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Solving Simultaneous Trigonometric Equations

If sinα+ si...

Question

If $sin α + sin β = a$ and $cos α + cos β = b$ , then the value of $sin α + β = \frac{m a b}{a^{2} + b^{2}}$ .Find $m$

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Solution

$sin α + sin β = a, cos α + cos β = b$
$a^{2} + b^{2} = {sin}^{2} α + {sin}^{2} β + {cos}^{2} α + {cos}^{2} β + 2 (sin α sin β + cos α cos β)$
$a^{2} + b^{2} = 2 + 2 cos (α - β)$
$∴ 1 + cos (α - β) = \frac{a^{2} + b^{2}}{2}$
$a b = sin α cos α + sin α cos β + sin β cos α + sin β cos β$
$= sin (α + β) + \frac{sin 2 α + sin 2 β}{2}$
$= sin (α + β) + sin (α + β) cos (α - β)$
$= sin (α + β) [1 + cos (α - β)]$
$a b = sin (α + β) (\frac{a^{2} + b^{2}}{2})$
$∴ sin (α + β) = \frac{2 a b}{a^{2} + b^{2}}$
Given,
$sin (α + β) = \frac{m a b}{a^{2} + b^{2}}$
$∴ m = 2$

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Similar questions

s i n α + s i n β = a

c o s α + c o s β = b,

s i n (α + β) = \frac{2 a b}{a^{2} + b^{2}}

c o s (α + β) = \frac{b^{2} - a^{2}}{b^{2} + a^{2}}

\sin α + \sin β = a and \cos α + \cos β = b

, prove that

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\sin (α + β) = \frac{2 a b}{a^{2} + b^{2}}

\cos (α - β) = \frac{a^{2} + b^{2} - 2}{2}

sin α + sin β = a

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