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[cos(a)cos(b)+sin(a)sin(b)][cos(a)cos(b)-sin(a)sin(b)]=?

 Apr 8, 2016

Best Answer 

 #1
avatar+26376 
+20

[cos(a)cos(b)+sin(a)sin(b)][cos(a)cos(b)-sin(a)sin(b)]=?

 

\(\begin{array}{rcll} && [~ \cos(a) \cos(b)+ \sin(a) \sin(b) ~] \cdot [~ \cos(a)\cos(b)- \sin(a) \sin(b) ~] \\ &=& \cos^2(a) \cos^2(b) - \cos(a) \cos(b)\sin(a) \sin(b)+ \sin(a) \sin(b)\cos(a)\cos(b) -\sin^2(a) \sin^2(b) \\ &=& \cos^2(a) \cos^2(b) + 0 -\sin^2(a) \sin^2(b) \\ &=& \cos^2(a) \cos^2(b) -\sin^2(a) \sin^2(b) \\ &=&[~ 1-\sin^2(a) ~] \cos^2(b) -\sin^2(a) [~ 1-\cos^2(b) ~] \\ &=& \cos^2(b) -\sin^2(a) \cos^2(b) -\sin^2(a) +\sin^2(a)\cos^2(b) \\ &=& \cos^2(b) -\sin^2(a) -\sin^2(a) \cos^2(b) +\sin^2(a)\cos^2(b) \\ &=& \cos^2(b) -\sin^2(a) +0 \\ &\mathbf{=}&\mathbf{ \cos^2(b) -\sin^2(a) }\\ \end{array}\)

 

laugh

 Apr 8, 2016
 #1
avatar+26376 
+20
Best Answer

[cos(a)cos(b)+sin(a)sin(b)][cos(a)cos(b)-sin(a)sin(b)]=?

 

\(\begin{array}{rcll} && [~ \cos(a) \cos(b)+ \sin(a) \sin(b) ~] \cdot [~ \cos(a)\cos(b)- \sin(a) \sin(b) ~] \\ &=& \cos^2(a) \cos^2(b) - \cos(a) \cos(b)\sin(a) \sin(b)+ \sin(a) \sin(b)\cos(a)\cos(b) -\sin^2(a) \sin^2(b) \\ &=& \cos^2(a) \cos^2(b) + 0 -\sin^2(a) \sin^2(b) \\ &=& \cos^2(a) \cos^2(b) -\sin^2(a) \sin^2(b) \\ &=&[~ 1-\sin^2(a) ~] \cos^2(b) -\sin^2(a) [~ 1-\cos^2(b) ~] \\ &=& \cos^2(b) -\sin^2(a) \cos^2(b) -\sin^2(a) +\sin^2(a)\cos^2(b) \\ &=& \cos^2(b) -\sin^2(a) -\sin^2(a) \cos^2(b) +\sin^2(a)\cos^2(b) \\ &=& \cos^2(b) -\sin^2(a) +0 \\ &\mathbf{=}&\mathbf{ \cos^2(b) -\sin^2(a) }\\ \end{array}\)

 

laugh

heureka Apr 8, 2016

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