If a line makes angles $α,β$ and $γ$ with positive x-axis, y-axis and z-axis respectively, then the value of $\sin^2\frac{α}{2}.\cos^2\frac{α}{2}+\sin^2\frac{β}{2}.\cos^2\frac{β}{2}+ \sin^2\frac{γ}{2}. \cos^2\frac{γ}{2}$ is |
1 $\frac{1}{2}$ 2 $\frac{1}{4}$ |
$\frac{1}{2}$ |
The correct answer is Option (2) → $\frac{1}{2}$ Given: $\alpha, \beta, \gamma$ are the direction angles of a line. Direction cosines: $l = \cos\alpha,\; m = \cos\beta,\; n = \cos\gamma$ satisfy: $l^2 + m^2 + n^2 = 1$ Expression to evaluate: $\sin^2\frac{\alpha}{2} \cdot \cos^2\frac{\alpha}{2} \;+\; \sin^2\frac{\beta}{2} \cdot \cos^2\frac{\beta}{2} \;+\; \sin^2\frac{\gamma}{2} \cdot \cos^2\frac{\gamma}{2}$ Use identity: $\sin^2\theta \cdot \cos^2\theta = \frac{\sin^2 2\theta}{4}$. So, each term becomes: $\sin^2\frac{\alpha}{2} \cdot \cos^2\frac{\alpha}{2} = \frac{\sin^2\alpha}{4}$ Similarly for $\beta, \gamma$. Therefore, the given sum is: $\frac{\sin^2\alpha + \sin^2\beta + \sin^2\gamma}{4}$ Since $\sin^2\alpha = 1 - \cos^2\alpha = 1 - l^2$, and similarly for $\beta, \gamma$: $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = (1 - l^2) + (1 - m^2) + (1 - n^2)$ $= 3 - (l^2 + m^2 + n^2)$ $= 3 - 1 = 2$ Thus, required value = $\frac{2}{4} = \frac{1}{2}$ |