הוכח: s i n ( α + β ) ∗ s i n ( α − β ) = s i n 2 α − s i n 2 β {\displaystyle sin(\alpha +\beta )*sin(\alpha -\beta )=sin^{2}\alpha -sin^{2}\beta }
s i n ( a + b ) s i n ( a − b ) = [ s i n a c o s b + s i n b c o s a ] [ s i n a c o s b − s i n b c o s a ] = ( a − b ) ( a + b ) = a 2 − b 2 ( s i n a c o s b ) 2 − ( s i n b c o s a ) 2 = s i n 2 a c o s 2 b − s i n 2 b c o s 2 a = s i n 2 a ( 1 − s i n 2 b ) − s i n 2 b ( 1 − s i n 2 a ) = s i n 2 a − s i n 2 a ∗ s i n 2 b − s i n 2 b + s i n 2 b ∗ s i n 2 a = s i n 2 a − s i n 2 b {\displaystyle {\begin{aligned}sin(a+b)sin(a-b)=[sinacosb+sinbcosa][sinacosb-sinbcosa]=\\(a-b)(a+b)=a^{2}-b^{2}\\(sinacosb)^{2}-(sinbcosa)^{2}=sin^{2}acos^{2}b-sin^{2}bcos^{2}a=\\sin^{2}a(1-sin^{2}b)-sin^{2}b(1-sin^{2}a)=sin^{2}a-sin^{2}a*sin^{2}b-sin^{2}b+sin^{2}b*sin^{2}a=sin^{2}a-sin^{2}b\end{aligned}}}
![{\displaystyle {\begin{aligned}sin(a+b)sin(a-b)=[sinacosb+sinbcosa][sinacosb-sinbcosa]=\\(a-b)(a+b)=a^{2}-b^{2}\\(sinacosb)^{2}-(sinbcosa)^{2}=sin^{2}acos^{2}b-sin^{2}bcos^{2}a=\\sin^{2}a(1-sin^{2}b)-sin^{2}b(1-sin^{2}a)=sin^{2}a-sin^{2}a*sin^{2}b-sin^{2}b+sin^{2}b*sin^{2}a=sin^{2}a-sin^{2}b\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/631dfd7b4f57018b114bc94987066abc395f73b6)
