Given that sin x + sin y = a and cos x + cos y =a, where a not equal to 0, express sin x + cos x in terms of a.

attemp:
sin x = a - sin y
cos x = a - cos y
sin x + cos x = 2A - (sin y + cos y)

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1. so sinx + siny = cosx + cosy
the trivial and obvious solution is x = y = 45° or π/4

another obvious one is 225° or 5π/4, because of the CAST rule

are there others , not multiples of 45°

recall:

sin a + sin b = 2 sin (1/2)(a+b) cos (1/2)(a-b)
cos a + cos b = 2 cos (1/2)(a+b) cos (1/2)(a-b)

so:
sinx + siny = 2 sin (1/2)(x+y) cos (1/2)(x-y)
cosx + cosy = 2 cos (1/2)(x+y) cos (1/2)(x-y)

2 sin (1/2)(x+y) cos (1/2)(x-y) = 2 cos (1/2)(x+y) cos (1/2)(x-y)
divide both sides by 2cos (1/2)(x-y)
sin (1/2)(x+y) = cos (1/2)(x+y)
divide by cos (1/2)(x+y)

tan (1/2)(x+y) = 1
we know tan 45° = 1
(1/2)(x+y) = 45
x+y = 90

so any pair of complimentary angles will work
e.g let x = 10°, y = 80°
LS = sin10° + sin80°
= cos80° + sin10
= RS

of course, I should have seen the property that
cosØ = sin(90°- Ø)

then sinx + siny = sin(90-x) + sin(90-y)
sinx + siny = sin90cosx - cos90sinx + sin90cosy - cos90siny
sinx + siny = (1)cosx - (0)sinx + (1)cosy - (0)siny
sinx + siny = cosx + cosx

back to the beginning

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