Question 869375
<pre>
cos&#952; + 2sin&#952;

We will strive to cause this to become a constant times
the right side of the identity 

sin(&#945;+&#946;) = sin&#945;·cos&#946; + cos&#945;·sin&#946;

Where &#946; = &#952;  and 

sin(&#945;+&#952;) = sin&#945;·cos&#952; + cos&#945;·sin&#952;

We wish to find k and &#945; so that 

cos&#952; + 2sin&#952; = k·sin&#945;·cos&#952; + k·cos&#945;·sin&#952; = k·sin(&#945;+&#952;)

So we must have 1 = k·sin&#945;  and 2 = k·cos&#945;

or              sin&#945; = {{{1/k}}}  and cos&#945; = {{{2/k}}}

Since {{{sine=opposite/hypotenuse}}} and {{{cosine=adjactent/hypotenuse}}}

we construct angle &#945; in a right triangle by choosing &#945;'s
opposite side to be 1 and its adjacent side to be 2,
and its hypotenuse to be k.

{{{drawing(200,400/3,-.5,2.5,-.5,1.5, triangle(0,0,2,0,2,1),
locate(2.05,.6,1),locate(1,0,2), locate(.4,.25,alpha),
locate(.9,.73,k) )}}}

By the Pythagorean theorem, k² = 2²+1² = 4+1 = 5, so k = {{{sqrt(5)}}}

So

cos&#952; + 2sin&#952; = k·sin&#945;·cos&#952; + k·cos&#945;·sin&#952; = k·sin(&#945;+&#952;) 
 
becomes

cos&#952; + 2sin&#952; = {{{sqrt(5)*expr(1/sqrt(5))}}}·cos&#952; + {{{sqrt(5)*expr(2/sqrt(5))}}}·cos&#945;·sin&#952; = {{{sqrt(5)}}}·sin(&#945;+&#952;) 

Since the amplitude of y = {{{sqrt(5)}}}·sin(&#945;+&#952;) is {{{sqrt(5)}}}, its

minimum value is {{{-sqrt(5)}}}

Edwin</pre>