Question

cos alpha + cos beta whole square + sin alpha + sin beta
whole square is equal to 4 cos square alpha minus Beta upon 2

Answer 1

here is your answer..........

Answer 2

Answer:

We have to show that:

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

Consider,

$LHS=(cos\,\alpha+cos\,\beta)^2+(sin\,\alpha+sin\,\beta)^2$

$=(2cos\,(\frac{\alpha+\beta}{2})\,cos\,(\frac{\alpha-\beta}{2}))^2+(2cos\,(\frac{\alpha+\beta}{2})\,sin\,(\frac{\alpha-\beta}{2}))^2$

$=2^2\times cos^2\,(\frac{\alpha+\beta}{2})(cos^2\,(\frac{\alpha-\beta}{2})+sin^2\,(\frac{\alpha-\beta}{2}))$

$=2^2\times cos^2\,(\frac{\alpha+\beta}{2})\times1$

$=4cos^2\,(\frac{\alpha+\beta}{2})$

$=RHS$

Hence Proved.