Given: \[ \sin \theta + \cos \theta = 1 \] Step 1: Square both sides \[ (\sin \theta + \cos \theta)^2 = 1^2 \] Expanding the left side using the identity \((a+b)^2 = a^2 + 2ab + b^2\), we get: \[ \sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta = 1 \] Step 2: Use the Pythagorean identity Recall that: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substitute this into the equation: \[ 1 + 2 \sin \theta \cos \theta = 1 \] Step 3: Simplify Subtract 1 from both sides: \[ 2 \sin \theta \cos \theta = 0 \] Divide both sides by 2: \[ \sin \theta \cos \theta = 0 \] Step 4: Verify possible values - If \(\sin \theta = 0\), then \(\cos \theta = 1\), and \(\sin \theta \cos \theta = 0 \times 1 = 0\). - If \(\cos \theta = 0\), then \(\sin \theta = 1\), and \(\sin \theta \cos \theta = 1 \times 0 = 0\). Both cases satisfy the equation.
Final answer: \[ \boxed{0} \]
| 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 \((\cos^3 \alpha + \cos^3 \beta + \cos^3 \gamma)^2 + (\sin^3 \alpha + \sin^3 \beta + \sin^3 \gamma)^2 =\)