# Prove that minimum of $\lambda \sin \theta + (1 - \lambda) \cos \theta \le -\frac1{\sqrt 2}$

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?

• You are almost there. What's the minimum of $2\lambda^2-2\lambda+1$? Hint: its graph is a parabola. Commented Jul 9, 2011 at 12:22
• I ran out of steam on this problem! Looks simple thinking about it as a parabola, Minimum value of $2\lambda^2-2\lambda+1$ is $-\dfrac{b}{2a} = \dfrac{1}{2}$, and hence maximum of $-\sqrt {2\lambda^2-2\lambda+1}$ is $-\dfrac{1}{\sqrt 2}$ Which implies $-2\lambda^2-2\lambda+1 \le - \dfrac{1}{\sqrt 2}$ Thanks @Jyrki. Commented Jul 9, 2011 at 13:13
• Mark this one solved. Well done! Commented Jul 9, 2011 at 13:20
• Took slightly more trouble than necessary. Your expression is equal to $\frac{-1}{\sqrt{2}}\sqrt{4\lambda^2-4\lambda+2}$, and $4\lambda^2-4\lambda+2=(2\lambda-1)^2+1$. Commented Jul 9, 2011 at 14:18
• @mathguy: Can you please add an answer an tick it? We don't want people (especially that Community fellow) to think this is still unanswered... Commented Jul 9, 2011 at 17:29

As per @Aryabhata's suggestion, adding the answer from the earlier comment to close this question.

The last part of the question resolves to,

Show that minimum of $-\sqrt {2\lambda^2 - 2\lambda + 1} \le -\dfrac{1}{\sqrt 2}$

$2\lambda^2 - 2\lambda + 1$ is a parabola whose minimum is at its vertex, $-\dfrac{b}{2a} = \dfrac{-(-2)}{2(2)} = \dfrac{1}{2}$

And the maximum of $-\sqrt {2\lambda^2 - 2\lambda + 1}$ is $-\dfrac{1}{\sqrt 2}$

Hence,

$$-\sqrt {2\lambda^2 - 2\lambda + 1} \le -\dfrac{1}{\sqrt 2}$$