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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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2 answers

  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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  2. not answered.

    Question asked: express sinx + cosx in term of a.

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