Question

sinx+siny=a;cosx+coy=b then tan(x+y÷2)=?;sin(x-y÷2)=?​

Answer 1

Answer:

i)

$\frac{sinx + siny}{cosx + cosy} = \frac{a}{b}$

$\frac{2 \sin( \frac{x + y}{2} ) \cos( \frac{x - y}{2} ) }{2 \cos( \frac{x + y}{2} ) \cos( \frac{x - y}{2} ) } = \frac{a}{b}$

$\tan( \frac{x + y}{2} ) = \frac{a}{b}$

ii)

$(sinx + siny) {}^{2} + (cosx + cosy) {}^{2} = {a}^{2} + {b}^{2}$

$2 + 2(cosx \: cosy + sinx \: siny) = {a}^{2} + {b}^{2}$

$2 \cos(x - y) = {a}^{2} + {b}^{2} - 2$

$\cos(x - y) = \frac{1}{2} ( {a}^{2} + {b}^{2} - 2)$

$1 - 2 {sin}^{2} ( \frac{x - y}{2} ) = \frac{ {a}^{2} + {b}^{2} - 2 }{2}$

$2 {sin}^{2} ( \frac{x - y}{2} ) = 1 - \frac{ {a}^{2} + {b}^{2} - 2 }{2} = \frac{4 - {a}^{2} - {b}^{2} }{2}$

${sin}^{2} ( \frac{x - y}{2} ) = \frac{4 - {a}^{2} - {b}^{2} }{4}$

$\sin( \frac{x - y}{2} ) = \frac{1}{2} \sqrt{4 - {a}^{2} - {b}^{2} }$