Question 1060384
Write 5sin(t)-12cos(t) in the form Asin(Bt+ϕ) using sum or difference formulas. 
<pre><b>
We can just take B as 1.

{{{5sin(t)-12cos(t)}}}{{{""=""}}}{{{A*sin(t+phi)}}}{{{""=""}}}{{{A*(sin(t)cos(phi)^""+cos(t)sin(phi)^"")}}} 

{{{5sin(t)-12cos(t)}}}{{{""=""}}}{{{A*(sin(t)cos(phi)^""+cos(t)sin(phi)^"")}}}

{{{5sin(t)-12cos(t)}}}{{{""=""}}}{{{A*sin(t)cos(phi)^""+A*cos(t)sin(phi)^"")}}}

So to make those identitically equal, we must have

(1)   {{{A*cos(phi)}}}{{{""=""}}}{{{5}}} and
(2)   {{{A*sin(phi)}}}{{{""=""}}}{{{-12}}}

Dividing equals by equals, using (2) and (1)

   {{{(A*sin(phi))/(A*cos(phi))}}}{{{""=""}}}{{{-12/5}}}
   {{{(sin(phi))/(cos(phi))}}}{{{""=""}}}{{{-12/5}}}
   {{{tan(phi)}}}{{{""=""}}}{{{-12/5}}}
   {{{phi}}}{{{""=""}}}{{{"-67.38013505°"}}}

To get a positive coterminal angle we add 360°

   {{{phi}}}{{{""=""}}}{{{"292.6198649°"}}}

   Squaring both sides of (1) and (2) and adding them:

{{{A^2*cos^2(phi)+A^2*sin^2(phi)}}}{{{""=""}}}{{{5^2+(-12)^2}}}

{{{A^2*(cos^2(phi)+sin^2(phi))}}}{{{""=""}}}{{{25+144}}}

{{{A^2*(1)}}}{{{""=""}}}{{{169}}}

{{{A^2}}}{{{""=""}}}{{{169}}}

{{{A}}}{{{""=""}}}{{{13}}}

Substituting:

{{{5sin(t)-12cos(t)}}}{{{""=""}}}{{{A*sin(t+phi)}}}{{{""=""}}}{{{13sin(t+"292.6198649°")}}}

Edwin</pre></b>