The maximum and minimum of $a\cos(\alpha+\theta)+b\cos(\beta+\theta)$

Question

The question is to prove:

$$-\sqrt{a^2+b^2+2ab\cos(\alpha-\beta)} < a\cos(\alpha+\theta)+b\cos(\beta+\theta)<\sqrt{a^2+b^2+2ab\cos(\alpha-\beta)}$$

I tried opening the brackets but I'm not succeeding. I even tried to differentiate the function. But I'm not getting any closer to the answer.

However, is the question itself correct?

Answer 1

$$a\cos(\alpha+\theta)+b\cos(\beta+\theta)=(a\cos\alpha+b\cos\beta)\cos\theta-(a\sin\alpha+b\sin\beta)\sin\theta\\ =c\cos\theta-s\sin\theta =\sqrt{c^2+s^2}\cos\left(\theta+\arctan\frac sc\right)$$

and

$$-1\le \cos\left(\theta+\arctan\frac sc\right)\le1.$$

You can finish.

Answer 2

Note that $$S=Re(ae^{\alpha + \theta} + be^{\beta + \theta})$$ You can imagine these two complex numbers as vectors in the Argand plane.The resultant of these two vectors, can be found from the rule of addition of vectors and would have magnitude equal to $\sqrt {a^2+b^2+2ab cos(\alpha-\beta)}$. Now, the real part of this resultant vector would be S. This would be its cosine component, which lies between $-1$ and $1$. Hence the result holds up.

Answer 3

This is equivalent to proving

${(a\cos(\alpha + \theta) + b\cos(\beta + \theta))}^2 \le a^2+b^2+2ab\cos(\alpha-\beta)$

$\iff a^2(1 - \cos^2(\alpha + \theta)) + b^2(1 - \cos^2(\beta + \theta)) + 2ab[\cos(\alpha-\beta) - \cos(\theta+\alpha)\cos(\theta+\beta)] \ge 0$

$\large{[Note: \cos(\alpha - \beta) = \cos((\theta+\alpha)-(\theta+\beta)) = \cos(\theta+\alpha)(\theta+\beta) + \sin(\theta+\alpha)(\theta+\beta)]}$

$\iff a^2\sin^2(\alpha + \theta) + b^2\sin^2(\beta + \theta) + 2ab\sin(\alpha + \theta)\sin(\beta + \theta) \ge 0$

$\iff (a\sin(\alpha + \theta) + b\sin(\beta + \theta))^2 \ge 0$, which is true

So, the given inequality holds

Answer 4

Imagine two rotating vectors of lengths $a,b$ and initial directions $\alpha,\beta$ (with constant difference $\alpha-\beta$). By the cosine rule in the triangle formed by these three vectors, the length of the sum is

$$\sqrt{a^2+b^2+2ab\cos(\alpha-\beta)}.$$

When the vectors rotate, the projection of the sum on the horizontal axis oscillates between plus and minus this value.