Can anyone help me with this trig question plz? (1 Viewer)

[CJ251]

Well hey everyone im back...
got another question...
how do we graph y = cosx - sinx

 

[pLuvia]

Draw the cosx and sinx graph, then subtract the cosx y ordinates from the sinx ordinate.
Or using auxliary method change y=cosx-sinx, into sqrt{2}cos(x+pi/4), then graph this curve, both will yeild the same answer

 

[CJ251]

Draw the cosx and sinx graph, then subtract the cosx y ordinates from the sinx ordinate.
Or using auxliary method change y=cosx-sinx, into sqrt{2}cos(x+pi/4), then graph this curve, both will yeild the same answer

Thanks heaps but i dont understand the auxiliary method. How to get sqrt 2 and how to graph it. If you could explain it to me please, thanks a million.

 

[pLuvia]

As explained on the other page
Given the equation y=acosx-bsinx
y=sqrt{a²+b²}cos(x-tan^-1(b/a))

asinx+bcosx=rsin(x+alpha)
asinx-bcosx=rsin(x-alpha)
acosx-bsinx=rcos(x+alpha)
acosx+bsinx=rcos(x-alpha)

r^2=a^2+b^2 but r>0
alpha=tan^(-1)(b/a)

More on this auxiliary method

So for your question
R=sqrt{1²+1²}
=sqrt{2}
alpha=tan^-11
=pi/4

 

[CJ251]

As explained on the other page
Given the equation y=acosx-bsinx
y=sqrt{a²+b²}cos(x-tan^-1(b/a))



More on this auxiliary method

So for your question
R=sqrt{1²+1²}
=sqrt{2}
alpha=tan^-11
=pi/4

Cewl thanks heaps again and again. But im sure ill be back. Thanks everyone or their help. Good luck for the Hsc.