Question

Question no. e. and f. please help me !!!!!​

Answer 1

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Answer 2

Answer: see proofs below

Step-by-step explanation:

Proofs are like a puzzle that you have to unravel.

You can manipulate the left side to get the right side (LHS → RHS)

                                                or

You can manipulate the right side to get the left side (RHS → LHS)

You will need to use Pythagorean Identities, Sum and Difference Identities, and/or Double Angle Identities when manipulating one of the sides.

Use the following Identities:

cos 2A = cos²A - sin²A

sin 2A = 2 cosA · sinA    Note that A can be substituted with 2A        

sin 3A = 3 sinA - 4sin³A

cos 3A = 4cos³A - 3cosA

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e) Prove: 4 sin A · cos³A - 4 cos A · sin³A = sin(4A)

LHS → RHS

Given:         4 sin A · cos³A - 4 cos A · sin³A

Factor:        4 sin A · cos x (cos²A - sin²A)

                  2(2 sin A · cos A)(cos²A - sin²A)

Double Angle Identity:  2(sin 2·A)(cos 2·A)

                                       2 sin (2A) · cos (2A)

Double Angle Identity:  sin (2·2A)

                                       sin 4A

Proven:  sin 4A = sin 4A   $\checkmark$

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f) Prove: 4 sin³A · cos 3A + 4 cos³A · sin 3A = 3 sin(4A)

LHS → RHS

Given:          4 sin³A · cos 3A + 4 cos³A · sin 3A

Triple Angle Identity:   4sin³A (4cos³A - 3 cosA) + 4cos³A (3sinA - 4sin³A)

Expand:      16sin³A · cos³A - 12sin³A · cosA + 12cos³A · sinA - 16sin³A · cos³A

Simplify:      12cos³A · sinA - 12sin³A · cosA

Expand:        6cos²A(2cosA · sinA) - 6sin²A(2cosA · sinA)

Double Angle Identity: 6cos²A(sin 2A) - 6sin²A(sin 2A)

Factor:       6sin 2A (cos²A - sin²A)

Double Angle Identity:  6sin 2A (cos 2A)

                                       3(2sin 2A · cos 2A)

Simplify:     3sin (2·2A)

                  3 sin 4A

Proven: 3sin 4A = 3sin4A   $\checkmark$