Question

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

Answer 1

Answer:

i) sinx+sing=a. & cosx+cosy=b then

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

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

ii)

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

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

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

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

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

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

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

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

Ans