Question

sin57град + sin41град = 2sinxcos8град

Answer 1

Ответ:

$sin57^\circ +sin41^\circ =2sinx\cdot cos8^\circ \\\\2\, sin\dfrac{57^\circ +41^\circ }{2}\cdot cos\dfrac{57^\circ -41^\circ }{2}=2\, sinx\cdot cos8^\circ \\\\2\, sin49^\circ \cdot cos8^\circ =2\, sinx\cdot cos8^\circ \\\\2\, sin49^\circ \cdot cos8^\circ -2\, sinx\cdot cos8^\circ =0\\\\2\, cos8^\circ \cdot (sin49^\circ -sinx)=0\\\\cos8^\circ \ne 0\ \ \ \Rightarrow \ \ \ \ sin49^\circ -sinx=0\\\\2\. sin\dfrac{49^\circ -x}{2}\cdot cos\dfrac{49^\circ +x}{2}=0$

$a)\ \ sin\dfrac{49^\circ -x}{2}=0\ \ ,\ \ \dfrac{49^\circ -x}{2}=180^\circ n\ ,\ \ 49^\circ -x=360^\circ n\ \ ,\\\\x=49^\circ -360^\circ n\ ,\ n\in Z\ \ \ ,\ \ \ \boxed {x=49^\circ +360^\circ n\ ,\ n\in Z}\\\\\\b)\ \ cos\dfrac{49^\circ +x}{2}=0\ \ ,\ \ \dfrac{49^\circ +x}{2}=90^\circ +180^\circ k\ \ ,\ \ 49^\circ +x}=180^\circ +360^\circ k\ \ ,\\\\\boxed {x=131^\circ +360^\circ k\ ,\ k\in Z}\\\\\\Otvet:\ \ x=49^\circ +360^\circ n\ ,\ \ x=131^\circ +360^\circ k\ ,\ \ n,k\in Z$