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MATHMAA

# 16th KVJ Solutions6

Here you can find the 16th KVJ Solutions6 in detail explanation. Feel Free to send your opinion  by clicking on  SHARE YOU ENORMOUS EFFORT AND SMART EXAMPLES HERE

6Q) If $$cos \alpha + cos \beta + cos \gamma = sin \alpha + sin \beta + sin \gamma = 0$$, Prove that

$$cos (2\alpha) + cos (2\beta) + cos (2\gamma)$$ =  $$sin (2\alpha) + sin (2\beta) + sin (2\gamma) =0$$

Solution:

Let x= $$cos ( \alpha ) + i sin (\alpha )$$

y = $$cos (\beta ) + i sin (\beta )$$

z= $$cos (\gamma) + i sin (\gamma )$$

x+y+z = $$cos( \alpha) + cos (\beta) + cos (\gamma) + i ( sin (\alpha) + sin (\beta) + sin (\gamma)$$

x+y+z = 0 + i 0 =0

$$(x+y+z)^{2} = 0$$

$$x^{2}+y^{2}+z^{2} = -2(xy+yz+zx)$$

$$x^{2} + y^{2} + z^{2} = -2xyz(\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$$ -----------(1)

$$\frac{1}{x}$$ = $$\frac{1}{cos (\alpha) + i sin (\alpha)}$$

$$\frac{1}{x} = cos( \alpha) - i sin( \alpha )$$

Similarly we get

$$\frac{1}{y} = cos (\beta) - i sin (\beta )$$

$$\frac{1}{z} = cos (\gamma) - i sin (\gamma )$$

$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = cos(\alpha)+cos(\beta)+cos(\gamma) - i(sin(\alpha)+sin(\beta)+sin(\gamma))$$

=0 - i (0) =0

Therefore from one we get

$$x^{2}+y^{2}+z^{2} =-2xyz(0)$$

$$x^{2} + y^{2} + z^{2} = 0$$ ----------(2)

Using De Moivre's Theorem we get

$$x^{2} = ( cos (\alpha) + i sin(\alpha))^{2}$$

= $$cos (2 \alpha) + i sin (2 \alpha)$$

Similarly we get

$$y^{2} = cos ( 2\beta ) + i sin (2 \beta)$$

$$z^{2} = cos (2 \gamma ) + i sin (2 \beta )$$

$$x^{2} + y^{2} + z^{2} =0$$

$$x^{2} + y^{2} + z^{2} = cos (2 \alpha ) + cos(2\beta)+cos(2\gamma) + i (sin(2\alpha)+sin(2\beta)+sin(2\gamma)) =0$$

$$cos(2\alpha) +cos(2 \beta ) + cos (2 \gamma) =0$$

$$sin (2 \alpha) + sin(2\beta) + sin(2 \gamma) = 0$$

Hence proved .

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