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How to proceed in Question 12?

I was thinking of thinking of using "family of____" by making it in a single parameter of cos or sin. So, I took the sin term on the right hand side and squared, but got stuck. How to proceed?

Avinash Bhawnani
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    Don't post pictures of text on this site. Type out the question (otherwise, the question is not readable using a search engine) – 5xum Jan 22 '18 at 10:32
  • Hint:

    Let $a=\cos A+i\sin A$ etc.

    Comparing the real & the imaginary parts, $$\sin^2(C-A)+\sin^2(A-B)=\sin^2(B-C)$$

    Using https://math.stackexchange.com/questions/345703/prove-that-cos-a-b-cos-a-b-cos-2a-sin-2b,

    $$1-\cos(B-C)\cos(2A-B-C)=\sin^2(B-C)$$

    $$\iff\cos(B-C)\cos(C-A)\cos(A-B)=0$$

     – lab bhattacharjee Jan 22 '18 at 10:44

2 Answers2

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Hint: $\;|z|=1 \iff \bar z = \dfrac{1}{z}\,$, so taking the conjugate of $\,a+b\cos \alpha +c\sin \alpha = 0 \;\;(1)\,$ gives:

$$\frac{1}{a} + \frac{1}{b} \cos\alpha + \frac{1}{c}\sin \alpha = 0 \quad\iff\quad bc+a(c\cos \alpha +b\sin \alpha) = 0 \tag{2}$$

Substituting $\,a=-(b\cos \alpha +c\sin \alpha)\,$ from $(1)$ into $(2)\,$:

$$\require{cancel} \begin{align} 0 &= bc-(b\cos \alpha + c\sin \alpha)(c\cos \alpha +b\sin \alpha) \\ &= \cancel{bc} - \big(\cancel{bc(\cos^2 \alpha +\sin^2 \alpha)} + \sin \alpha \cos \alpha \,(b^2+c^2)\big) \\ &= \sin \alpha \cos \alpha \,(b^2+c^2) \end{align} $$

Given that $\,\sin \alpha, \cos \alpha \ne 0\,$, it follows that $\,b^2+c^2=0=(b+ic)(b-ic)\,$.

dxiv
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From the initial equation, draw $$\|-a\|^2=\|b\cos\alpha+c\sin\alpha\|^2=\|b\|^2\cos^2\alpha+\|c\|^2\sin^2\alpha+2\Re(bc^*)\sin\alpha\cos\alpha.$$

After simplification, we have that $bc^*$ is purely imaginary, hence (a).