Trigonometric Equations Question 374
Question: If the solution for $ \theta $ of $ \cos p\theta +\cos q\theta =0,\ p>0,\ q>0 $ are in A.P., then the numerically smallest common difference of A.P. is
[Kerala (Engg.) 2001]
Options:
A) $ \frac{\pi }{p+q} $
B) $ \frac{2\pi }{p+q} $
C) $ \frac{\pi }{2(p+q)} $
D) $ \frac{1}{p+q} $
Show Answer
Answer:
Correct Answer: B
Solution:
- Given $ \cos p\theta =-\cos q\theta =\cos (\pi +q\theta ) $
Þ $ p\theta =2n\pi \pm (\pi +q\theta ),n\in I $
Þ $ \theta =\frac{(2n+1)\pi }{p-q} $ or $ \frac{(2n-1)\pi }{p+q},n\in I $ Both the solutions form an A.P. $ \theta =\frac{(2n+1)\pi }{p-q} $ gives us an A.P. with common difference $ \frac{2\pi }{p-q} $ and $ \theta =\frac{(2n-1)\pi }{p+q} $ gives us an A.P. with common difference $ =\frac{2\pi }{p+q} $ . Certainly, $ \frac{2\pi }{p+q}<,| ,\frac{2\pi }{p-q}, | $ .
BETA
