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Properties of Conjugate of a Complex Number

If X = sin ...

Question

If $X = sin (θ + \frac{7 π}{12}) + sin (θ - \frac{π}{12}) + sin (θ + \frac{3 π}{12}), Y = cos (θ + \frac{7 π}{12}) + cos (θ - \frac{π}{12}) + cos (θ + \frac{3 π}{12}),$ then prove that $\frac{X}{Y} - \frac{Y}{X} = 2 tan 2 θ .$

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Solution

Given that,

$X = sin (θ + \frac{7 π}{12}) + sin (θ - \frac{π}{12}) + sin (θ + \frac{3 π}{12})$

$Y = cos (θ + \frac{7 π}{12}) + cos (θ - \frac{π}{12}) + cos (θ + \frac{3 π}{12})$

Then,

$X = sin (θ + \frac{7 π}{12}) + sin (θ - \frac{π}{12}) + sin (θ + \frac{3 π}{12})$

$= 2 sin \frac{(θ + \frac{7 π}{12} + θ - \frac{π}{12})}{2} cos \frac{(θ + \frac{7 π}{12} - θ + \frac{π}{12})}{2} + sin (θ + \frac{3 π}{12})$

$= 2 sin \frac{(2 θ + \frac{6 π}{12})}{2} cos \frac{8 π}{24} + sin (θ + \frac{3 π}{12})$

$= 2 sin (θ + \frac{π}{4}) cos \frac{π}{3} + sin (θ + \frac{3 π}{12})$

$= 2 \times \frac{1}{2} sin (θ + \frac{π}{4}) + sin (θ + \frac{3 π}{12})$

$= sin (θ + \frac{π}{4}) + sin (θ + \frac{3 π}{12})$

$X = 2 sin (θ + \frac{π}{4})$

Similarly,

$Y = cos (θ + \frac{7 π}{12}) + cos (θ - \frac{π}{12}) + cos (θ + \frac{3 π}{12})$

$= 2 cos \frac{(θ + \frac{7 π}{12} + θ - \frac{π}{12})}{2} cos \frac{(θ + \frac{7 π}{12} - θ + \frac{π}{12})}{2} + cos (θ + \frac{3 π}{12})$

$= 2 cos (θ + \frac{π}{4}) cos \frac{π}{3} + cos (θ + \frac{3 π}{12})$

$= cos (θ + \frac{π}{4}) + cos (θ + \frac{π}{4})$

$Y = 2 cos (θ + \frac{π}{4})$

Put the value of $X & Y$ in LHS we get,

LHS

$\frac{X}{Y} - \frac{Y}{X}$

$= \frac{2 sin (θ + \frac{π}{4})}{2 cos (θ + \frac{π}{4})} - \frac{2 cos (θ + \frac{π}{4})}{2 sin (θ + \frac{π}{4})}$

$= \frac{sin (θ + \frac{π}{4})}{cos (θ + \frac{π}{4})} - \frac{cos (θ + \frac{π}{4})}{sin (θ + \frac{π}{4})}$

$= tan (θ + \frac{π}{4}) - cot (θ + \frac{π}{4})$

$= \frac{tan θ + t a n \frac{π}{4}}{1 - tan θ tan \frac{π}{4}} - \frac{c o θ cot \frac{π}{4} - 1}{c o θ + cot \frac{π}{4}}$

$= \frac{1 + tan θ}{1 - tan θ} - \frac{cot θ - 1}{cot θ + 1}$

$= \frac{1 + tan θ}{1 - tan θ} - \frac{\frac{1}{tan θ} - 1}{\frac{1}{tan θ} + 1}$

$= \frac{1 + tan θ}{1 - tan θ} - \frac{1 - tan θ}{1 + tan θ}$

$= \frac{{(1 + tan θ)}^{2} - (1 - {tan}^{2} θ)}{1 - {tan}^{2} θ}$

$= \frac{1 + {tan}^{2} θ + 2 tan θ - 1 - {tan}^{2} θ + 2 tan θ}{1 - {tan}^{2} θ}$

$= \frac{4 tan θ}{1 - {tan}^{2} θ}$

$= \frac{2 \times 2 tan θ}{1 - {tan}^{2} θ}$

$= 2 tan 2 θ$
$= R H S$

Hence, proved.

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