JEE Main
- Exam
- JEE Main
- Subject
- Mathematics
- Chapter
- Trigonometric Ratios & Identities
JEE Main · Mathematics · Trigonometric Ratios & Identities
The value of(1 + cos(π)/(8)) ( 1 + cos(3π)/(8))( 1 + cos(5π)/(8))...
The value of\left(1 + cos\dfrac{\pi}{8}\right) \left(1 + cos\dfrac{3\pi}{8}\right)\left(1 + cos\dfrac{5\pi}{8}\right) \left(1 + cos\dfrac{7\pi}{8}\right)is equal to
Options
- A
- B
- C
- D
Correct answer
C.
Step-by-step solution
Given expression can be written as
\left(1 + cos \dfrac{\pi}{8}\right) \left(\dfrac{7\pi}{8}\right)\left(1 + cos \dfrac{3\pi}{8}\right) \left(\dfrac{5\pi}{8}\right)
= \left(1 + \text\cos\dfrac{π}{8}\right) \left(1 - \text\cos\dfrac{π}{8}\right)\left(1 + cos\dfrac{3\pi}{8}\right) \left(1 - \text\cos\dfrac{3\pi}{8}\right)<br/><br/>= \left(1 - \cos^{2}\dfrac{π}{8}\right)\left(\dfrac{3\pi}{8}\right)
= \sin^{2} \dfrac{\pi}{8} \cdot \cos^{2} \dfrac{\pi}{8}<br/><br/>=\dfrac{1}{4} {\left(2 \text\sin \dfrac{\pi}{8} \dfrac{\text\cos \pi}{8}\right)}^{2}<br/><br/>=\dfrac{1}{4} \sin^{2} \dfrac{\pi}{4}<br/><br/>= \dfrac{1}{8}
\left(1 + cos \dfrac{\pi}{8}\right) \left(\dfrac{7\pi}{8}\right)\left(1 + cos \dfrac{3\pi}{8}\right) \left(\dfrac{5\pi}{8}\right)
= \left(1 + \text\cos\dfrac{π}{8}\right) \left(1 - \text\cos\dfrac{π}{8}\right)\left(1 + cos\dfrac{3\pi}{8}\right) \left(1 - \text\cos\dfrac{3\pi}{8}\right)<br/><br/>= \left(1 - \cos^{2}\dfrac{π}{8}\right)\left(\dfrac{3\pi}{8}\right)
= \sin^{2} \dfrac{\pi}{8} \cdot \cos^{2} \dfrac{\pi}{8}<br/><br/>=\dfrac{1}{4} {\left(2 \text\sin \dfrac{\pi}{8} \dfrac{\text\cos \pi}{8}\right)}^{2}<br/><br/>=\dfrac{1}{4} \sin^{2} \dfrac{\pi}{4}<br/><br/>= \dfrac{1}{8}