Sinx+siny+sin(x+y)=4sinx/2cosx/2cosx-y/2

Предыдущий
вопрос Следующий
вопрос

To prove the given identity:

Start with the left-hand side of the equation, and apply the sum-to-product formula for sine to simplify the expression:

sinx + siny + sin(x+y)

= sinx + siny + sinxcosy + cosxsiny (using the sum-to-product formula)

= sinx + sinxcosy + siny + cosxsiny

= sinx(1+cosy) + siny(1+cosx)

= 2sin(x/2)cos(y/2) + 2sin(y/2)cos(x/2) (using the double-angle formula)

= 4(sin(x/2)cos(y/2)sin(y/2)*cos(x/2)) / 2

= 4sin(x/2)cos(x/2)cos(y/2)*sin(y/2) / 2

= 4sin(x/2)cos(x/2) / 2cos(y/2)*sin(y/2)

= 4sin(x/2)cos(x/2) / 2cos(x/2-y/2)

= 4sin(x/2)*cos(x/2) / 2cos(x/2-y/2)

Which is equal to the right-hand side of the equation. Therefore, we have shown that sinx+siny+sin(x+y)=4sinx/2cosx/2cosx-y/2.