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How do I solve $\tan x = \sin(x +45^{\circ})$?

This is how far I have come: $\sqrt{2}\sin x = \sin x\cdot\cos x + \cos^2 x$

quid
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guest
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  • You can express $\cos x=\sqrt{1-\sin(x)^2}$ and solve the equation. – user1537366 Jan 04 '15 at 14:35
  • it is wrong you must set $\cos(x)=\pm\sqrt{1-\sin(x)^2}$ – Dr. Sonnhard Graubner Jan 04 '15 at 14:48
  • Substitute $x=2\arctan(t)$, it usually helps for some reason. (Note: $\sin(x)=\frac{2t}{t^2+1}$, $\cos(x)=\frac{1-t^2}{t^2+1}$, and $\tan(x)=\frac{2t}{1-t^2}$.) – Akiva Weinberger Jan 04 '15 at 16:23
  • There are two solution of $x$ with $|x| \le 180^\circ$: one is the obvious $\pi/4 = 45^\circ$. The other one is

    $$\tan^{-1}\left( \frac{\sqrt[3]{6\sqrt{78}-8}}{6} - \frac{7}{3\sqrt[3]{6\sqrt{78}-8}} - \frac13 \right) + \pi\ \approx 2.764013611898439 \approx 158.3663144784912^\circ$$ What is sad is that only the most downvoted answer here seems to get the second solution.

     – achille hui Jan 18 '15 at 04:02

3 Answers3

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we will show that the only solution to $\tan x = \sin(x+45^\circ)$ is $x = 45^\circ$

here is another way to solve $\tan t = \sin (t + 45^\circ).$ i have changed the independent variable to $t$ because i want to use $x = \cos t$ and $y = \sin t.$ using the addition formula for $\sin$ we can rewrite the equation as two equations in $x,y$ in the following way: $${y \over x} = {x + y\over \sqrt 2}, x^2 + y^2 = 1$$ this simplifies to the unit circle and a hyperbola $$x^2 + y^2 = 1, x^2 + xy -\sqrt 2y = 0 $$ the unit circle and hyperbola may cut at most four points. one such point is $(\sqrt 2/2,\sqrt 2/2)$ using quadratic formula we can solve write the equation for hyperbola explicitly as $$2x = -y \pm \sqrt{y^2 + 4\sqrt 2y} $$ which shows that $-\infty < y \le -4\sqrt2$ or $0 \le y <\infty$ therefore the lower branch of the hyperbola cannot cut the unit circle.so we are only looking for the intersection of $$ x^2 + y^2 = 1, 2x = \sqrt{y^2 + 4\sqrt 2 y}-y, y>0$$ the only solution for this is one we already guessed that $$x = y = {\sqrt 2 \over 2} \text{ which corresponds to } t = 45^\circ$$

i wish i could add the graphs of the unit circle and the graphs of $2x = -y \pm \sqrt{y^2 + 4\sqrt 2y} $

abel
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  • I don't understand what you are doing! Please, could you try to do calculate the question in a more simplified way – guest Jan 04 '15 at 17:11
  • yx=x+y/√2 ,x^2+y^2=1 I understand the part before the comma, but then after the comma I don't understand where those numbers come from – guest Jan 04 '15 at 17:14
  • @guest, have you used unit circle to resolve trigonometric equation? that is thinking $\cos t, \sin t$ as the $x$ and $y$ coordinate on the unit circle and $t$ as the angle between the positive $x$-axis and the radius $(0,0)$ to $(x,y)$ – abel Jan 04 '15 at 17:16
  • yes I have. I really appreciate your effort in trying to help me – guest Jan 04 '15 at 17:18
  • you don't have $yx = (x+y)/\sqrt 2$ but $y/x = (x+y)/\sqrt 2$ because $\tan = \sin / \cos$ – abel Jan 04 '15 at 17:19
  • yes I understand that. The part I don't understand is when you add x^2 + y^2 and equals everything to 1 – guest Jan 04 '15 at 17:21
  • remember the identity $\sin^2 t + \cos^2 t= 1$ for all angles $t.$ that is the reason we can take the $x,y$-coordinates as $x = \cos t, y = \sin t.$ when you make this change, all trig becomes basically dealing with quadratics and lines. – abel Jan 04 '15 at 17:24
  • yes I know that but how come you added x^2 + y^2 to the equation? – guest Jan 04 '15 at 17:26
  • that is the link between the angle $t$ and the $x, y$. once we find $x$ and $y$ we still have to back to the unit circle and find the angle/radius whose terminal point is that $x,y$ for example we have the solution $x = y = \sqrt 2 /2$ the angle that it correspond to is $45^\circ$ plus multiples of $360^\circ$ – abel Jan 04 '15 at 17:29
  • @guest, have you figured out? – abel Jan 04 '15 at 17:47
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Hint: use that $\sin^2x + \cos^2x =1$.

Martín-Blas Pérez Pinilla
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Set $x=y+45^\circ$ to get $\tan(y+45^\circ)=\sin(45^\circ+45^\circ+y)=\cos y$

Clearly, $y=0$ is a solution

Now $-1\le\cos y\le1\implies-1\le\tan(y+45^\circ)\le1$

If we consider $-180^\circ<y\le180^\circ,-45^\circ\le y+45^\circ\le45^\circ\iff-90^\circ\le y\le0\implies\cos y\ge0$

$\implies0\le\tan(y+45^\circ)\le1\implies0\le y+45^\circ\le45^\circ\iff-45^\circ\le y\le0$ $\implies\cos y\ge\cos45^\circ=\dfrac1{\sqrt2}$

$\implies\dfrac1{\sqrt2}\le\tan(y+45^\circ)\le1$

But as $\dfrac1{\sqrt2}>\dfrac1{\sqrt2+1}=\sqrt2-1=\tan15^\circ,$

$\implies15^\circ<y+45^\circ\le45^\circ\iff-30^\circ<y\le0$

$\implies\cos y\ge\cos30^\circ=\dfrac{\sqrt3}2$

If we follow this method, the range will continually shrink and eventually we shall reach at $0\le y\le0\iff y=0$

lab bhattacharjee
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