NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Mathematics
- Chapter
- Trigonometric Ratios & Identities
NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities
The value of (1 + cos(π)/(8))(1 + cos (3π)/(8))(1 + cos(5π)/(8))(1...
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(1 + cos \dfrac{7\pi}{8}\right)\left(1 + cos \dfrac{3\pi}{8}\right) \left(1 + cos \dfrac{5\pi}{8}\right)<br /><br />= \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(\text{cos}\dfrac{3\pi}{8}\right)
= \sin^{2}
\left(1 + cos \dfrac{\pi}{8}\right) \left(1 + cos \dfrac{7\pi}{8}\right)\left(1 + cos \dfrac{3\pi}{8}\right) \left(1 + cos \dfrac{5\pi}{8}\right)<br /><br />= \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(\text{cos}\dfrac{3\pi}{8}\right)
= \sin^{2}