Question

Prove that cos alpha + cos theta ka whole square + sin alpha minus sin beta ka whole square is equal to 4 cos square alpha + beta by 2

Answer 1

actually my marker is not working but you just need to add the sign of+ at blank space of formula

Answer 2

Answer:

$L.H.S.=(cos\alpha+cos\beta )^2+(sin\alpha-sin\beta)^2$

$=[2 cos((\alpha+\beta)/2).cos((\alpha-\beta)/2) ]^2+[2 cos((\alpha+\beta)/2).sin((\alpha-\beta)/2)]^2$

$[Since\quad cosC+cosD=2 cos((C+D)/2).cos(((C-D)/2)) and\quad sinC-sinD =2 cos((C+D)/2).sin((C-D)/2)]$

$=4 cos^2 ((\alpha+\beta)/2).cos^2((\alpha-\beta)/2) +4cos^2((\alpha+\beta)/2).sin^2((\alpha-\beta)/2)$

$=4 cos^2 ((\alpha+\beta)/2){cos^2((\alpha-\beta)/2) +sin^2((\alpha-\beta)/2)}$

$=4 cos^2 ((\alpha+\beta)/2).1 [∵cos^2\theta+sin^2\theta=1]$

$=4 cos^2 ((\alpha+\beta)/2)=L.H.S.$