Solving $\sin^{2015}x+\cos^{2015}x=\frac12$

Question

Find all the roots of $$\sin^{2015}x+\cos^{2015}x=\frac12\tag{1}$$

I'm a high school student, and this is my homework. This's my try:

Let $\displaystyle t=\tan \frac x2\Rightarrow \sin x=\frac{2t}{1+t^2}, \ \ \cos x=\frac{1-t^2}{1+t^2}$

We'll have $\displaystyle\color{Red}{(1)} \Leftrightarrow \frac{(2t)^{2015}+\sum\limits_{k=0}^{2015}\binom{2015}k\cdot \left ( -1 \right )^{2015-k}\cdot t^{2k }}{\left (1+t^2 \right )^{2015}}=\frac12$

But I think it is very complicated. I have no ideas for it, please help me.

What have you been taught about the sine and cosine functions. — Mark Viola, Aug 10 '15 at 01:50
Hint: you need either $\sin x$ or $\cos x$ very close to $1$, else the left hand side is very close to $0$ (and can't equal $\frac 12$). — vadim123, Aug 10 '15 at 02:02
Are you looking for exact solutions or numerical solutions? Perhaps you're supposed to find an approximate solution? — Ben Grossmann, Aug 10 '15 at 02:08
@Omnomnomnom I would like to reiterate that this is my homework. I think an exact solution would be better. But either exact solutions or numerical solutions are both good. — mja, Aug 10 '15 at 02:17
You can just neglect one of the terms and work only with the other one like solve $cos(x)^{2015}=\frac{1}{2}$. Do the same for sin(x) and you get all the answers. It works because the other number will be a number close to 0 which will almost banish when taking the 2015th power. — Martín Forsberg Conde, Aug 10 '15 at 02:38
$x\simeq\arccos\dfrac1{\sqrt[2015]2}$ or $x\simeq\arcsin\dfrac1{\sqrt[2015]2}$ — Lucian, Aug 10 '15 at 03:02
@DDK are you sure you typed the question correctly? If this is homework, then it is likely you are missing a factor of $i$ : $$\cos(x)^{2015} + {\color{red}i}~\sin(x)^{2015} = \frac12$$ — DanielV, Sep 02 '15 at 17:15

score 2 · Accepted Answer · answered Aug 16 '15 at 14:23

To get an approximation, note that either $\cos x$ or $\sin x$ must be very close to $1$. The other will be tiny, so we ignore it. Looking for the root just above $0$, you can use the Taylor series for the cosine. $(1-\frac {x^2}2)^{2015}=\frac 12$ This is a quadratic in $x$

Solving $\sin^{2015}x+\cos^{2015}x=\frac12$

1 Answers1