Given that sinx + siny = a, and cosx + cosy = a, where a is NOT equivalent to zero. Find sin x + cosx in terms of a
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1) sin(x) + sin(y) = a
==> 2sin{(x+y)/2}*cos{(x-y)/2} = a [By sum/product relation]
2) Similarly, cos(x) + cos(y) = a, ==> 2cos{(x+y)/2}*cos{(x-y)/2} = a
3) Dividing (1) by (2),
tan{(x + y)/2} = 1
==> Either (x+y)/2 = 45° or 225°
So, x + y = either 90° or 450° [Since trigonometric functions are periodic functions, 450° is same as 90°]; thus x + y = 90°
4) We have sin(x) + sin(y) = a
Substituting for y = 90° - x from the above,
sin(x) + sin(90° - x) = a
==> sin(x) + cos(x) = a
==> 2sin{(x+y)/2}*cos{(x-y)/2} = a [By sum/product relation]
2) Similarly, cos(x) + cos(y) = a, ==> 2cos{(x+y)/2}*cos{(x-y)/2} = a
3) Dividing (1) by (2),
tan{(x + y)/2} = 1
==> Either (x+y)/2 = 45° or 225°
So, x + y = either 90° or 450° [Since trigonometric functions are periodic functions, 450° is same as 90°]; thus x + y = 90°
4) We have sin(x) + sin(y) = a
Substituting for y = 90° - x from the above,
sin(x) + sin(90° - x) = a
==> sin(x) + cos(x) = a