Question

Need hints on this question.
Express \(\sqrt{3}\sin(x)+\cos(x)\) in the form \(R\sin(x+\alpha)\).

Answer 1

\[
R\sin(x+\alpha)=R\left(\sin(x)\cos(\alpha)+\sin(\alpha)\cos(x)\right)
\]
What to do next?

Answer 2

Divide amd multiply each term by 2

Answer 3

*and

Answer 4

I am sure that was what Ganeshi was going to say , most probably :)

Answer 5

yes :)

\[\large \sqrt{3}\sin(x)+\cos(x) = \langle \cos x , \sin x \rangle \bullet \langle 1, \sqrt{3} \rangle \]

Answer 6

See ! ^^^ :D

Answer 7

You will get this on dividing and multiplying by 2:

\[\large{2(\cfrac{\sqrt{3}}{2} \sin x + \cfrac{1}{2} \cos x)}\]

Notice the sqrt(3)/2 and 1/2

Answer 8

Seem familiar ? @thomas5267

Answer 9

cos(30 deg) = sqrt(3)/2

sin(30 deg) = 1/2

Answer 10

The expression would thus become:

\[\large{2(\cos \cfrac{\pi}{6} \sin x + \sin \cfrac{\pi}{6} \cos x)}\]

Compare this with :

\(\color{blue}{\text{Originally Posted by}}\) @thomas5267
\[
R\sin(x+\alpha)=R\left(\sin(x)\cos(\alpha)+\sin(\alpha)\cos(x)\right)
\]
What to do next?
\(\color{blue}{\text{End of Quote}}\)