The Value of \( \frac{\sin5\alpha-\sin3\alpha}{\cos5\alpha+2\cos4\alpha+\cos3\alpha} \)
Question
Find the value of
\[ \frac{\sin5\alpha-\sin3\alpha} {\cos5\alpha+2\cos4\alpha+\cos3\alpha} \]
(a) \(\cot\frac{\alpha}{2}\)
(b) \(\cot\alpha\)
(c) \(\tan\frac{\alpha}{2}\)
(d) none of these
Solution
Using the identity
\[ \sin C-\sin D = 2\cos\frac{C+D}{2} \sin\frac{C-D}{2} \]
the numerator becomes
\[ \sin5\alpha-\sin3\alpha = 2\cos4\alpha\sin\alpha \]
Now simplify the denominator:
\[ \cos5\alpha+\cos3\alpha = 2\cos4\alpha\cos\alpha \]
Therefore,
\[ \cos5\alpha+2\cos4\alpha+\cos3\alpha = 2\cos4\alpha\cos\alpha + 2\cos4\alpha \]
\[ = 2\cos4\alpha(1+\cos\alpha) \]
Hence,
\[ \frac{2\cos4\alpha\sin\alpha} {2\cos4\alpha(1+\cos\alpha)} = \frac{\sin\alpha} {1+\cos\alpha} \]
Using the standard identity
\[ \frac{\sin\alpha}{1+\cos\alpha} = \tan\frac{\alpha}{2} \]
Therefore,
\[ \frac{\sin5\alpha-\sin3\alpha} {\cos5\alpha+2\cos4\alpha+\cos3\alpha} = \tan\frac{\alpha}{2} \]
Final Answer
\[ \boxed{\tan\frac{\alpha}{2}} \]
Hence, the correct option is (c) \(\tan\frac{\alpha}{2}\).