Question

if cos x + cosy = 1/3, sin x + siny = 1/4 cos(x +y) =?​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{cos(x) + cos(y) = \dfrac{1}{3}\,\,\,\,\,and\,\,\,\,\sin(x) + sin(y) = \dfrac{1}{4}}$

$\sf{\implies\,2cos\bigg(\dfrac{x+ y}{2}\bigg) cos\bigg(\dfrac{x- y}{2}\bigg)= \dfrac{1}{3}\,\,\,\,\,and\,\,\,\,\,2sin\bigg(\dfrac{x+ y}{2}\bigg)cos\bigg(\dfrac{x-y}{2}\bigg) = \dfrac{1}{4}}$

Divide the above equations,

$\sf{\dfrac{2cos\bigg(\dfrac{x+ y}{2}\bigg) cos\bigg(\dfrac{x- y}{2}\bigg)}{2sin\bigg(\dfrac{x+ y}{2}\bigg) cos\bigg(\dfrac{x- y}{2}\bigg)}=\dfrac{\dfrac{1}{3}}{\dfrac{1}{4}}}$

$\sf{\implies\dfrac{cos\bigg(\dfrac{x+ y}{2}\bigg) }{sin\bigg(\dfrac{x+ y}{2}\bigg) }=\dfrac{4}{3}}$

$\sf{\implies\,cot\bigg(\dfrac{x+ y}{2}\bigg) =\dfrac{4}{3}}$

$\sf{\implies\,tan\bigg(\dfrac{x+ y}{2}\bigg) =\dfrac{3}{4}}$

Now,

$\sf{cos(x+y)}$

$\sf{=\dfrac{1-tan^{2}\bigg(\dfrac{x+y}{2}\bigg)}{1+tan^{2}\bigg(\dfrac{x+y}{2}\bigg)}}$

$\sf{=\dfrac{1-\bigg(\dfrac{3}{4}\bigg)^2}{1+\bigg(\dfrac{3}{4}\bigg)^2}}$

$\sf{=\dfrac{1-\dfrac{9}{16}}{1+\dfrac{9}{16}}}$

$\sf{=\dfrac{\dfrac{16-9}{16}}{\dfrac{16+9}{16}}}$

$\sf{=\dfrac{16-9}{16+9}}$

$\sf{=\dfrac{7}{25}}$

$\boxed{\sf{\blue{cos(x+y)=\dfrac{7}{25}}}}$