Step 1: Use the given equations.
$ \cos A + \cos B = -\cos C $
$ \sin A + \sin B = -\sin C $
Step 2: Square and add the equations.
$ (\cos A + \cos B)^2 + (\sin A + \sin B)^2 = (-\cos C)^2 + (-\sin C)^2 $ $ (\cos^2 A + 2 \cos A \cos B + \cos^2 B) + (\sin^2 A + 2 \sin A \sin B + \sin^2 B) = \cos^2 C + \sin^2 C $
Step 3: Simplify using trigonometric identities.
$ (\cos^2 A + \sin^2 A) + (\cos^2 B + \sin^2 B) + 2 (\cos A \cos B + \sin A \sin B) = 1 $
$ 1 + 1 + 2 \cos(A - B) = 1 $
Step 4: Solve for $ \cos(A - B) $.
$ 2 + 2 \cos(A - B) = 1 $
$ 2 \cos(A - B) = -1 $
$ \cos(A - B) = -\frac{1}{2} $
Step 5: Conclusion.
The value of $ \cos(A - B) $ is $ -\frac{1}{2} $.
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |
If \(\cos \alpha + \cos \beta + \cos \gamma = \sin \alpha + \sin \beta + \sin \gamma = 0,\) then evaluate \((\cos^3 \alpha + \cos^3 \beta + \cos^3 \gamma)^2 + (\sin^3 \alpha + \sin^3 \beta + \sin^3 \gamma)^2 =\)