Express asin(x) + bcos(x) in the form Rsin(x+c), where c is a non-zero constant.
The trick to solving this is to use the trig identity sin(a+b) = sin(a)cos(b) + sin(b)cos(a) From the identity above, we can write rewrite Rsin(x+c) as follows: Rsin(x+c) = R[sin(x)cos(c) + sin(c)cos(x)] Expanding out the bracket gives: Rsin(x+c) = Rsin(x)cos(c) + Rsin(c)cos(x) Comparing the right-hand side of the above equation to the form given in the question: Rcos(c)sin(x) + Rsin(c)cos(x) = asin(x) + bcos(x) Then we can equate the coefficients of sin(x) and cos(x), giving 2 equations: 1: Rsin(c) = b 2: Rcos(c) = a Divinding equation 1 by equation 2, and using the identity sin(x) / cos(x) = tan(x): Rsin(c)/Rcos(c) = b/a (Note the Rs will cancel) tan(c) = b/a c = tan-1(b/a) The value of c can then be subsituted into either of equations 1 or 2 to give the value of R.
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