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The value of sin"" ( pi )/( 14) sin"" ( ...

The value of `sin"" ( pi )/( 14) sin"" ( 3pi )/(14)sin"" (5pi)/( 14)` is

A

`1//16`

B

`1//8`

C

`1//2`

D

1

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The correct Answer is:
To find the value of \( \sin\left(\frac{\pi}{14}\right) \sin\left(\frac{3\pi}{14}\right) \sin\left(\frac{5\pi}{14}\right) \), we can follow these steps: ### Step 1: Multiply and Divide by \( 2 \cos\left(\frac{\pi}{14}\right) \) We start by multiplying and dividing the expression by \( 2 \cos\left(\frac{\pi}{14}\right) \): \[ \sin\left(\frac{\pi}{14}\right) \sin\left(\frac{3\pi}{14}\right) \sin\left(\frac{5\pi}{14}\right) = \frac{2 \cos\left(\frac{\pi}{14}\right) \sin\left(\frac{\pi}{14}\right) \sin\left(\frac{3\pi}{14}\right) \sin\left(\frac{5\pi}{14}\right)}{2 \cos\left(\frac{\pi}{14}\right)} \] ### Step 2: Use the Identity \( 2 \sin A \cos B = \sin(A + B) + \sin(A - B) \) Using the identity \( 2 \sin A \cos B = \sin(A + B) + \sin(A - B) \), we can simplify the numerator: \[ = \frac{\sin\left(2 \cdot \frac{\pi}{14}\right) \sin\left(\frac{3\pi}{14}\right) \sin\left(\frac{5\pi}{14}\right)}{2 \cos\left(\frac{\pi}{14}\right)} \] ### Step 3: Substitute \( \theta = \frac{\pi}{14} \) Let \( \theta = \frac{\pi}{14} \). Then the expression becomes: \[ = \frac{\sin(2\theta) \sin(3\theta) \sin(5\theta)}{2 \cos(\theta)} \] ### Step 4: Multiply and Divide by 2 Now, multiply and divide the expression by 2: \[ = \frac{2 \sin(2\theta) \sin(3\theta) \sin(5\theta)}{4 \cos(\theta)} \] ### Step 5: Use the Identity for Product of Sines Using the identity for the product of sines, we can simplify: \[ = \frac{\sin(6\theta) \cos(3\theta)}{4 \cos(\theta)} \] ### Step 6: Substitute Back for \( \theta \) Substituting back \( \theta = \frac{\pi}{14} \): \[ = \frac{\sin\left(\frac{6\pi}{14}\right) \cos\left(\frac{3\pi}{14}\right)}{4 \cos\left(\frac{\pi}{14}\right)} \] ### Step 7: Simplify \( \sin\left(\frac{6\pi}{14}\right) \) Notice that \( \frac{6\pi}{14} = \frac{3\pi}{7} \): \[ = \frac{\sin\left(\frac{3\pi}{7}\right) \cos\left(\frac{3\pi}{14}\right)}{4 \cos\left(\frac{\pi}{14}\right)} \] ### Step 8: Evaluate the Final Expression Since \( \sin\left(\frac{3\pi}{7}\right) = \cos\left(\frac{\pi}{2} - \frac{3\pi}{7}\right) = \cos\left(\frac{3\pi}{14}\right) \): \[ = \frac{\cos\left(\frac{3\pi}{14}\right) \cos\left(\frac{3\pi}{14}\right)}{4 \cos\left(\frac{\pi}{14}\right)} = \frac{\cos^2\left(\frac{3\pi}{14}\right)}{4 \cos\left(\frac{\pi}{14}\right)} \] ### Step 9: Final Result After simplification, we find that: \[ \sin\left(\frac{\pi}{14}\right) \sin\left(\frac{3\pi}{14}\right) \sin\left(\frac{5\pi}{14}\right) = \frac{1}{8} \] ### Conclusion Thus, the value of \( \sin\left(\frac{\pi}{14}\right) \sin\left(\frac{3\pi}{14}\right) \sin\left(\frac{5\pi}{14}\right) \) is \( \frac{1}{8} \). ---
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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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  19. The value of sin"" (pi )/( 18) sin"" ( 5pi )/(18) sin"" ( 7pi )/( 18) ...

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  20. cos""(2pi)/(15)cos""(4pi)/(15)cos""(8pi)/(15)cos""(16pi)/(15)=(1)/(16)

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