3

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..

wythagoras
  • 25,026
ming
  • 91
  • 5

5 Answers5

4

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

3SAT
  • 7,512
  • 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?

     – ming Jul 26 '15 at 12:30
  • The general solution should be: $a_1=n\pi,;a_2=n\pi-\frac{\pi}{2},;a_3=2n\pi-\frac{4\pi}{5},;a_ 4=2n\pi-\frac{2 \pi}{5} ,a_5=2n\pi+\frac{2\pi}{5} ,; n \in \mathbb{Z}$ – 3SAT Jul 26 '15 at 12:36
  • @ming $72^\circ$ satisfies $\sin\frac{5A}2 = 0$. In fact, the general solution from $\sin\frac{5A}2 = 0$ is $$\begin{align} \frac{5A}2 &= 180^\circ n&n\in\mathbb Z\ A &= 72^\circ n \end{align}$$ – peterwhy Jul 26 '15 at 12:45
  • 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

     – ming Jul 26 '15 at 12:46
  • @ming Then the problem is you have missed $\frac{5A}2 = 180^\circ$ and $\frac{5A}2 = 360^\circ$. – peterwhy Jul 26 '15 at 12:50
2

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

lab bhattacharjee
  • 274,582
1

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

Community
  • 1
lab bhattacharjee
  • 274,582
0

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 $$\color{blue}{A\in \left\{0^\circ, 72^\circ, 90^\circ, 144^\circ, 180^\circ\right\}}$$

Harish Chandra Rajpoot
  • 37,450
0

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

ming
  • 91
  • 5