Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
3  
You are almost there. What's the minimum of $2\lambda^2-2\lambda+1$? Hint: its graph is a parabola. –  Jyrki Lahtonen Jul 9 '11 at 12:22
1  
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. –  mathguy80 Jul 9 '11 at 13:13
1  
Mark this one solved. Well done! –  Jyrki Lahtonen Jul 9 '11 at 13:20
1  
Took slightly more trouble than necessary. Your expression is equal to $\frac{-1}{\sqrt{2}}\sqrt{4\lambda^2-4\lambda+2}$, and $4\lambda^2-4\lambda+2=(2\lambda-1)^2+1$. –  André Nicolas Jul 9 '11 at 14:18
4  
@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... –  Aryabhata Jul 9 '11 at 17:29
show 2 more comments

1 Answer

up vote 3 down vote accepted

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,

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.