I have this expression to convert in trig form: $z=-\cos\frac{\pi}{7}+ i\sin\frac{\pi}{7}$
Tried to move the minus sine inside the $\cos$ function, but that just wouldn't work, then with the help of $\sin x=\cos(90-x))$ tried to change the expression to: $$ z=-\sin\frac{5\pi}{14}+i\cos\frac{5\pi}{14} $$
Here I moved the minus sine inside the sine function since it's allowed: $$ z=\sin\dfrac{-5\pi}{14}+i\cos\frac{5\pi}{14} $$
Again using $\cos x=\sin(90-x))$ to convert the real part to $\cos$: $$ \sin\frac{-5\pi}{14}=\cos\left(90+\frac{5\pi}{14}\right) $$
$$ \sin\frac{-5\pi}{14}=\cos\left(\frac{6\pi}{7}\right) $$
and the imaginary part to: $$ \cos\frac{5\pi}{14}=\sin\left(90-\frac{5\pi}{14}\right) $$
$$ \cos\frac{5\pi}{14}=\sin\left(\frac{\pi}{7}\right) $$
now I would get different angles for sine and cosine, which I think doesn't fit the general trig form: $z=r(\cos x+i\sin x)$
I would really appreciate the help for this problem!
$\endgroup$ 12 Answers
$\begingroup$What you need is write this expression in the form $\cos\theta+i\sin\theta$ with the same value of $\theta$ and without the minus sign in front of the cosine. Note that we have the following identities: $$\cos(\pi-\alpha)=-\cos\alpha \quad \text{and} \quad \sin(\pi-\alpha)=\sin\alpha.$$ Therefore: $$z=-\cos\frac{\pi}{7}+ i\sin\frac{\pi}{7}=\cos\left(\pi-\frac{\pi}{7}\right)+ i\sin\left(\pi-\frac{\pi}{7}\right)=\cos\frac{6\pi}{7}+ i\sin\frac{6\pi}{7}.$$
EDIT. Of course, in fact you need to write it in the form $r(\cos\theta+i\sin\theta)$. But after performing the step explained above, it's quite clear that $r=1$.
$\endgroup$ $\begingroup$If $$r(\cos y+i\sin y)=-\cos\dfrac\pi7+i\sin\dfrac\pi7,$$ where $r>0,y$ are real
Equate the real & the imaginary parts
to find $\dfrac{r\sin y}{r\cos y}=-\tan\dfrac\pi7=\tan\left(-\dfrac\pi7\right)$
using this
$\implies y=\pi+\left(-\dfrac\pi7\right)$
Square & add the real & the imaginary parts to find $r=1$
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