Q.If $\cos\alpha + \cos\beta = 0 = \sin\alpha + \sin\beta$, then prove that $\cos 2\alpha + \cos 2\beta = -2\cos(\alpha + \beta)$.

The key idea is to use sum-to-product trigonometric identities to simplify the given conditions and the expression to be proved.

1. From the given conditions, $\cos\alpha + \cos\beta = 0$ and $\sin\alpha + \sin\beta = 0$, we apply the sum-to-product formulas:

   $2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right) = 0 \quad \text{(i)}$

   $2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right) = 0 \quad \text{(ii)}$

2. For both equations (i) and (ii) to hold, and knowing that $\cos x$ and $\sin x$ cannot be simultaneously zero (since $\cos^2 x + \sin^2 x = 1$), it must be that $\cos\left(\frac{\alpha-\beta}{2}\right) = 0$.

3. This implies $\frac{\alpha-\beta}{2} = (2n+1)\frac{\pi}{2}$ for some integer $n$. Therefore, $\alpha - \beta = (2n+1)\pi$.

   Consequently, $\cos(\alpha-\beta) = \cos((2n+1)\pi) = -1$. …

The problem asks us to prove a trigonometric identity given two conditions. The core idea is to extract a fundamental relationship between $\alpha$ and $\beta$ from the given conditions, and then use this relationship to simplify the expression $\cos 2\alpha + \cos 2\beta$ until it matches the right-hand side, $-2\cos(\alpha + \beta)$.

The conditions $\cos\alpha + \cos\beta = 0$ and $\sin\alpha + \sin\beta = 0$ are quite strong. They tell us that the sum of the cosine values is zero, and the sum of the sine values is also zero. This implies that the angles $\alpha$ and $\beta$ must be related in a specific way.

One powerful technique when you have sums of sines and cosines equal to zero is to square both equations and add them. This often leads to an identity involving $\cos(\alpha - \beta)$ or $\cos(\alpha + \beta)$.

1. Derive a relationship from the given conditions.

   We are given:

   (1) $\cos\alpha + \cos\beta = 0$

   (2) $\sin\alpha + \sin\beta = 0$

   Square both equations:

   From (1): $(\cos\alpha + \cos\beta)^2 = 0^2$

   $\cos^2\alpha + \cos^2\beta + 2\cos\alpha\cos\beta = 0 \quad (*)$

   From (2): $(\sin\alpha + \sin\beta)^2 = 0^2$

   $\sin^2\alpha + \sin^2\beta + 2\sin\alpha\sin\beta = 0 \quad (**)$

   Now, add equations $(*)$ and $(**)$:

   $(\cos^2\alpha + \sin^2\alpha) + (\cos^2\beta + \sin^2\beta) + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) = 0 + 0$

   Using the fundamental identity $\sin^2 x + \cos^2 x = 1$:

   $1 + 1 + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) = 0$

   The cosine addition formula is $\cos(A - B) = \cos A \cos B + \sin A \sin B$.

   Applying this formula to the term in the parenthesis:

   $2 + 2\cos(\alpha - \beta) = 0$

   $2\cos(\alpha - \beta) = -2$

   $\cos(\alpha - \beta) = -1$

   This is a crucial relationship. It tells us that the difference between angles $\alpha$ and $\beta$ must be an odd multiple of $\pi$. For example, $\alpha - \beta = \pi, 3\pi, -\pi$, etc.

   Tip

   Alternatively, you could use sum-to-product formulas for the initial conditions:

   $\cos\alpha + \cos\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right) = 0$

   $\sin\alpha + \sin\beta = 2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right) = 0$

   For both to be zero, either $\cos\left(\frac{\alpha-\beta}{2}\right) = 0$ or both $\cos\left(\frac{\alpha+\beta}{2}\right) = 0$ and $\sin\left(\frac{\alpha+\beta}{2}\right) = 0$. The latter is impossible since $\sin^2 x + \cos^2 x = 1$.

   Thus, we must have $\cos\left(\frac{\alpha-\beta}{2}\right) = 0$.

   This implies $\frac{\alpha-\beta}{2} = (2n+1)\frac{\pi}{2}$ for some integer $n$.

   So, $\alpha - \beta = (2n+1)\pi$. …