Rsin(x+a) problem

**200001** · April 17th 2010, 12:00 AM

Hi
I am stuck on the last bit were I am asked to state the values of x, their maximum and where the occur

do i just set the other side to -1 and multiply out to solve?

**running-gag** · April 17th 2010, 01:38 AM

Hi

$a \sin x + b \cos x = \sqrt{a^2+b^2} \cdot \left(\frac{a}{\sqrt{a^2+b^2}} \sin x + \frac{b}{\sqrt{a^2+b^2}} \cos x \right)$

And since $\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2 + \left(\frac{b}{\sqrt{a^2+b^2}}\right)^2 = 1$

there exists $\phi$ such that

$\frac{a}{\sqrt{a^2+b^2}} = \cos \phi$

$\frac{b}{\sqrt{a^2+b^2}} = \sin \phi$

Therefore $a \sin x + b \cos x = \sqrt{a^2+b^2} \cdot \left(\sin x \cos \phi + \cos x \sin \phi \right) = \sqrt{a^2+b^2} \cdot \sin\left(x+\phi\right)$

For instance

$5 \sin x + 12 \cos x = 13 \sin\left(x+\phi\right)$

The minimum is -13 and the maximum is 13

**200001** · April 17th 2010, 09:18 AM

Thanks for the lengthy response

Im cool with that but not where that 30 has come from at the end?

**running-gag** · April 17th 2010, 11:05 AM

Well the minimum of $5 \sin x + 12 \cos x$ is -13 and the maximum is 13

Therefore
the minimum of $5 \sin x + 12 \cos x +17$ is -13+17=4 and the maximum is 13+17=30

the minimum of $\frac{1}{5 \sin x + 12 \cos x +17}$ is 1/30 and the maximum is 1/4

the minimum of $\frac{30}{5 \sin x + 12 \cos x +17}$ is 30/30 = 1 and the maximum is 30/4 = 7.5

**200001** · April 24th 2010, 11:45 PM

great! makes sense now

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