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Definition of Functions

If α and ...

Question

If $α$ and $β$ are the $s o l^{n}$ of $a cos θ + b sin θ = c,$ Then show that $c o s (α - β) = \frac{2 c^{2} - (a^{2} + b^{2})}{a^{2} + b^{2}}$

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Solution

$a cos θ + b sin θ = c ⟹ b sin θ = c - a cos θ$
Squaring on both sides
$b^{2} {sin}^{2} θ = c^{2} - 2 a c cos θ + a^{2} {cos}^{2} θ$
$(a^{2} + b^{2}) {cos}^{2} θ - 2 a c cos θ + (c^{2} - b^{2}) = 0$
$cos α cos β = \frac{c^{2} - b^{2}}{a^{2} + b^{2}}$
$a cos θ + b sin θ = c ⟹ a cos θ = c - b sin θ$
Squaring on both sides
$a^{2} {cos}^{2} θ = c^{2} - 2 b c sin θ + b^{2} {sin}^{2} θ$
$(a^{2} + b^{2}) {sin}^{2} θ - 2 b c sin θ + (c^{2} - a^{2}) = 0$
$sin α sin β = \frac{c^{2} - a^{2}}{a^{2} + b^{2}}$
$cos (α - β) = cos α cos β + sin α sin β = \frac{2 c^{2} - (a^{2} + b^{2})}{a^{2} + b^{2}}$

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