# Thread: trig arithmetic series..

1. ## trig arithmetic series..

The first three terms of an arithmetic series are √3cosθ, sin(θ−30°) and sin θ, where θ is acute. Find the value of θ.
So far I have expanded sin(θ−30°)
to get;

$\displaystyle \sqrt{3}cos\theta + \frac{\sqrt{3}}{2}sin\theta -\frac{1}{2}cos\theta + sin\theta$

how do I find theta? Is there a formula?

2. Originally Posted by Tweety
So far I have expanded sin(θ−30°)
to get;

$\displaystyle \sqrt{3}cos\theta + \frac{\sqrt{3}}{2}sin\theta -\frac{1}{2}cos\theta + sin\theta$

how do I find theta? Is there a formula?
You can use the usual arithmetic sequence formulae. It will require simultaneous equations though

$\displaystyle U_3 = U_2 + d = a + 2d$

$\displaystyle U_2 = a + d$

$\displaystyle a = \sqrt{3}cos\theta$

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$\displaystyle \sqrt{3}cos\theta + \frac{\sqrt{3}}{2}sin\theta -\frac{1}{2}cos\theta + sin\theta = \sqrt{3}cos\theta + 2d$

$\displaystyle \therefore \: \: \frac{2+\sqrt{3}}{2}\,sin\theta -\frac{1}{2}cos\theta = d$..........[1]

$\displaystyle sin \theta = \sqrt{3}cos \theta + 2d \:\: \rightarrow \:\: d = \frac{1}{2}(sin \theta - \sqrt{3}cos\theta)$............[2]

sub [1] into [2]

$\displaystyle \frac{2+\sqrt{3}}{2}\,sin\theta -\frac{1}{2}cos\theta = \frac{1}{2}(sin \theta - \sqrt{3}cos\theta) = \frac{1}{2}sin\theta - \frac{\sqrt3}{2}cos\theta$

$\displaystyle \frac{1+\sqrt3}{2}sin\theta + \frac{\sqrt3 -1}{2}cos\theta=0$

$\displaystyle (1+\sqrt3)sin\theta = -(\sqrt3 -1)cos\theta = (1-\sqrt3)cos\theta$

$\displaystyle \frac{sin\theta}{cos\theta} = tan\theta = \frac{1-\sqrt3}{1+\sqrt3}$

$\displaystyle \theta = arctan \left(\frac{1-\sqrt3}{1+\sqrt3}\right) = -15^{\circ}$

3. HGello, Tweety!

The first three terms of an arithmetic sequence are: .$\displaystyle \sqrt{3}\cos\theta,\;\sin(\theta-30^o),\;\sin\theta$
where $\displaystyle \theta$ is acute. . Find the value of $\displaystyle \theta.$
The first three terms are: .$\displaystyle a,\;a+d,\;a+2d$

We have: .$\displaystyle \begin{Bmatrix}a &=& \sqrt{3}\cos\theta & [1] \\ a + d &=& \sin(\theta-30^o) & [2] \\ a+2d &=& \sin\theta & [3] \end{Bmatrix}$

Subtract [2] from [1]: .$\displaystyle d \:=\:\sin(\theta-30^o) - \sqrt{3}\cos\theta \;\;[4]$

Subtract [3] from [2]: .$\displaystyle d \;=\;\sin\theta - \sin(\theta-30^o)\;\;[5]$

Equate [4] and [5]: .$\displaystyle \sin(\theta-30^o) - \sqrt{3}\cos\theta \;=\;\sin\theta - \sin(\theta-30^o)$ . $\displaystyle \Rightarrow\quad 2\sin(\theta-30^o) \;=\;\sin\theta + \sqrt{3}\cos\theta$

Divide by 2: .$\displaystyle \sin(\theta-30^o) \;=\;\tfrac{1}{2}\sin\theta + \tfrac{\sqrt{3}}{2}\cos\theta$

Since $\displaystyle \sin30^o = \tfrac{1}{2}\,\text{ and }\,\cos30^o = \tfrac{\sqrt{3}}{2}$

. . we have: .$\displaystyle \sin(\theta-30^o) \;=\;\sin30^o\sin\theta + \cos30^o\cos\theta$

. . . That is: .$\displaystyle \sin(\theta-30^o) \;=\;\cos(\theta -30^o) \quad\Rightarrow\quad \frac{\sin(\theta-30^o)}{\cos(\theta-30^o)} \;=\;1$

Therefore: .$\displaystyle \tan(\theta-30^o) \:=\:1 \quad\Rightarrow\quad \theta-30^o \:=\:45 \quad\Rightarrow\quad\boxed{ \theta \:=\:75^o}$