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

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

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

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

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

[edited from yeongil's post]
\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}

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

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

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

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

$0

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

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

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

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

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

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

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

\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}

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

\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:

\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}

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

\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!