4
$\begingroup$

I've tried using factor formula but still did not manage to get the answer, not sure if factor formula is the right method.

I rearrange to $\sin 4A + \sin 2A + \sin 3A + \sin A = 0$,

and after applying factor formula,

$2 \sin 3A \cos A + 2 \sin 2A \cos A = 0$

$2 \cos A ( \sin 3A + \sin 2A) = 0$

$2 \cos A ( \sin \frac{5}{2} A \cos \frac{1}{2} A) = 0$

Then I'm stuck..

$\endgroup$
2

5 Answers 5

5
$\begingroup$

From the point that you stopped

$$2 \cos A ( \sin \frac{5}{2} A \cos \frac{1}{2} A) = 0$$

$$2\cos(A/2)\cos(A)\sin(\frac{5A}{2})=0 \bigg/:2$$

$$\cos(A/2)\cos(A)\sin(\frac{5A}{2})=0$$

$\cos(A/2)=0\;$ or $\cos(A)=0\;$ or $\sin(\frac{5A}{2})=0$

$\endgroup$
5
  • $\begingroup$ I solve and got 0, 90, 144, 180 but im still missing 72(answer sheet shows that 72 is one of the answer). Can you see if you have 72 as one of your answer? $\endgroup$
    – ming
    Commented Jul 26, 2015 at 12:30
  • $\begingroup$ The general solution should be: $a_1=n\pi,\;a_2=n\pi-\frac{\pi}{2},\;a_3=2n\pi-\frac{4\pi}{5},\;a_ 4=2n\pi-\frac{2 \pi}{5} ,a_5=2n\pi+\frac{2\pi}{5} ,\; n \in \mathbb{Z}$ $\endgroup$
    – 3SAT
    Commented Jul 26, 2015 at 12:36
  • $\begingroup$ @ming $72^\circ$ satisfies $\sin\frac{5A}2 = 0$. In fact, the general solution from $\sin\frac{5A}2 = 0$ is $$\begin{align*} \frac{5A}2 &= 180^\circ n&n\in\mathbb Z\\ A &= 72^\circ n \end{align*}$$ $\endgroup$
    – peterwhy
    Commented Jul 26, 2015 at 12:45
  • $\begingroup$ to be honest I actually dont understand what are y'all doing, maybe a little but mostly no. here's how I approach, Let basic angle = a cos A = 0 sin 5/2 A = 0 cos 1/2 A = 0 a = 90 a = 0 a = 90 $\endgroup$
    – ming
    Commented Jul 26, 2015 at 12:46
  • $\begingroup$ @ming Then the problem is you have missed $\frac{5A}2 = 180^\circ$ and $\frac{5A}2 = 360^\circ$. $\endgroup$
    – peterwhy
    Commented Jul 26, 2015 at 12:50
3
$\begingroup$

You've done the hard work

Now the product of three multiplicands is zero

so at least one of them must be equal to zero

If $\sin B=0,B=n180^\circ$

If $\cos C=0, C=(2m+1)90^\circ$ where $m,n$ are integers

$\endgroup$
1
$\begingroup$

Had the number of summands been larger, we could employ the method dsecribed in How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?

If $\sin\dfrac A2=0,$ given expression holds true

and $\dfrac A2=n180^\circ\iff A=n360^\circ$ where $n$ is any integer

Else $A\ne n360^\circ$ and $\sin A+\sin2A+\sin3A+\sin4A=\cdots=\dfrac{\cos\dfrac A2-\cos\dfrac{9A}2}{\sin\dfrac A2}$

So we need $\cos\dfrac A2-\cos\dfrac{9A}2=0\iff\cos\dfrac{9A}2=\cos\dfrac A2$

$\implies\dfrac{9A}2=360^\circ m\pm\dfrac A2$ where $m$ is any integer, but $A\ne n360^\circ$

Consider the +, - separately

$\endgroup$
0
$\begingroup$

To be honest I actually don't understand what are y'all doing, maybe a little but mostly no.

here's how I approach,

Let basic angle = a

cos A = 0
a = 90
A = 90

sin 5/2 A = 0 a = 0
5/2 A = 0, 360
A = 0, 144

cos 1/2 A = 0

a = 90

1/2 A = 90

A = 180

Therefore, A = 0, 90, 144, 180

But where's the 72 T.T

$\endgroup$
2
  • $\begingroup$ Then the problem is you have missed $\frac{5A}2 = 180^\circ$. $\endgroup$
    – peterwhy
    Commented Jul 26, 2015 at 12:51
  • $\begingroup$ oh my gawd.. thanks so much! $\endgroup$
    – ming
    Commented Jul 26, 2015 at 12:54
0
$\begingroup$

we have, $$\sin A+\sin 2A+\sin 3A+\sin 4A=0$$ $$(\sin A+\sin 3A)+(\sin 2A+\sin 4A)=0$$ $$2\sin\left(\frac{A+3A}{2}\right)\cos\left(\frac{A-3A}{2}\right)+2\sin\left(\frac{2A+4A}{2}\right)\cos \left(\frac{A-3A}{2}\right)=0$$ $$2\sin 2A\cos A+2\sin 3A\cos A=0$$ $$2\cos A(\sin 2A+\sin 3A)=0$$ $$\implies \cos A=0\implies \color{blue}{A=2n\pi+\frac{\pi}{2}}$$ $$\implies \sin 2A+\sin 3A=0\implies \sin 3A=-\sin 2A$$ $$\implies 3A=2n\pi-2A\implies \color{blue}{A=\frac{2n\pi}{5}}$$ Or $$3A=(2n+1)\pi-(-2A)$$ $$\implies \color{blue}{A=(2n+1)\pi}$$ Where, $n$ is any integer.

Edit: For $0\leq A\leq 180^\circ$ put $n=0$, $n=1$ & $n=2$ in the solutions, we get the following $$\color{blue}{A\in \left\{0, \frac{2\pi}{5}, \frac{\pi}{2}, \frac{4\pi}{5}, \pi\right\}}$$ or $$\color{blue}{A\in \left\{0^\circ, 72^\circ, 90^\circ, 144^\circ, 180^\circ\right\}}$$

$\endgroup$
4
  • $\begingroup$ $$\begin{align*}\sin A + \sin 3A &= 2\sin 2A \color{red}\cos A\\ \sin 2A + \sin 4A &= 2 \sin 3A \color{red}\cos A\end{align*}$$ which is what OP has written. $\endgroup$
    – peterwhy
    Commented Jul 26, 2015 at 15:30
  • $\begingroup$ @peterwhy: So, could you please explain if there is something wrong in the solution, if you don't mind? $\endgroup$ Commented Jul 26, 2015 at 15:34
  • $\begingroup$ Then the $5$th line should be $$\implies \cos A = 0 \implies \color{blue}{A= n\pi + \frac\pi2} $$ $\endgroup$
    – peterwhy
    Commented Jul 26, 2015 at 15:37
  • $\begingroup$ Oh, sorry I couldn't spot the error. you are right. thanks! $\endgroup$ Commented Jul 26, 2015 at 15:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .