# Trigonometric identities

Determine a sequence of transformations which maps the graph of y=sinx onto the graph of y=(root3) sinx-3cosx+4
Original post by 01sbrals
Determine a sequence of transformations which maps the graph of y=sinx onto the graph of y=(root3) sinx-3cosx+4

Express the transformed graph in the form $R\sin(x+\alpha) + A$ then its obvious.
Original post by RDKGames
Express the transformed graph in the form $R\sin(x+\alpha) + A$ then its obvious.

I worked it out to be 2root3 sin (x-30). what would i do after that?
Original post by 01sbrals
I worked it out to be 2root3 sin (x-30). what would i do after that?

What happened to the +4 ?
Original post by RDKGames
What happened to the +4 ?

2root3 sin (x-30) +4
sorry
Original post by 01sbrals
2root3 sin (x-30) +4
sorry

I'm just going to assume it's correct.

So then rearrange into $\dfrac{y-4}{2\sqrt{3}} = \sin(x-30)$.

What transformations have taken place to get from $y=\sin x$ ??

Hint: there is one horizontal transformation, and two vertical ones.
Original post by RDKGames
I'm just going to assume it's correct.

So then rearrange into $\dfrac{y-4}{2\sqrt{3}} = \sin(x-30)$.

What transformations have taken place to get from $y=\sin x$ ??

Hint: there is one horizontal transformation, and two vertical ones.

vertical stretch by 2root 3
translation by 30 to the right
and translation 4 up?
Original post by 01sbrals
vertical stretch by 2root 3
translation by 30 to the right
and translation 4 up?

Thats right.

Always remember to write the order of transformations when talking about two or more transformations happening along one axis.

For instance, you are correct if we stretch vertically first before applying the translation, but if we do the translation first and then the stretch then the magnitudes of these are not necessarily the same.
I think it’s wrong no, the translation is correct but shouldn’t it be 60 instead of 30Tan^-1 of sqrt3/3 =60? Its tan^-1B/a = alpha
(edited 3 months ago)
I think it’s wrong no, the translation is correct but shouldn’t it be 60 instead of 30Tan^-1 of sqrt3/3 =60? Its B/a = alpha