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The task is to manipulate $$\cos(x) + \sqrt{3}\sin(x)$$

into the form $$A\sin(x+ \theta)$$

My question is : why is $\pi$ in the numerator and denominator both divided by $6$?

I am familiar with most of the question, however I get stuck in a certain area. The following image shows the question as a whole:- enter image description here

I am unsure why $\pi$ in the numerator and denominator are both divided by $6$?

Please can someone explain how this is so ?

Many thanks.

Aztec warrior
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    I recommend you type your steps out instead of giving us a picture for your future questions. – Arbuja Jan 31 '16 at 02:26
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    Because $30^\circ={\pi\over6}$ radian. – Jack's wasted life Jan 31 '16 at 02:26
  • @Jack'swastedlife, that's what I initially thought, however I set my calculator to Rad and inputted π/6 and the result was 0.5235. – Aztec warrior Jan 31 '16 at 02:30
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    Your calculator is giving you numerical value of ${\pi\over 6}$, it's not doing unit conversion. – Jack's wasted life Jan 31 '16 at 02:31
  • You can have a look at some very similar posts asked in the past: http://math.stackexchange.com/questions/363222/expressing-cos-theta-sqrt3-sin-theta-r-sin-theta-alpha http://math.stackexchange.com/questions/201399/solving-sin-x-sqrt-3-cos-x-1-is-my-solution-correct http://math.stackexchange.com/questions/213545/solving-trigonometric-equations-of-the-form-a-sin-x-b-cos-x-c http://math.stackexchange.com/questions/397984/identity-for-a-weighted-sum-of-sines-sines-with-different-amplitudes – Martin Sleziak Jan 31 '16 at 07:09

1 Answers1

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First expand $\sin(x+\theta)$:

$$\sin(x+\theta)=\cos(\theta)\sin(x)+\sin(\theta)\cos(x)$$

Since You want to find $A$ such that

$$A\sin(x+\theta)=\cos(x)+\sqrt{3}\sin(x)$$

You obtain the following conditions for $\theta$:

$$ \begin{cases} \sin(\theta)=1/A\\ \cos(\theta)=\sqrt{3}/A \end{cases} $$

Now using the Pythagoras theorem one gets $1=\sin^2(\theta)+\cos^2(\theta)=4/A^2$ from where $A=2$. Finally from

$$ \begin{cases} \sin(\theta)=1/2\\ \cos(\theta)=\sqrt{3}/2 \end{cases} $$

we get $\theta=\pi/6$.

Marcin Malogrosz
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  • Thankyou for your response, I can say I follow most of the process. Please may you explain in a little more detail where you have got $$4/A^2$$ , and how does A = 2 ? Sorry for the inconvenience. – Aztec warrior Jan 31 '16 at 02:58
  • $1=\sin^2(\theta)+\cos^2(\theta)=(1/A)^2+(\sqrt{3}/A)^2=4/A^2$. Now multiply by $A^2$, and take the square root. – Marcin Malogrosz Jan 31 '16 at 03:44