Thread: Turning asinx+bcosx to a multiplication.

Hi everyone.
I have been absent the past week in school because I was ill and I missed a about 5 days worth of new stuff that my classmates learned in maths. I also missed the exam for the chapter of trigonometry.
So while I was revising every single page I was absent to, one lesion made me very confused and I am having a hard time understanding. I have to turn "asinx+bcosx" to a multiplication. I am not a native English speaker, so I'll try to translate my book as good as i can. About this, it says:

Solve the equation: 3sinx-sqroot of3*cosx=3
First, we divide both parts of the equation by 3, and we get: sinx-(sqrootof3/3*cosx)=1
which is sinx-tan30*cosx=1, so [sin(x-30)]/cos30=1, so sin(x-30)=cos30=sqroot3/2.
The angle is 60 degrees and we have x-30=k*360+60 or x-30=k*360+120.
So in conclusion, we have x=k*360+90 or x=k*360+150

Sorry if Im making you confused. I don't understand the part in bold. I understand what he did with writing tan30 as Sin30/cos30, but where did the "cosx" go, that was right before this step? The one that is underlined and in narrow letters?

On the other hand, I have a couple of other exercises that I am not able to solve, due to the absence last week ...
The first one:
-Find the highest and the lowest value:

y=sin3x+ sin(90-3x)
-Prove that:
cotan2x+1/sin2x=cotanx
-Turn this into a form of multiplication (from the lesion that I don't understand):
cos(x-45) + sin (x+ 45)
-And the last one: sinx-sqrt2=0 (Solve the equation).

Thank you very much for your help, would appreciate any assistance.

Hey satycorn.

I understand that you have been sick or have missed school for another reason, but can you show any ideas or partial attempts for these questions? It's a lot easier for everyone if you do this.