47.2k views
2 votes
Sin4xcos2x =

1/2sin6x + 1/2sin2x
1/2sin6x - 1/2sin2x
1/2cos2x + 1/2cos6x
1/2cos2x - 1/2cos6x

User BlitzKrieg
by
4.3k points

2 Answers

5 votes

Answer: 1/2sin(6x) + 1/2sin(2x)

Explanation:

took the test and got a 100

User Nijraj Gelani
by
4.9k points
5 votes

Answer:

1/2sin(6x) + 1/2sin(2x)

Explanation:

You can look up the formulas for the product identities for sine and cosine, or you can guess and check using a graphing calculator. I did the calculator solution first (see the first attachment), then looked up the identities so I can tell you what they are (see the second attachment).

__

These identities are based on the sum and difference angle identities:

sin(α+β) +sin(α-β) = (sin(α)cos(β) +sin(β)cos(α)) + (sin(α)cos(β) -sin(β)cos(α))

= 2sin(α)cos(β)

Dividing by 2 gives the identity of interest in this problem:

sin(4x)cos(2x) = (1/2)(sin(4x +2x) +sin(4x -2x))

sin(4x)cos(2x) = (1/2)(sin(6x) +sin(2x))

Sin4xcos2x = 1/2sin6x + 1/2sin2x 1/2sin6x - 1/2sin2x 1/2cos2x + 1/2cos6x 1/2cos2x-example-1
Sin4xcos2x = 1/2sin6x + 1/2sin2x 1/2sin6x - 1/2sin2x 1/2cos2x + 1/2cos6x 1/2cos2x-example-2
User Sterlingalston
by
4.7k points