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Trigonometric Ratios for Sum of Two Angles

If sinα-sinβ=...

Question

If $s i n α - s i n β = a$ and $c o s α + c o s β = b,$ then write the value of $c o s (α + β)$ .

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Solution

We have,
$s i n α - s i n β = a$ and $c o s α + c o s β = b$
Now, $a^{2} + b^{2} = (s i n α - s i n β)^{2} + (c o s α + c o s β)^{2}$
$= s i n^{2} α + s i n^{2} β - 2 s i n α s i n β + c o s^{2} α c o s^{2} β + 2 c o s α c o s β$
$= (s i n^{2} α + c o s^{2} α) + (s i n^{2} β + c o s^{2} β) + 2 [c o s α c o s β - s i n α s i n β]$
$= 1 + 1 + 2 c o s (α + β)$
$= 2 + 2 c o s (α + β)$
$\Rightarrow a^{2} + b^{2} = 2 + 2 c o s (α + β)$
$\Rightarrow a^{2} + b^{2} - 2 = 2 c o s (α + β)$
$\Rightarrow 2 c o s (α + β) = a^{2} + b^{2} - 2$
$\Rightarrow c o s (α + β) = \frac{a^{2} + b^{2} - 2}{2}$

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Trigonometric Ratios for Sum of Two Angles

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