Question
Thu August 14, 2008 By:

trigo..plz..help...

Expert Reply
Mon December 15, 2008

sin4A / a +cos4A / b = 1 / a+b

Cubing both the sides we have:

sin12A/a3 + cos12A/b3 +3sin4Acos4A/ab(sin4A / a +cos4A / b)=1/(a+b)3

 Hence sin8A/a3 (sin4a + acos4A/b)+cos8A/b3 (cos4a + bsin4A/a)=1/(a+b)3

So the given condition is not true in general .It is true only when :sin4a + acos4A/b=cos4a + bsin4A/a=1

 

Sun March 26, 2017

solve

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