Q. 22

Prove that the following identities :

2(sin6 θ – cos6 θ) – 3(sin4 θ + cos4 θ ) + (sin2 θ + cos2 θ)

Answer :

Taking LHS

= 2(sin6 θ – cos6 θ) – 3(sin4 θ+ cos4 θ ) + (sin2 θ + cos2 θ)


= 2[(sin2 θ)3 – (cos2 θ)3 ] – 3[(sin2 θ)2 + (cos2 θ)2 ]+1 [ cos2 θ + sin2 θ = 1]


Now, we use these identities, (a3 – b3)= (a + b)3 – 3ab(a+b) and (a2 + b2) = (a +b)2 – 2ab]


= 2[(sin2 θ + cos2 θ)3 – 3sin2θ cos2θ (sin2 θ+ cos2 θ)] –3[(sin2 θ + cos2 θ) – 2 sin2 θ cos2 θ]+ 1


=2[(1) – 3sin2θ cos2θ (1)] – 3[(1) – 2 sin2 θ cos2 θ] + 1 [ cos2 θ + sin2 θ = 1]


=2(1 – 3 sin2 θ cos2 θ )– 3 + 6sin2 θ cos2 θ+ 1


= 2– 6sin2θ cos2θ – 2 + 6sin2 θ cos2 θ


=0


=RHS


Hence Proved


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Trigonometric Identities33 mins
Champ Quiz | Trigonometric Identities33 mins
NCERT | Trigonometric Identities52 mins
Quiz | Task on Trigonometric Ratios46 mins
Trigonometric Identities44 mins
Solving NCERT Questions on Trigonometric Identities56 mins
Algebraic Identities48 mins
Quiz | Practice Important Questions on Trigonometrical Identities46 mins
Quiz on Trigonometric Ratios31 mins
T- Ratios of Specified Angles58 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses