Question

I looking for direction on this question: 1A-6: Express in the form Asin(x+c). For a) sinx +√3cosx & b) sinx-cosx. I'm thinking I need the Sine sum formula: sin(a+b) = sina cosb + sinb cosa, but I haven't trouble getting there.

Answer 1

To arrive at
\[Csin(x+\delta)\]
from
\[Asinx+Bcosx\]
or for this problem
\[sinx+\sqrt{3}cosx\]
First Obtain A B and C
so in this case its
\[A = 1;B=\sqrt{3}\]
\[C = \sqrt{A^2+B^2}\]
\[C = \sqrt{1^2+\sqrt{3}^2} = \sqrt{1+3} = 2; \]
Now to find
\[\delta\]
\[sin(\delta)= B/C\]
\[ cos(\delta)= \frac{A}{C}\]
In this case
\[sin(\delta)=\frac{\sqrt{3}}{2}; \delta =\frac{\pi}{3}\]
\[cos(\delta)=\frac{1}{2}; \delta = \frac{\pi}{3}\]
or
\[tan(\delta) = \sqrt{3}\]
So
\[\delta = \frac{\pi}{3}\]
Now just assemble the equation and
\[Csin(x+\delta)\]
becomes
\[2sin(\frac{\pi}{3}+x)\]
This is the basic identity, if you would like a little more rigor there are numerous examples on the web or in textbooks, however this will suffice for rewriting this type of identity.

Answer 2

Thanks...I was making the problem harder then it had to be.