Prove that: 4sin³A.cos3A +4cos³A.sin3A = 3sin4A

This is a class 10 Question From Values of Trigonometric Ratios chapter of Unit Trigonometry. All the steps for the solutions are mentioned in the description below. If that's hard for you to navigate, you can always visit the facebook link given at the end of every posts.

Solution:

Taking LHS,

= 4sin³A.cos3A +4cos³A.sin3A

= 4sin³A (4cos³A-3cosA) +4cos³A(3sinA-4sin³A)

= 16sin³Acos³A -12sin³AcosA +12sinAcos³A -16sin³Acos³A

= 12sinAcos³A -12sin³AcosA

= 6 (2sinAcos³A -2sin³AcosA)

= 6 (2sinAcosA) (cos²A-sin²A)

= 6 (sin2A) (cos2A)

= 3 (2sin2Acos2A)

= 3 sin(2A+2A)

= 3 sin4A

= RHS

Explanation to this solution:

Step 1: Write the LHS same from the question.

Step 2: We use the formula of multiple angles of sin3A and cos3A. [sin3A = 3sinA-4sin³A] and [cos3A = 4cos³A-3cosA]

Step 3: Write the expression by multiply the factors. 

Step 4: 16sin³Acos³A - 16sin³Acos³A= 0. Now, write the remaining expression.

Step 5: Take 6 as common factor in the two terms. 

Step 6: Again take 2sinAcosA common from the other terms. 

Step 7: Now, write [2sinAcosA = sin2A] and [cos²A-sin²A = cos2A].

Step 8: Break 6 into 2*3 and write 2sin2Acos2A.

Step 9: 2sin2Acos2A can be written as sin(2A+2A) using the identity [sin(A+A) = 2sinAcosA.]

Step 10: Now, write 3sin4A as our answer. 

Here is the Facebook link to the solution of this question in image. 

Related Notes:

Link: Introduction To Trigonometry
Link: Values of Trigonometric Ratios
Link: Compound Angles

Question: Prove that: 4sin³Acos3A +4cos³Asin3A = 3sin4A | Trigonometric Identities | SciPiPupil

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