Differentiation Question 369
Question: If $ y={{\sin }^{2}}\alpha +{{\cos }^{2}}(\alpha +\beta )+2\sin \alpha \sin \beta \cos (\alpha +\beta ) $ , then $ \frac{d^{3}y}{d{{\alpha }^{3}}} $ is, (keeping $ \beta $ as constant)
Options:
A) $ \frac{{{\sin }^{3}}(\alpha +\beta )}{\cos \alpha } $
B) $ \cos (\alpha +3\beta ) $
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ y={{\sin }^{2}}\alpha +{{\cos }^{2}}(\alpha +\beta )+2\sin \alpha \sin \beta \cos (\alpha +\beta ) $
$ ={{\sin }^{2}}\alpha +\cos (\alpha +\beta ){\cos (\alpha +\beta )+2\sin \alpha \sin \beta } $
$ ={{\sin }^{2}}\alpha +\cos (\alpha +\beta )\cos (\alpha -\beta ) $
$ ={{\sin }^{2}}\alpha +\frac{1}{2}(\cos 2\alpha +\cos 2\beta ) $
$ ={{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha -\frac{1}{2}+\frac{\cos 2\beta }{2} $
Therefore $ y= $ constant
Therefore $ \frac{d^{3}y}{d{{\alpha }^{3}}}=0 $
Trick: Let $ \beta =180{}^\circ $ {since $ \beta $ is constant}
$ \therefore y={{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1\Rightarrow \frac{d^{3}y}{d{{\alpha }^{3}}}=0 $ .
BETA
