Question 936612
{{{3sin(theta)-4cos(theta)=5}}} find the value of {{{3sin(theta)+4cos(theta)}}}
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<pre>
{{{3sin(theta)-4cos(theta)=5}}}

Divide through by 5

 {{{expr(3/5)sin(theta)-expr(4/5)cos(theta)=1}}}

Let {{{sin(alpha)=3/5}}} and {{{cos(alpha)=4/5}}} using this right triangle:

{{{drawing(100,2000/27,-.5,4.4,-.5,3.5,triangle(0,0,4,0,4,3),
locate(2,0,4), locate(4.2,1.7,3),locate(.6,.9,alpha), locate(1.3,2.1,5)

 )}}}

 {{{sin(alpha)sin(theta)-cos(alpha)cos(theta)=1}}}

Multiply through by -1 and reverse terms on the left:

 {{{cos(alpha)cos(theta)-sin(alpha)sin(theta)=-1}}}

Now we recognize the left side as the identity for {{{cos(alpha+theta)}}}.

 {{{cos(alpha+theta)=-1}}}

So {{{alpha+theta}}}{{{""=""}}}{{{matrix(1,5,any,odd,multiple,of,pi) }}}

   {{{alpha+theta}}}{{{""=""}}}{{{k*pi}}}, k an odd number

   {{{theta}}}{{{""=""}}}{{{k*pi-alpha}}}

We want to find the value of  {{{3sin(theta)+4cos(theta)}}}, so

 {{{3sin(theta)+4cos(theta)}}}{{{""=""}}}{{{3sin(k*pi-alpha)+4cos(k*pi-alpha)}}}{{{""=""}}}

{{{3(sin(k*pi)cos(alpha)^""-cos(k*pi)sin(alpha))+4(cos(k*pi)cos(alpha)^""+sin(k*pi)sin(alpha))}}}

           {{{sin(k*pi)=0}}} and {{{cos(k*pi)=-1}}} since k is odd.  So the above becomes:

{{{3((0)*cos(alpha)^""-(-1)sin(alpha))+4((-1)cos(alpha)^""+(0)sin(alpha))}}}{{{""=""}}}

{{{3sin(alpha)-4cos(alpha)}}}

And by the right triangle above that equals

{{{3(3/5)-4(4/5)=9/5-16/5=-7/5}}}

Edwin</pre>