Question

How do you convert this equation?

Answer 1

How to convert (4 sin (2x-38) + 3sin (2x+22)-6) to asin (bx+c)+d form? PLEASE HELP

Answer 2

7sin (4x-16)-6? i'm not sure if it's right but I just added the like terms

Answer 3

the answer is supposed to be 6.1 sin(2x-12.7)-6. I dont know how to get that.

Answer 4

Use the formulas for sin of a sum and sin of a difference, then convert what you have to the desired form.

Answer 5

Formulas to help:\[\sin(\alpha + \beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta\]\[\sin(\alpha - \beta)=\sin \alpha \cos \beta-\cos \alpha \sin \beta\]

Answer 6

\[4 \sin (2x-38) + 3\sin (2x+22)-6\]\[=4(\sin2xcos38-\cos2xsin38)+3(\sin2xcos22+\cos2xsin22)-6\]

Answer 7

How would your answer get to 6.1sin(2x-12.7)-6, AnimalAin?

Answer 8

Not really sure. I got a step or two further, but sort of ran out of ideas and ambition at the same time. I think the next step is to factor out sin 2x and cos 2x, then simplify the remaining terms. From there, you should (might) be able to put it back together in the form you want.

Answer 9

\[4(\sin2xcos38−\cos2xsin38)+3(\sin2xcos22+\cos2xsin22)−6\]\[=\sin 2x(4\cos38+3\cos22)+\cos2x(3\sin22-4\sin38)-6\]

Answer 10

\[=3\sin2x(\cos38+\cos22)+3\cos2x(\sin22−\sin38)+\]\[\sin2xcos38-\cos2xsin38−6 \]

Answer 11

\[=3[\sin2x(2\cos30\cos16)+\cos2x(2\cos30\sin16)]+ \sin(2x-38)−6 \]\[=6\cos30[\sin2x(\cos16)+\cos2x(\sin16)]+\sin(2x−38)−6\]\[=6\cos30[\sin(2x+16)]+\sin(2x−38)−6 \]

Answer 12

I've tinkered here for a while, and think my work is ok, but really don't know how I would get from here to where your text says I should end up. Sorry, but I'm afraid I need to practice my trig identities a bit more.