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?
Hi
$\displaystyle a \sin x + b \cos x = \sqrt{a^2+b^2} \cdot \left(\frac{a}{\sqrt{a^2+b^2}} \sin x + \frac{b}{\sqrt{a^2+b^2}} \cos x \right)$
And since $\displaystyle \left(\frac{a}{\sqrt{a^2+b^2}}\right)^2 + \left(\frac{b}{\sqrt{a^2+b^2}}\right)^2 = 1$
there exists $\displaystyle \phi$ such that
$\displaystyle \frac{a}{\sqrt{a^2+b^2}} = \cos \phi$
$\displaystyle \frac{b}{\sqrt{a^2+b^2}} = \sin \phi$
Therefore $\displaystyle a \sin x + b \cos x = \sqrt{a^2+b^2} \cdot \left(\sin x \cos \phi + \cos x \sin \phi \right) = \sqrt{a^2+b^2} \cdot \sin\left(x+\phi\right)$
For instance
$\displaystyle 5 \sin x + 12 \cos x = 13 \sin\left(x+\phi\right)$
The minimum is -13 and the maximum is 13
Well the minimum of $\displaystyle 5 \sin x + 12 \cos x$ is -13 and the maximum is 13
Therefore
the minimum of $\displaystyle 5 \sin x + 12 \cos x +17$ is -13+17=4 and the maximum is 13+17=30
the minimum of $\displaystyle \frac{1}{5 \sin x + 12 \cos x +17}$ is 1/30 and the maximum is 1/4
the minimum of $\displaystyle \frac{30}{5 \sin x + 12 \cos x +17}$ is 30/30 = 1 and the maximum is 30/4 = 7.5