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The value of cos ( pi //5) cos ( 2pi //5...

The value of `cos ( pi //5) cos ( 2pi //5) cos ( 4pi //5) cos ( 8pi //5)` is

A

`1//16`

B

0

C

`-1//8`

D

`-1//16`

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The correct Answer is:
To find the value of \( \cos \left( \frac{\pi}{5} \right) \cos \left( \frac{2\pi}{5} \right) \cos \left( \frac{4\pi}{5} \right) \cos \left( \frac{8\pi}{5} \right) \), we can follow these steps: ### Step 1: Rewrite the expression We start with the expression: \[ \cos \left( \frac{\pi}{5} \right) \cos \left( \frac{2\pi}{5} \right) \cos \left( \frac{4\pi}{5} \right) \cos \left( \frac{8\pi}{5} \right) \] ### Step 2: Use the identity We can use the identity \( 2 \sin x \cos x = \sin(2x) \). We will multiply and divide by \( 2 \sin \left( \frac{\pi}{5} \right) \): \[ = \frac{2 \sin \left( \frac{\pi}{5} \right) \cos \left( \frac{\pi}{5} \right) \cos \left( \frac{2\pi}{5} \right) \cos \left( \frac{4\pi}{5} \right) \cos \left( \frac{8\pi}{5} \right)}{2 \sin \left( \frac{\pi}{5} \right)} \] This gives us: \[ = \frac{\sin \left( \frac{2\pi}{5} \right) \cos \left( \frac{2\pi}{5} \right) \cos \left( \frac{4\pi}{5} \right) \cos \left( \frac{8\pi}{5} \right)}{2 \sin \left( \frac{\pi}{5} \right)} \] ### Step 3: Simplify further Now, we can apply the same identity again to \( \sin \left( \frac{2\pi}{5} \right) \): \[ = \frac{\sin \left( \frac{4\pi}{5} \right) \cos \left( \frac{4\pi}{5} \right) \cos \left( \frac{8\pi}{5} \right)}{4 \sin \left( \frac{\pi}{5} \right)} \] ### Step 4: Continue simplifying Next, we apply the identity to \( \sin \left( \frac{4\pi}{5} \right) \): \[ = \frac{\sin \left( \frac{8\pi}{5} \right) \cos \left( \frac{8\pi}{5} \right)}{8 \sin \left( \frac{\pi}{5} \right)} \] ### Step 5: Final simplification Now, we know that \( \sin \left( \frac{8\pi}{5} \right) = \sin \left( 3\pi + \frac{\pi}{5} \right) = -\sin \left( \frac{\pi}{5} \right) \): \[ = \frac{-\sin \left( \frac{\pi}{5} \right) \cos \left( \frac{8\pi}{5} \right)}{8 \sin \left( \frac{\pi}{5} \right)} \] The \( \sin \left( \frac{\pi}{5} \right) \) cancels out: \[ = \frac{-\cos \left( \frac{8\pi}{5} \right)}{8} \] ### Step 6: Evaluate \( \cos \left( \frac{8\pi}{5} \right) \) Since \( \cos \left( \frac{8\pi}{5} \right) = \cos \left( 3\pi + \frac{3\pi}{5} \right) = -\cos \left( \frac{3\pi}{5} \right) \): \[ = \frac{-(-\cos \left( \frac{2\pi}{5} \right))}{8} \] Thus, we have: \[ = \frac{1}{16} \] ### Final Answer The value of \( \cos \left( \frac{\pi}{5} \right) \cos \left( \frac{2\pi}{5} \right) \cos \left( \frac{4\pi}{5} \right) \cos \left( \frac{8\pi}{5} \right) \) is: \[ \boxed{-\frac{1}{16}} \]
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ML KHANNA-TRIGONOMETRY RATIOS AND IDENTITIES-PROBLEM SET (5) ( MULTIPLE CHOICE QUESTIONS)
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  2. The value of sin"" ( pi )/( 14) sin"" ( 3pi )/(14)sin"" (5pi)/( 14) is

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  3. The value of cos ( pi //5) cos ( 2pi //5) cos ( 4pi //5) cos ( 8pi //5...

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  4. sin 20^(@) sin 40^(@) sin 60^(@) sin 80^(@) =(3)/(16)

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  5. The value of cos 20^(@) cos 40^(@)cos 60^(@) cos 80^(@) is equal to

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  6. tan 20^(@)" " tan 40^(@)" " tan 60^(@)" " tan 80^(@)=

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  7. cos^6pi/9-33tan^4pi/9+27tan^2pi/9 is equal to 0 (b) sqrt(3) (c) 3 ...

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  8. Prove that"sin"(pi)/(14)"sin"(3pi)/(14)"sin"(5pi)/(14)"sin"(7pi)/(14)=...

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  9. The value of sin""(pi)/(14).sin""(3pi)/(14).sin""(5pi)/(14).sin""(7pi)...

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  10. sin 12^(@) sin 48^(@) sin 54^(@) is equal to

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  11. sin 6^(@) sin 42^(@) sin66^(@) sin 78^(@) =

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  12. Prove that cos 6^(@) cos 42^(@) cos 66^(@) cos 78^(@) =(1)/(16)

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  13. sin 36^(º)sin 72^(º) + 108^(º) sin 144^(º) =

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  14. cos "" (pi )/( 65) cos "" ( 2pi )/( 65)cos"" ( 4pi )/( 65) cos "" ( 8 ...

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  15. If A=tan6^(@) tan 42^(@) and B= cot 66^(@)cot 78^(@) then-

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  16. The value of tan 6^(@) tan 42^(@) tan 66^(@) tan 78^(@) is

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  17. cos 60^(@) cos 36^(@) cos 42^(@) cos 78^(@) is

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  18. Value of sin 47^(@) + sin 61^(@)- sin 11 ^(@)-sin 25 ^(@) is

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  19. The value of sin"" (pi )/( 18) sin"" ( 5pi )/(18) sin"" ( 7pi )/( 18) ...

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