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?

  • 3
    $\begingroup$ Setting $a=b=0$ gives a counterexample. $\endgroup$
    – Wojowu
    Jun 6, 2016 at 6:13
  • 1
    $\begingroup$ What is the source of the problem? Can it be $\ge1$? $\endgroup$ Jun 6, 2016 at 6:16
  • 1
    $\begingroup$ @labbhattacharjee It's from a poorly handwritten notes on isoperimetric inequality... $\endgroup$
    – 3x89g2
    Jun 6, 2016 at 6:17
  • $\begingroup$ 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. $\endgroup$
    – Narasimham
    Jun 6, 2016 at 6:35
  • $\begingroup$ $\frac{\sqrt3}{4}\sin\frac{\pi}{6}-\frac14\cos\frac{\pi}{6}=0$ but ... $\endgroup$
    – Math-fun
    Jun 6, 2016 at 8:10

3 Answers 3


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.


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:


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:


$a^2+b^2\leq1$, 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 $a^2+b^2=1$.


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

  • $\begingroup$ 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! $\endgroup$
    – user21820
    Jun 7, 2016 at 1:58

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