Question 388621


{{{3cos(x)+5sin(x)}}}
<pre>
Draw a right triangle with those coefficients as legs:

horizontal leg = 3 and vertical leg = 5

{{{drawing(2000/7,400,-1,4,-1,6,

triangle(0,0,3,0,3,5), locate(1.5,-.1,3), locate(3.1,2.5,5),
locate(.35,.4,theta)

 )}}} 

Calculate the hypotenuse using the Pythagorean theorem

{{{c^2=a^2+b^2}}}
{{{c^2=3^2+5^2}}}
{{{c^2=9+25}}}
{{{c^2=34}}}
{{{c=sqrt(34)}}}

{{{drawing(2000/7,400,-1,4,-1,6,

triangle(0,0,3,0,3,5), locate(1.5,-.1,3), locate(3.1,2.5,5),
locate(.35,.4,theta), locate(.7,2.5,sqrt(34))

 )}}} 

{{{sin(theta) = 5/sqrt(34)}}} so {{{5=sqrt(34)sin(theta)}}}

{{{cos(theta) = 3/sqrt(34)}}} so {{{3=sqrt(34)cos(theta)}}}

Substitute {{{sqrt(34)cos(theta)}}} for 3 and
substitute {{{sqrt(34)sin(theta)}}} for 5 in


{{{3cos(x)+5sin(x)}}} =

{{{sqrt(34)cos(theta)cos(x)+sqrt(34)sin(theta)sin(x)}}} =

factor out {{{sqrt(34)}}} on the left side:

{{{sqrt(34)(cos(theta)cos(x)+sin(theta)sin(x))}}} =

Now we remember the formula {{{cos(alpha-beta)=cos(alpha)cos(beta)+sin(alpha)sin(beta)}}}, and the above can be written:

{{{sqrt(34)(cos(theta)cos(x)+sin(theta)sin(x))}}} =

{{{sqrt(34)cos(theta-x)}}}

Since {{{cos(alpha) = cos(-alpha)}}}, the above becomes:

{{{sqrt(34)cos(-(theta-x))}}}

{{{sqrt(34)cos(-theta+x)}}}

{{{sqrt(34)cos(x-theta)}}}


This is in the form  {{{k*cos("x°"-"a°")}}}

where {{{k=sqrt(34)}}} and {{{theta=arctan(5/3) = "59.03624347°"}}}
 
So we have

{{{3cos(x)+5sin(x)}}} = {{{k*cos("x°"-"a°")}}} = {{{sqrt(34)*cos("x°"-"59.03624347°")}}}

Edwin</pre>