CBSE Class 11-science Answered
periodicity of trigonometric functions
Asked by | 15 Jul, 2008, 08:12: PM
Expert Answer
we know that a3-b3 = (a-b) (a2+b2+ab)
put cos2A = a and sin2A =b
= (cos2A-sin2A) (cos4A+sin4A+cos2Asin2A)
now we know that (cos2A-sin2A) = cos2A
=cos2A (cos4A+sin4A+2cos2Asin2A - cos2Asin2A)
=cos2A ( (cos2A+sin2A )2 - (1/4) (2 sinA cosA )2 )
since cos2A+sin2A=1 and 2 sinA cosA=sin2A
we get cos2A ( 1 - (1/4) sin22A )
Answered by | 11 Aug, 2008, 06:29: PM
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