Question 1043975
<pre>
(1)     -5.9sin(x)+6.6cos(x) in the form ksin(x+&#981;)

That makes us think of the right side of the identity:

cos(x+&#981;) = cos(x)cos(&#981;) - sin(x)sin(&#981;)

multiplied by some constant k:

(2)     k*cos(x+&#981;) = k[cos(x)cos(&#981;) - sin(x)sin(&#981;)]

Let's get 

(1)     -5.9sin(x)+6.6cos(x)

in the form of the right side of (2)

        k[cos(x)cos(&#981;) - sin(x)sin(&#981;)]  

(1)     -5.9sin(x)+6.6cos(x) =
        
        cos(x)(6.6) - sin(x)(5.9) 

Let A be some positive non-zero constant. Then by multiplying
then dividing by A

       {{{A(cos(x)(6.6/A)^"" - sin(x)(5.9/A))}}} 

We find A and &#981; such that

{{{6.6/A=cos(phi)}}} and {{{5.9/A=sin(phi)}}}

We see that &#981; must be in QIV because that's where the
cosine is positive and the sine is positive.  We draw
an angle in QI to represent &#981;.  

{{{drawing(200,1000/7,-3,10,-3,7,red(locate(1.3,1,phi)),
line(-8,0,8,0),line(0,-8,0,8), line(0,0,6.6,5.9),locate(6.9,2.8,5.9),
locate(3,0,6.6),locate(2.8,4,A),
line(6.9,0,6.9,5.9), red(arc(0,0,2.5,-2.5,0,42)) )}}} 

We can find A by the Pythagorean theorem:

{{{A^2=(6.6)^2+(-5.9)^2)}}}
{{{A^2=43.56+34.81}}}
{{{A^2=78.37}}}
{{{A=8.852683209}}}

We can find &#981; from 

{{{tan(phi)=5.9/6.6}}}

using the inverse tangent feature on a calculator
in radian mode:

&#981; = 0.7294565922

So 
       {{{A(cos(x)(6.6/A)^"" - sin(x)(5.9/A))}}} 

now becomes

       {{{A(cos(x)cos(phi) - sin(x)sin(phi))}}}

which is equivalent to

      {{{A*cos(x+phi)}}}

or

      {{{8.852683209*cos(x+0.7294565922)}}}

Edwin</pre>