Question 1139613
<pre>Let this be any triangle:

{{{drawing(250,250,-1,5,-1,5, 
triangle(0,0,3,4,4,0),
locate(.5,.5,x), locate(3.5,.5,y),locate(2.76,3.5,z),
locate(3.7,2,a), locate(1,2,b),locate(2,0,c) 



)}}}

By the law of cosines:

(1)     {{{c^2=a^2+b^2-2*a*b*cos(z)}}}

By the law of sines:

        {{{a/sin(x)=b/sin(y)=c/sin(z)}}}

Let those equal to k

        {{{a/sin(x)=b/sin(y)=c/sin(z)=k}}}

Then we have 

        {{{matrix(1,7,a=k*sin(x),",","",b=k*sin(y),",","",c=k*sin(z))}}} 

Substitute in (1)

        {{{(k*sin(z))^2=(k*sin(x))^2+(k*sin(y))^2-2*(k*sin(x))*(k*sin(y))*cos(z)}}}

        {{{k^2*sin^2(z)=k^2*sin^2(x)+k^2*sin(y))^2-2*k*sin^2(x)*k*sin(y)*cos(z)}}}

        {{{k^2*sin^2(z)=k^2*sin^2(x)+k^2*sin^2(y))^2-2*k^2*sin(x)*sin(y)*cos(z)}}}

Divide through by k<sup>2</sup>:

        {{{sin^2(z)=sin^2(x)+sin^2(y))-2*sin(x)*sin(y)*cos(z)}}}

Since x + y + z = 180°, z = 180°-x-y = 180°-(x+y)

Use the identities {{{sin(180^"o"-theta)=sin(theta)}}}, and {{{cos(180^"o"-theta)=-cos(theta)}}}:

        {{{sin^2(x+y)=sin^2(x)+sin^2(y)-2*sin(x)*sin(y)*(-cos(x+y)^"")}}}  

        {{{sin^2(x+y)=sin^2(x)+sin^2(y)+2*sin(x)*sin(y)*cos(x+y)}}}

        {{{sin^2(x)+sin^2(y)+2*sin(x)*sin(y)*cos(x+y)=sin^2(x+y)}}}

Edwin</pre>