mar 31 '16 at 17:27
  • $\begingroup$ @Airman01 You're welcome! I'm glad that I could help $\endgroup$ – Jan Mar 31 '16 at 17:28
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    $\begingroup$ @JanEerland: Nice approach. (+1) $\endgroup$ – Markus Scheuer Mar 31 '16 at 18:45
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we have after squaring $$\sin(x)^2+\cos(x)^2+\sin(2x)=\frac{1}{9}$$ or $$\sin(2x)=-\frac{8}{9}$$ the answer is $$c_1\in \mathbb{Z}\land \left(x=2 \pi c_1+2 \tan ^{-1}\left(\frac{1}{10} \left(9-\sqrt{161}\right)\right)\lor x=2 \pi c_1+2 \tan ^{-1}\left(\frac{1}{10} \left(9+\sqrt{161}\right)\right)\right)$$

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  • $\begingroup$ Thank you this is such a good approach, much easier than mine or any other that comes up to my mind. $\endgroup$ – Airman01 Mar 31 '16 at 17:24
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    $\begingroup$ Squaring introduces extraneous solutions. $\endgroup$ – egreg Mar 31 '16 at 17:31
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Others have proposed other correct methods for solving the equation. I propose to complete your chosen (ultimately correct) method.

There are two problems with your solution:

  1. You have taken a square root of both sides in the second line of your solution, and then squared both sides in the fourth line. Both these steps lose information about the sign of expressions, and introduce the possibility of finding roots that are, in fact, not correct roots, but extraneous roots. Thus, in these cases,(maybe all cases!) all possible solutions should be verified by substitution into the original equation.
  2. Whenever you use an inverse trigonometric function to find an angle, you must remember that there are always two angles in the range from $0$ to $2\pi$ that satisfy the given trigonometric equation. (Except for edge cases like $90^o$)

You have found that: $$ z_1 = -0.520517 $$ My calculator gives me $121.37^o$ as the solution. But since the cosine function is negative in the second and third quadrants, the other solution is $238.63^o$.

Similarly,$$ z_2 = 0.8538509 $$ gives $31.37^o$. But cosine is positive in the first and fourth quadrant, so $328.63^o$ is another possible answer.

Checking these four values into the original equation will give the values (only one in this case) that apply to the original equation.

Apart from these two steps, your solution is correct.

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$$(\sin(x) + \cos(x))^2 = 1/9$$ $$\sin^2(x) + \cos^2(x) + 2\sin(x)\cos(x) = 1/9$$ $$1 + \sin(2x) = 1/9$$ $$\sin(2x) = -8/9$$ $$x = \arcsin(-8/9) / 2$$ $$ x = -0.547457... $$

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Another way to solving this by the tangent half angle substitution $t = \tan \frac{x}{2}$ (or $x = 2 \arctan t$)

$$ \begin{align} \cos x & \rightarrow \frac{1-t^2}{1+t^2} \\ \sin x & \rightarrow \frac{2 t}{1+t^2} \end{align} $$

Your equation becomes

$$ \frac{2 t}{1+t^2} + \frac{1-t^2}{1+t^2} = \frac{1}{3} $$ which has two solutions

$$ \left. \begin{aligned} t & = \frac{3}{4} + \frac{\sqrt{17}}{4} \\ t & = \frac{3}{4} - \frac{17}{4} \end{aligned} \right\} \begin{aligned} \cos x & = \frac{1}{6} - \frac{17}{6} & \sin x & = \frac{1}{6} + \frac{17}{6} \\ \cos x & = \frac{1}{6} + \frac{17}{6} & \sin x & = \frac{1}{6} - \frac{17}{6} \end{aligned} $$

The two solutions are then

$$ \begin{aligned} x & = \arctan \left( \frac{\sqrt{17}-9}{8} \right) + 2 \pi n \\ x & = \frac{\pi}{2} - \arctan \left( \frac{\sqrt{17}-9}{8} \right) + 2 \pi n \end{aligned} $$

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An even simpler way is to substitute $x \rightarrow y - \frac{\pi}{4}$

You equation is now $$\sqrt{2} \sin(y) = \frac{1}{3}$$

The solutions are

$$\begin{aligned} x & = \arcsin \left( \frac{\sqrt{2}}{6} \right) - \frac{\pi}{4} + 2\pi n \\ x & = \arccos \left( \frac{\sqrt{2}}{6} \right) - \frac{\pi}{4} + 2 \pi n \end{aligned}$$

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