I need a little nudge to the finish for the last bit of this problem.
Express $\lambda \sin \theta + (1 - \lambda) \cos \theta$ in the form $R \sin (\theta + \phi)$, where $R(R>0)$ and $\tan \phi$ are to be given in terms of $\lambda$.
Write down an expression in terms of $\lambda$ for the minimum value of $\lambda \sin \theta + (1 - \lambda) \cos \theta$ as $\theta$ varies.
Show that, for all $\lambda$, this minimum is less than or equal to $-\dfrac{1}{\sqrt 2}$.
The first part needs expansion like the $a\cos x + b\sin x$ formula.
Let, $ \begin{align} \lambda \sin \theta + (1 - \lambda) \cos \theta &\equiv R \sin (\theta + \phi) \\ &\equiv R[\sin \theta \cos \phi + \cos \theta \sin \phi] \\ &\equiv (R \cos \phi)\sin \theta + (R \sin \phi) \cos \theta \\ &\equiv a \sin \theta + b \cos \theta \end{align} $
Where,
$a = R \cos \phi = \lambda$
$b = R \sin \phi = (1 - \lambda)$
$\tan \phi = \dfrac{b}{a} = \dfrac{1 - \lambda}{\lambda}$
$R = \sqrt {2\lambda^2 - 2\lambda + 1}$
Thus the expression in terms of $\lambda$ is,
$\sqrt {2\lambda^2 - 2\lambda + 1}\Big(\sin (\theta + \phi)\Big)$
Since $\sin$ has minimum value of $-1$, the minimum value of the expression is $-\sqrt {2\lambda^2 - 2\lambda + 1}$
This is as far I have gotten. I don't understand the -$1/\sqrt 2$ part? I thought may be the root $\ge$ 0 would help, and I tried solving that quadratic, but it has no real roots. How do you go about proving this?
Thanks for your help!