# Thread: Find the value between 0 and 2pi

1. ## Find the value between 0 and 2pi

Find the values of A in the range between 0 and 2pi for which
sinA+sin3A=cosA+cos3A

I have been trying for two days and still can't get it yet. Any help is greatly appreciated.

2. Originally Posted by arze
Find the values of A in the range between 0 and 2pi for which
sinA+sin3A=cosA+cos3A

I have been trying for two days and still can't get it yet. Any help is greatly appreciated.
$\displaystyle \sin C+\sin D=2\sin\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)$

$\displaystyle \cos C+\cos D=2\cos\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)$

$\displaystyle \sin A+\sin 3A=2\sin2A\cos A$

$\displaystyle \cos A+\cos 3A=2\cos 2A\cos A$

$\displaystyle \Rightarrow 2\sin2A\cos A=2\cos 2A\cos A$

[edited from yeongil's post]
\displaystyle \begin{aligned} 2\sin2A\cos A &= 2\cos 2A\cos A \\ \sin2A\cos A &= \cos 2A\cos A \\ \sin2A\cos A - \cos 2A\cos A &= 0 \\ \cos A(\sin2A - \cos 2A) &= 0 \end{aligned}

\displaystyle \begin{aligned} \cos A &= 0 \\ A &= \frac{\pi}{2},\;\frac{3\pi}{2} \end{aligned}

$\displaystyle \Rightarrow \sin 2A=\cos 2A$

$\displaystyle \Rightarrow \sin 2A=\sin\left(\frac\pi{2}-2A\right)$

$\displaystyle \Rightarrow 2A=n\pi+(-1)^n\left(\frac\pi{2}-2A\right)$

$\displaystyle 0<A<2\pi$

Possible values $\displaystyle \frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{8},\frac{ 7\pi}{2},\frac{9\pi}{8},\frac{11\pi}{2},\frac{13\p i}{8}$

3. Originally Posted by great_math
$\displaystyle \sin C+\sin D=2\sin\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)$

$\displaystyle \cos C+\cos D=2\cos\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)$

$\displaystyle \sin A+\sin 3A=2\sin2A\cos A$

$\displaystyle \cos A+\cos 3A=2\cos 2A\cos A$

$\displaystyle \Rightarrow 2\sin2A\cos A=2\cos 2A\cos A$

$\displaystyle \Rightarrow {\color{red}\sin 2A=\cos 2A}$
It's not a good idea to divide both sides by cos A like that. You lose roots that way. Instead, do this:

\displaystyle \begin{aligned} 2\sin2A\cos A &= 2\cos 2A\cos A \\ \sin2A\cos A &= \cos 2A\cos A \\ \sin2A\cos A - \cos 2A\cos A &= 0 \\ \cos A(\sin2A - \cos 2A) &= 0 \end{aligned}

\displaystyle \begin{aligned} \cos A &= 0 \\ A &= \frac{\pi}{2},\;\frac{3\pi}{2} \end{aligned}

\displaystyle \begin{aligned} \sin2A - \cos 2A &= 0 \\ \sin 2A &= \cos 2A \end{aligned}
Then proceed as in the previous post.

01

4. Originally Posted by yeongil
It's not a good idea to divide both sides by cos A like that. You lose roots that way. Instead, do this:

\displaystyle \begin{aligned} 2\sin2A\cos A &= 2\cos 2A\cos A \\ \sin2A\cos A &= \cos 2A\cos A \\ \sin2A\cos A - \cos 2A\cos A &= 0 \\ \cos A(\sin2A - \cos 2A) &= 0 \end{aligned}

\displaystyle \begin{aligned} \cos A &= 0 \\ A &= \frac{\pi}{2},\;\frac{3\pi}{2} \end{aligned}

\displaystyle \begin{aligned} \sin2A - \cos 2A &= 0 \\ \sin 2A &= \cos 2A \end{aligned}
Then proceed as in the previous post.

01
Oops! I keep making that mistake again and again ....

5. Thank you so much!