periodicity of trigonometric functions

Asked by  | 15th 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  | 11th Aug, 2008, 06:29: PM

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