cos6a-sin6a=cos2a(1-1/4sin22a)

(cos2a)3 – (sin2a)3= (cos2a – sin2a )(cos4a + sin4a + cos2a*sin2a)= (cos2a)(1 – 2 cos2a*sin2a + cos2a*sin2a)= (cos2a)(1 – cos2a*sin2a )= (cos2a)(1 – 4cos2a*sin2a/4 )= cos2a(1-1/4sin22a)

BACK

Grade 11Trigonometry

cos⁶a-sin⁶a=cos2a(1-1/4sin²2a)

PRATUSH KUMAR MISHRA

11 Years agoGrade 11

1 Answer

Nishant Vora

11 Years ago

(cos²a)³ – (sin²a)³
= (cos²a – sin²a )(cos⁴a + sin⁴a + cos²a*sin²a)
= (cos2a)(1 – 2 cos²a*sin²a + cos²a*sin²a)
= (cos2a)(1 – cos²a*sin²a )
= (cos2a)(1 – 4cos²a*sin²a/4 )
= cos2a(1-1/4sin²2a)