Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The question is to solve: $$ \sqrt{3} \sin \left ( \phi - \frac \pi 6 \right )= \sin \phi $$

I tried to turn it into $a \sin \phi-b \cos\phi= \sin\phi$,

Then I got $ \frac 32 \sin\phi - \frac 3 4 \cos \phi= \sin\phi$,

Therefore, $ \frac 12 \sin\phi- \frac 3 4 \cos\phi=0$.

Then I turned it back into $R\sin(\phi-a)=0$,

$\implies (√13/4)\sin(ø-0.983)=0$

So I think the solution should be $\phi=n\pi+0.983$

Is that correct? The answer on my book is $n\pi+\frac \pi 3$.(I made a mistake here, It should be π/3.)

Update: The reason why I got the wrong answer is because I forgot to √ that 4/3.

Thanks for pointing that mistake out :D.

share|improve this question
    
The coefficient of $-3/4$ in front of $\cos\phi$ is incorrect. It should be $-\sqrt{3}/2$. –  anon Jun 27 '12 at 7:31

2 Answers 2

up vote 1 down vote accepted

It may be directly written that: $$\sqrt3 \cos\phi= \sin \phi \implies \tan\phi = \sqrt{3} \implies \phi = n \pi + \dfrac{\pi}3 \text{ where }n\in \mathbb{Z}$$

Q.E.D.

share|improve this answer

$$\sqrt{3} \sin(\phi - \pi/6) = \sin(\phi) \implies \sqrt{3} \left( \sin(\phi) \cos(\pi/6) - \cos(\phi) \sin(\pi/6)\right) = \sin(\phi)$$ This gives us $$\sqrt{3}\sin(\phi) \times \sqrt{3}/2 - \sqrt{3} \cos(\phi)\times1/2 = \sin(\phi) \implies 3/2 \sin(\phi) - \sin(\phi) = \sqrt{3}/2 \cos(\phi)\\ \implies \dfrac{\sin(\phi)}2 = \dfrac{\sqrt{3}}2 \cos(\phi) \implies \tan(\phi) = \sqrt{3} \implies \phi = n \pi + \dfrac{\pi}3 \text{ where }n\in \mathbb{Z}$$ The mistake you have made is that your $b$ should be $\sqrt{3}/2$ and not $3/4$ as you have written. Also, the answer given in your book answer is incorrect, assuming you have stated the problem correctly.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.