Question

If $ \alpha $, $ \beta $, and $ \gamma $ are the angles made by a vector with the $ x $-, $ y $-, and $ z $-axes respectively, then find the value of $ \sin^2\alpha + \sin^2\beta $.

• A. $ \sin^2\gamma $
• B. $ \cos^2\gamma $
• C. $ 1 + \cos^2\gamma $
• D. $ 1 + \sin^2\gamma $

Hint:

When dealing with direction cosines, always use the fundamental property $ \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 $ and the Pythagorean identity $ \sin^2\theta + \cos^2\theta = 1 $.

Answer 1

The Correct Option is C

Step 1: Use the Pythagorean identity.
For any angle $ \theta $: $$ \sin^2\theta + \cos^2\theta = 1 $$ Thus: $$ \sin^2\alpha = 1 - \cos^2\alpha $$ $$ \sin^2\beta = 1 - \cos^2\beta $$ Step 2: Sum $ \sin^2\alpha $ and $ \sin^2\beta $.
$$ \sin^2\alpha + \sin^2\beta = (1 - \cos^2\alpha) + (1 - \cos^2\beta) $$ Simplify: $$ \sin^2\alpha + \sin^2\beta = 2 - (\cos^2\alpha + \cos^2\beta) $$ Step 3: Use the direction cosine property.
From the property of direction cosines: $$ \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 $$ Rearrange: $$ \cos^2\alpha + \cos^2\beta = 1 - \cos^2\gamma $$ Step 4: Substitute back.
Replace $ \cos^2\alpha + \cos^2\beta $ in the expression: $$ \sin^2\alpha + \sin^2\beta = 2 - (1 - \cos^2\gamma) $$ Simplify: $$ \sin^2\alpha + \sin^2\beta = 2 - 1 + \cos^2\gamma $$ $$ \sin^2\alpha + \sin^2\beta = 1 + \cos^2\gamma $$ Final Answer: $ \boxed{1 + \cos^2\gamma} $