Question 1090465
<pre><font size=4><b>
{{{20sin(x)+21cos(x)}}}

We draw a right triangle with legs having the lengths of those
two coefficients.  We put the 20 on the adjacent side to y
and 21 on the opposite side, so that we can use the given 
sin(x+y) formula.  [If we did it the other way we'd have to 
use the cos(x-y) formula.]

{{{drawing(200,200,-4,24,-4,24,

triangle(0,0,20,0,20,21),locate(10,0,20),locate(20.5,11,21),
red(arc(0,0,15,-15,0,47),locate(4,3,y)) )}}} 

Note here that y is the angle whose tangent is 21/20, and we
state that mathematically by the equation y = tan<sup>-1</sup>(21/20).

We calculate the hypotenuse by the Pythagorean theorem:

{{{c^2=a^2+b^2}}}
{{{c^2+20^2+21^2}}}
{{{c^2=400+441}}}
{{{c^2=841}}}
{{{c=sqrt(841)}}}
{{{c=29}}}

{{{drawing(200,200,-4,24,-4,24,
locate(7.7,12,29),
triangle(0,0,20,0,20,21),locate(10,0,20),locate(20.5,11,21),
red(arc(0,0,15,-15,0,47),locate(4,3,y)) )}}} 

We multiply the given expression, which is 

{{{20sin(x)+21cos(x)}}}

by {{{29*(1/29)}}} which is the same as multiplying
by 1, and does not change the value:

{{{29(1/29)(20sin(x)^""+21cos(x))}}}

We distribute the {{{(1/29)}}}

{{{29(expr(20/29)sin(x)+expr(21/29)cos(x))}}}

We use the right triangle above to substitute trig
ratios for the fractions:

{{{29(cos(y)^""sin(x)+sin(y)cos(x))}}}

We rearrange the factors in the parentheses so that 
what's inside the parentheses will look like the right 
side of the given formula sin(x+y)=sin(x)cos(y)+sin(y)cos(x)

{{{29(sin(x)^""cos(y)+sin(y)cos(x))}}}

We replace the parentheses using the given formula:

{{{29sin(x+y)}}}, where y = tan<sup>-1</sup>(21/20).

Edwin</pre></font></b>