3
$\begingroup$

I have been beating my head against the following problem and would like a gentle nudge in the right direction.

The question states, by writing $1 + i$ and $\sqrt3 + i$ in polar form, deduce that

$$\cos (\frac{\pi}{12}) = \frac{\sqrt3 + 1}{2\sqrt2}, \sin(\frac{\pi}{12}) = \frac{\sqrt3 - 1}{2\sqrt2}$$

so I have written them in polar form

EDIT (as polar forms were incorrect): $$1 + i = \sqrt2e^{i\pi/4}, \sqrt3 + i = 2e^{i\pi/6}$$

another part of the question also asks you to put $\frac{1 + i}{\sqrt3 + i}$ into form $x + yi$ which I figured is

$$\frac{\sqrt3 + 1}{4} + \frac{1 - \sqrt3}{4}i$$

I just can't seem to connect it all together unfortunately so any help would be greatfully received.

$\endgroup$
  • 1
    $\begingroup$ $1+i = \sqrt{2}e^{\frac{i\pi}{4}}$ and $\sqrt{3}+i = 2e^{\frac{i\pi}{6}}$. $\endgroup$ – John Habert Mar 20 '14 at 14:01
  • 1
    $\begingroup$ Your polar forms are wrong. For example: $1+i=\sqrt{2}e^{i\frac{\pi}{4}}$. $\endgroup$ – Michael Hoppe Mar 20 '14 at 14:02
  • 2
    $\begingroup$ I don't understand the downvote on this. Writing a wrong expression in an answer deserves a downvote, in a question the expressions can be wrong. That is why the OP is asking! $\endgroup$ – Guy Mar 20 '14 at 14:10
3
$\begingroup$

Hint: $\frac14-\frac16=\frac{3}{12}-\frac{2}{12}=\frac{1}{12}$.

$\endgroup$
  • $\begingroup$ This is great, I felt this might be the case but as above I had gotten the polar forms wrong which confused me. I am accepting as this is core to the problem. $\endgroup$ – Matt Martineau Mar 20 '14 at 14:06
3
$\begingroup$

Notice

$$1 + i = \sqrt2e^{i\frac{\pi}{4}}$$ $$\sqrt{3} - i = 2e^{-i\frac{\pi}{6}}$$

Multiply both and we get

$$(1 + i)(\sqrt{3} - i) = \sqrt2e^{i(\frac{\pi}{4} - \frac{\pi}{6})}$$ $$(\sqrt3 + 1) + (\sqrt3-1)i= 2\sqrt{2}e^{i\frac{\pi}{12}}$$ $$(\sqrt3 + 1) + (\sqrt3-1)i= 2\sqrt2\left(\sin{\frac{\pi}{12}} + i\cos\frac{\pi}{12}\right)$$ $$\frac{\sqrt3 + 1}{2\sqrt2} + \frac{\sqrt3 - 1}{2\sqrt2}i = \cos{\frac{\pi}{12}} + i\sin\frac{\pi}{12}$$

Now, compare both real and imaginary parts to deduce:

$$\sin{\frac{\pi}{12}} = \frac{\sqrt3 + 1}{2\sqrt2}$$ $$\cos{\frac{\pi}{12}} = \frac{\sqrt3 - 1}{2\sqrt2}$$

$\endgroup$
  • $\begingroup$ Awesome, very thorough solution to the problem. I used this to check once I had given it a stab myself. $\endgroup$ – Matt Martineau Mar 20 '14 at 14:40
  • $\begingroup$ Hope it helped you, this appeared on my homework as well! $\endgroup$ – Yiyuan Lee Mar 20 '14 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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