If sin-sin-a and cos-cos-b- show that- i sin -2 a b-a2 b2ii cos -b2-a2-b2-a2

(i) sin(α+ β)= √(1 cos^2(α+β)) b^2+a^2=(cos α+ cos β)^2+ (sinα+ sin β)^2 ⇒ b^2+a^2=(cos^2α+sin^2α)+(cos^2β+sin^2β)+2(cos α cos β+sin α sin β) ⇒ b^2+a^2=1+1+2cos ...

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Byju's Answer

Standard XII

Mathematics

Solving Simultaneous Trigonometric Equations

If sinα+sinβ=...

Question

If $s i n α + s i n β = a$ and $c o s α + c o s β = b,$ show that
(i) $s i n (α + β) = \frac{2 a b}{a^{2} + b^{2}}$
(ii) $c o s (α + β) = \frac{b^{2} - a^{2}}{b^{2} + a^{2}}$

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Solution

(i) $s i n (α + β) = \sqrt{1 - c o s^{2} (α + β)}$
$b^{2} + a^{2} = (c o s α + c o s β)^{2} + (s i n α + s i n β)^{2}$
$\Rightarrow b^{2} + a^{2} = (c o s^{2} α + s i n^{2} α) + (c o s^{2} β + s i n^{2} β) + 2 (c o s α c o s β + s i n α s i n β)$
$\Rightarrow b^{2} + a^{2} = 1 + 1 + 2 c o s (α - β) = 2 + 2 c o s (α - β)$
and, $b^{2} - a^{2} = (c o s α + c o s β)^{2} - (s i n α + s i n β)^{2}$
$b^{2} - a^{2} = c o s^{2} α + c o s^{2} β - s i n^{2} α - s i n^{2} β + 2 (c o s α c o s β - s i n α s i n β)$
$\Rightarrow b^{2} - a^{2} = (c o s^{2} α - s i n^{2} β) + (c o s^{2} β - s i n α) + 2 c o s (α + b e t a)$
$\Rightarrow b^{2} - a^{2} = c o s (α + β) c o s (α - β) + c o s (β + α) c o s (β - α) + 2 c o s (α + β)$
$\Rightarrow b^{2} - a^{2} = c o s (α + β) 2 c o s (α - β) = 2$
$\Rightarrow b^{2} - a^{2} = c o s (α + β) (b^{2} + a^{2})$
Thus, $b^{2} - a^{2} = (b^{2} + a^{2}) c o s (α + β)$
$\Rightarrow c o s (α + β) = \frac{b^{2} - a^{2}}{b^{2} + a^{2}}$
$\Rightarrow s i n (α + β) = \sqrt{1 - {(\frac{b^{2} - a^{2}}{b^{2} - a^{2}})}^{2}} = \sqrt{\frac{4 a^{2} b^{2}}{(a^{2} + b^{2})}} = \frac{2 a b}{a^{2} + b^{2}}$
(ii) $c o s (α + β) = \frac{b^{2} - a^{2}}{b^{2} + a^{2}}$
$b^{2} + a^{2} = (c o s α + c o s β)^{2} + (s i n α + s i n β)^{2}$
$\Rightarrow b^{2} + a^{2} = (c o s^{2} α + s i n^{2} α) + (c o s^{2} β + s i n^{2} β) + 2 (c o s α c o s β + s i n α s i n β)$
$\Rightarrow b^{2} + a^{2} = 1 + 1 + 2 c o s (α - β) = 2 + 2 c o s (α - β)$
and, $b^{2} - a^{2} = (c o s α + c o s β)^{2} - (s i n α + s i n β)^{2}$
$b^{2} - a^{2} = (c o s^{2} α + c o s^{2} β - s i n^{2} α - s i n^{2} β + 2 (c o s α c o s β - s i n α s i n β))$
$\Rightarrow b^{2} - a^{2} = (c o s^{2} α - s i n^{2} β) + (c o s^{2} β - s i n^{2} α) + 2 c o s (α + β)$
$\Rightarrow b^{2} - a^{2} = c o s (α + β) c o s (α - β) + c o s (β + α) c o s (β - α) + 2 c o s (α + β)$
$\Rightarrow b^{2} - a^{2} = 2 c o s (α + β) c o s (α - β) + 2 c o s (α + β)$
$\Rightarrow b^{2} - a^{2} = c o s (α + β) 2 c o s (α - β) = 2$
$\Rightarrow b^{2} - a^{2} = c o s (α + β) (b^{2} + a^{2})$
Thus, $b^{2} - a^{2} = (b^{2} + a^{2}) c o s (α + β)$
$\Rightarrow c o s (α + β) = \frac{b^{2} - a^{2}}{b^{2} + a^{2}}$

Suggest Corrections

If sinα+sinβ=a and cosα+cosβ=b, show that (i)sin(α+β)=2aba2+b2 (ii)cos(α+β)=b2−a2b2+a2

If $s i n α + s i n β = a$ and $c o s α + c o s β = b,$ show that
(i) $s i n (α + β) = \frac{2 a b}{a^{2} + b^{2}}$
(ii) $c o s (α + β) = \frac{b^{2} - a^{2}}{b^{2} + a^{2}}$