Matematyka.pl

Należy udowodnić, że
\(\displaystyle{ \sin x \cos^3 y + \cos x \sin^3 y = \sin x \cos x = \frac{1}{2} \sin 2x}\)

ale
\(\displaystyle{ \sin x \cos^3 y + \cos x \sin^3 y = (\cos y \sin x + \cos x \sin y)(\cos^2 y + \sin^2 y) - \sin y \cos y (\cos y \sin x + \cos x \sin y) = \sin(x+y) (1+ \cos(x-y))}\)

a z założenia \(\displaystyle{ \sin(x+y)= -\frac{1}{2}\sin(2y)}\)

czy \(\displaystyle{ 1+ \cos(x-y)= - \frac{\sin 2x}{\sin 2y}}\)
?

Ta równość jest gdyż:
\(\displaystyle{ - \cos(x-y) = \frac{2\sin(x+y) \cos(x-y)}{\sin 2y}= \frac{\sin 2x + \sin 2y}{\sin 2y}}\)
z założenia

cbdo.

istotnie \(\displaystyle{ \frac{1}{2}sin 2y}\) Wyjątkowo pomysłowe zadanie...

\(\displaystyle{ \dfrac{\cos x}{\cos y}+\dfrac{\sin x}{\sin y}=-1---(1),}\)
\(\displaystyle{ \dfrac{\cos^3 y}{\cos x}+\dfrac{\sin^3 y}{\sin x}=1---(2)}\)
\(\displaystyle{ a=\dfrac{cos\, y}{cos\, x}}\)
\(\displaystyle{ b=\dfrac{sin\, y}{sin\, x}}\)
jesli (2) to:\(\displaystyle{ \cos^2\,y=\dfrac {1-b}{a-b}}\)
oraz (1)+(2)\(\displaystyle{ =a\,cos^2\,y+b\,sin^2\,y+\dfrac{1}{a}+\dfrac {1}{b}=0}\)
\(\displaystyle{ =cos^2\,y(a-b)+b+\dfrac{1}{a}+\dfrac{1}{b}=0}\)
\(\displaystyle{ 1-b+b+\dfrac{1}{a}+\dfrac{1}{b}=0}\)
\(\displaystyle{ \dfrac{1}{a}+\dfrac{1}{b}=-1}\)