cosxcosy = sin²y
sinxsiny = cos²y
cosx*cosy + sinx*siny = sin²y+ cos²y
cos(x-y)=1
вычитаем
cosx*cosy - sinx*siny = sin²y- cos²y
cos(x+y)=-cos2y
{cos(x-y)=1 ⇒ [b] x-y=2πk, k ∈ Z[/b]
{cos(x+y)=-cos2y ⇒ cos(x+y)+cos2y=0
cos(x+y)+cos2y=0
2cos[m]\frac{x+y+2y}{2}[/m]*cos[m]\frac{x+y-2y}{2}[/m]=0
2cos[m]\frac{x+3y}{2}[/m]*cos[m]\frac{x-y}{2}[/m]=0
cos[m]\frac{x+3y}{2}[/m]=0 или cos[m]\frac{x-y}{2}[/m]=0
[m]\frac{x+3y}{2}=\frac{\pi}{2}+\pi n, [/m]n ∈ Z или [m]\frac{x-y}{2}=\frac{\pi}{2}+\pi m, [/m]m∈ Z
Две системы:
{[m]\frac{x+3y}{2}=\frac{\pi}{2}+\pi n, [/m]n ∈ Z
{[b]x-y=2πk, k ∈ Z[/b]
или
{[m]\frac{x-y}{2}=\frac{\pi}{2}+\pi m, [/m]m∈ Z
{[b]x-y=2πk, k ∈ Z[/b]
их решаем и находим х и y
{[m]x+3y=\pi+ 2\pi n, [/m]n ∈ Z
{[b]x-y=2πk, k ∈ Z[/b]
Умножаем второе на 3
{[m]x+3y=\pi+ 2\pi n, [/m]n ∈ Z
{[b]3x-3y=6πk, k ∈ Z[/b]
и складываем
[m]4x=\pi+ 2\pi n+6\pi k[/m]n, k ∈ Z
[m]x=\frac{\pi}{4}+2\pi \cdot (3k+n)[/m]
[m]y=x-2\pi k = \frac{\pi}{4}+2\pi \cdot (3k+n)-2\pi k[/m]
[m]x=\frac{\pi}{4}+2\pi \cdot m[/m], m=3k+n
[m]y=x-2\pi k = \frac{\pi}{4}+2\pi \cdot (2k+n)[/m]
Вторая система:
{[m]\frac{x-y}{2}=\frac{\pi}{2}+\pi m, [/m]m∈ Z
{[b]x-y=2πk, k ∈ Z[/b]
{[m]x-y=\pi+2\pi m, [/m]m∈ Z
{[b]x-y=2πk, k ∈ Z[/b]