Question

Answer this question soon! Thanks in advance! ​

Answer 1

Required Answer:-

Given:

$\sf \dfrac{ \sin(x) }{ \sin({}^{x}/ _{8} ) } = \lambda \cos({}^{x}/_{8}) \cos({}^{x}/_{4}) \cos({}^{x}/_{2})$

To Find:

• The value of λ.

Answer:

• λ = 8

Solution:

We know that,

$\sf \implies \sin(2x) = 2 \sin(x) \cos(x)$

Therefore,

$\sf \implies \dfrac{ \sin(x) }{ \sin({}^{x}/ _{8} ) } = \lambda \cos({}^{x}/_{8}) \cos({}^{x}/_{4}) \cos({}^{x}/_{2})$

$\sf \implies \dfrac{ \sin(2 \times {}^{x}/_{2}) }{ \sin({}^{x}/ _{8} ) } = \lambda \cos({}^{x}/_{8}) \cos({}^{x}/_{4}) \cos({}^{x}/_{2})$

$\sf \implies \dfrac{ 2\sin({}^{x}/_{2}) \cos({}^{x}/_{2}) }{ \sin({}^{x}/ _{8} ) } = \lambda \cos({}^{x}/_{8}) \cos({}^{x}/_{4}) \cos({}^{x}/_{2})$

Cancelling out cos(x/2) from both sides, we get,

$\sf \implies \dfrac{ 2\sin({}^{x}/_{2})}{ \sin({}^{x}/ _{8} ) } = \lambda \cos({}^{x}/_{8}) \cos({}^{x}/_{4})$

Again, applying same formula, we get,

$\sf \implies \dfrac{ 2\sin(2 \times {}^{x}/_{4})}{ \sin({}^{x}/ _{8} ) } = \lambda \cos({}^{x}/_{8}) \cos({}^{x}/_{4})$

$\sf \implies \dfrac{ 2 \times 2\sin({}^{x}/_{4}) \cos({}^{x}/_{4}) }{ \sin({}^{x}/ _{8} ) } = \lambda \cos({}^{x}/_{8}) \cos({}^{x}/_{4})$

Cancelling out cos(x/4) from both sides, we get,

$\sf \implies \dfrac{4\sin({}^{x}/_{4})}{ \sin({}^{x}/ _{8} ) } = \lambda \cos({}^{x}/_{8})$

Applying same formula again, we get,

$\sf \implies \dfrac{4\sin(2 \times {}^{x}/_{8})}{ \sin({}^{x}/ _{8} ) } = \lambda \cos({}^{x}/_{8})$

$\sf \implies \dfrac{4 \times 2\sin({}^{x}/_{8}) \cos({}^{x}/ _{8}) }{ \sin({}^{x}/ _{8} ) } = \lambda \cos({}^{x}/_{8})$

$\sf \implies \dfrac{8 \cancel{\sin({}^{x}/_{8})} \cos({}^{x}/ _{8}) }{ \cancel{\sin({}^{x}/ _{8}} ) } = \lambda \cos({}^{x}/_{8})$

$\sf \implies 8 \cos({}^{x}/ _{8}) = \lambda \cos({}^{x}/_{8})$

$\sf \implies 8 = \lambda$

$\sf \implies \lambda = 8$

→ Hence, λ = 8.

Answer:

• λ = 8.

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