Suppose $-a\sin(s) - b\cos(s) = 0$, then $a^2 + b^2 = 1$?

Question

Suppose $-a\sin(s) - b\cos(s) = 0$ and given that $a^2 + b^2 \le 1$, then $a^2 + b^2 = 1$?

I am having trouble getting the above identity. I vaguely recall that

$$a\sin(s) + b\cos(s)=\sqrt{a^2+b^2}\sin(s+\theta)$$

where $$\tan(\theta) = \frac{b}{a}$$

But I still don't see why it's obvious?

@labbhattacharjee It's from a poorly handwritten notes on isoperimetric inequality... — 3x89g2, Jun 06 '16 at 06:17
Whether $\sqrt{a^2+b^2}$ is to be set to unity in trig mode or leave it as it is in such simplification is perhaps what you are unable to recall. — Narasimham, Jun 06 '16 at 06:35
$\frac{\sqrt3}{4}\sin\frac{\pi}{6}-\frac14\cos\frac{\pi}{6}=0$ but ... — Math-fun, Jun 06 '16 at 08:10
You can find the proof of the result you vaguely recall, for example, here (and probably in other posts on this site). It is also mentioned on Wikipedia. — Martin Sleziak, Jun 06 '16 at 08:50

score 1 · Answer 1 · answered Jun 06 '16 at 06:14

I don't see how the identity could be true. In the identity that you "vaugely recall", the only way for the right side to be identically zero is for the part in front of $sin$ to be zero. This requires that both $a$ and $b$ are zero, so $a^2+b^2$ would be zero.

user21820 · Answer 2 · 2016-06-07T10:42:10.380

0

It makes no sense for a mathematical claim to have a condition of the form "$-P-Q = 0$", so maybe the intended claim is:

Given reals $a,b,t$ such that $1 - a \sin(t) - b \cos(t) = 0$ and $a^2+b^2 \le 1$, then $a^2+b^2 = 1$.

And indeed your method proves this.

For fun, here is an alternative proof that uses weighted AM-GM:

Take reals $a,b,t$ such that $1 - a \sin(t) - b \cos(t) = 0$. Note that $a,b$ are not both zero.

Then $\dfrac1{a^2+b^2} = \dfrac{ a^2 \frac1a \sin(t) + b^2 \frac1b \cos(t) }{a^2+b^2} \le \sqrt{\dfrac{a^2 ( \frac1a \sin(t) )^2 + b^2 ( \frac1b \cos(t) )^2 }{a^2+b^2}} = \dfrac1{\sqrt{a^2+b^2}}$.

Thus $a^2+b^2 \ge 1$.

edited Jun 07 '16 at 10:42

answered Jun 06 '16 at 06:39

user21820

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Why on earth do two people think that this answer is wrong? It's the only one at the time of posting that attempted to figure out what was intended, where adding a single extra symbol "$1$" makes the claim correct! – user21820 Jun 07 '16 at 01:58

score 0 · Answer 3 · edited Apr 13 '17 at 12:20

Let us see the identity you vaguely recall:

$$a\sin(s) + b\cos(s)=\sqrt{a^2+b^2}\sin(s+\theta)$$

with $\tan\theta=\frac ba$. To get it, we factor out the square root, then set $\sin\theta=\frac{b}{\sqrt{a^2+b^2}}$ and $\cos\theta=\frac{a}{\sqrt{a^2+b^2}}$, which is licit since the squares of those numbers sum to 1. Now we have:

$$a\sin s+b\cos s=\sqrt{a^2+b^2}(\cos\theta\sin s+\sin\theta\cos s),$$

and $\sin(s+\theta)$ expands to just that using addition formula.

So the identity at question top now becomes:

$$\sqrt{a^2+b^2}\sin(s+\theta)=0.$$

I have, of course, changed signs, which doesn't alter the equation. This can never be true for all $s$ unless $a=b=0$, in which case obviously $a^2+b^2=0\neq1$.

Another answer suggests the problem might have a $1$ at the start of the equations, and that the identity works only for some $t$. Then we would have, for some $t$, that:

$$1-\sqrt{a^2+b^2}\sin(t+\theta)=0\iff\sin(t+\theta)=\frac{1}{\sqrt{a^2+b^2}}.$$

$a^2+b^2\leq1$, so the RHS is at least 1, but the LHS is at most 1 in absolute value, so for the equation to hold you need $a^2+b^2=1$.

Suppose $-a\sin(s) - b\cos(s) = 0$, then $a^2 + b^2 = 1$?

3 Answers3