sin a - sin b + sin y = 4*sin a/2 * cos b/2 * sin y/2

[Гост]

sin a - sin b + sin y = 4*sin a/2 * cos b/2 * sin y/2 където a,b,y са ъгли в триъгълник
y е гама,не знаех как да го напиша.задачата е за домашно,спешно за утре.Ако ми се обясни подробно като на идиот ще е най-добре :)

[Knowledge Greedy]

[tex]sin\alpha-sin\beta+sin\gamma=[/tex]

[tex]=(sin {\alpha}-sin\beta)+sin\gamma=[/tex]

[tex]=2sin{\frac{\alpha-\beta}{2}}cos{\frac{\alpha+\beta}{2}}+sin\gamma=[/tex]

[tex]=2sin{\frac{\alpha-\beta}{2}}cos{\frac{180^\circ-\gamma}{2}}+sin\gamma=[/tex]

[tex]=2sin{\frac{\alpha-\beta}{2}}cos\left ( 90^\circ-\frac{\gamma}{2}\right )+2sin{\frac{\gamma}{2}}cos{\frac{\gamma}{2}}=[/tex]

[tex]=2sin{\frac{\alpha-\beta}{2}}sin{\frac{\gamma}{2}}+2sin{\frac{\gamma}{2}}cos{\frac{\gamma}{2}}=[/tex]

[tex]=2sin{\frac{\gamma}{2}}\left ( sin{\frac{\alpha-\beta}{2}}+cos{\frac{\gamma}{2}}\right )=[/tex]

[tex]=2sin{\frac{\gamma}{2}}\left [ sin{\frac{\alpha-\beta}{2}}+sin \left ( 90^\circ -{\frac{\gamma}{2}}\right ) \right ]=[/tex]

[tex]=2sin{\frac{\gamma}{2}} 2 sin{\frac{\frac{\alpha-\beta}{2}+90^\circ-\frac{\gamma}{2}}{2}}.cos{\frac{\frac{\alpha-\beta}{2}-90^\circ+\frac{\gamma}{2}}{2}}=[/tex]

[tex]=4sin\frac{\gamma}{2}.sin{\frac{\alpha+180^\circ - \beta - \gamma}{4}}.cos{\frac{\alpha+\gamma+180^\circ - \beta }{4}}=[/tex]

[tex]=4sin\frac{\gamma}{2}.sin{\frac{\alpha}{2}}.cos { \frac {-\beta}{2}} =4sin{\frac{\alpha}{2}}.cos { \frac {\beta}{2}}sin\frac{\gamma}{2}.[/tex]

[Гост]

Благодаря изключително много.Помощта е чудесна :)