Using formula (1) and (2), show that sin(a-b) = sin(a)cos(b) - cos(a)-sin(b).
* cos(a-b) = cos(a)cos(b) + sin(a)sin(b)..........(1)
* cos(π/2 - b) = sin(b)..................................…
* cos(a-b) = cos(a)cos(b) + sin(a)sin(b)..........(1)
* cos(π/2 - b) = sin(b)..................................…
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The issue is that you the actual problem should read
sin(a-b) = sin(a)cos(b) - cos(a)sin(b)
If we let (a - b) = c then using (2)
cos(π/2 - c) = sin(c) = sin(a - b)
if we let π/2 + a = d then using (1)
cos(d - b) = cos(d)cos(b) +sin(d)sin(b) = cos(π/2+a)cos(b) + sin(π/2+a)sin(b) = sin(a)cos(b) - cos(a)sin(b)
and since d - b = π/2 - c we can say that cos(π/2 - c) = cos(d - b)
so sin(a - b) = sin(a)cos(b) - cos(a)sin(b)
sin(a-b) = sin(a)cos(b) - cos(a)sin(b)
If we let (a - b) = c then using (2)
cos(π/2 - c) = sin(c) = sin(a - b)
if we let π/2 + a = d then using (1)
cos(d - b) = cos(d)cos(b) +sin(d)sin(b) = cos(π/2+a)cos(b) + sin(π/2+a)sin(b) = sin(a)cos(b) - cos(a)sin(b)
and since d - b = π/2 - c we can say that cos(π/2 - c) = cos(d - b)
so sin(a - b) = sin(a)cos(b) - cos(a)sin(b)