Logarithm + Trigonometry + Products

$\Rightarrow \log _{ \frac { 1 }{ 2 } }{ \left( \sin({ 1 }^{ \circ })\times \sin(3^{ \circ })\times \sin(5^{ \circ })\times \sin(7^{ \circ })...............\sin(89^{ \circ }) \right) } \\ \Longrightarrow \log _{ \frac { 1 }{ 2 } }{ \left( \frac { \sin({ 1 }^{ \circ })\times \sin(2^{ \circ })\times \sin(3^{ \circ })\times \sin(4^{ \circ })\times \sin(5^{ \circ })...............\sin(89^{ \circ }) }{ \sin({ 2 }^{ \circ })\times \sin(4^{ \circ })\times \sin({ 6 }^{ \circ })....................\sin(88^{ \circ }) } \right) } \\ \Longrightarrow \log _{ \frac { 1 }{ 2 } }{ \left( \frac { \sin({ 1 }^{ \circ })\times \sin(2^{ \circ })\times \sin(3^{ \circ })...............\sin(89^{ \circ }) }{ { 2 }^{ 44 }\times \sin({ 1 }^{ \circ })\times \cos(1^{ \circ })\times \sin(2^{ \circ })\times \cos({ 2 }^{ \circ })\times \sin(3^{ \circ })\times \cos(3^{ \circ })\times ....................\sin(44^{ \circ })\times \cos(44^{ \circ }) } \right) } \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad [\because \quad \sin(2x)=2\sin(x)\cos(x)]\\ \\ \Longrightarrow \log _{ \frac { 1 }{ 2 } }{ \left( \frac { \sin({ 1 }^{ \circ })\times \sin(2^{ \circ })\times \sin(3^{ \circ })\times \sin(4^{ \circ })\times \sin(5^{ \circ })\times \sin(6^{ \circ })\times \sin(7^{ \circ })...............\cos(3^{ \circ })\times \cos(2^{ \circ })\times \cos(1^{ \circ }) }{ { 2 }^{ 44 }\times \sin({ 1 }^{ \circ })\times \cos(1^{ \circ })\times \sin(2^{ \circ })\times \cos({ 2 }^{ \circ })\times \sin(3^{ \circ })\times \cos(3^{ \circ })\times ....................\sin(44^{ \circ })\times \cos(44^{ \circ }) } \right) } \quad \\ \\ \quad \quad \quad \quad [\because \quad \sin(90-x)=\cos(x)]\\ \\ \Longrightarrow \quad all\quad terms\quad would\quad cancel\quad except\quad \rightarrow \quad \sin(45^{ \circ })\\ \Longrightarrow \log _{ \frac { 1 }{ 2 } }{ \left( \frac { \sin(45^{ \circ }) }{ { 2 }^{ 44 } } \right) } \Longrightarrow \log _{ \frac { 1 }{ 2 } }{ \left( \frac { 1 }{ { 2 }^{ 44 }\sqrt { 2 } } \right) } \Longrightarrow \log _{ \frac { 1 }{ 2 } }{ { \left( \frac { 1 }{ 2 } \right) }^{ 44.5 }=\quad 44.5 } \\$

Apply the double angle formula $\sin(2A) = 2\sin(A) \cos(A)$ , the product equals to

$\begin{aligned} && \sin(1^\circ) \sin(3^\circ) \sin(5^\circ)\times \cdots \times \sin(89^\circ) \\ &= & \bigg [\sin(1^\circ) \sin(89^\circ) \bigg] \times \bigg[\sin(3^\circ) \times \sin(87^\circ) \bigg ] \times \cdots \times \bigg[\sin(43^\circ) \times \sin(47^\circ) \bigg ] \times \sin(45^\circ) \\ &= & \bigg [\sin(1^\circ) \cos(1^\circ) \bigg] \times \bigg[\sin(3^\circ) \times \cos(3^\circ) \bigg ] \times \cdots \times \bigg[\sin(43^\circ) \times \cos(43^\circ) \bigg ] \times 2^{-1/2} \\ &= & \frac12 \sin(2^\circ) \times \frac12 \sin(6^\circ)\times \cdots \times \frac12 \sin(86^\circ) \times 2^{-1/2} \\ &= & \left(\frac12\right)^{22} \cdot 2^{-1/2} \times \bigg[ \sin(2^\circ) \times\sin(6^\circ)\times \cdots \times\sin(86^\circ) \bigg ] \\ &= & 2^{-22.5} \times \bigg[ \cos(88^\circ) \times \cos(84^\circ)\times \cdots \times \cos(4^\circ) \bigg ] \\ \end{aligned}$

Let $C = \cos(88^\circ) \times \cos(84^\circ)\times \cdots \times \cos(4^\circ)$ , then $C> 0$ because it is the product of all positive terms.

Consider the equation $\cos(90x) = 1$ , then $90x = n\cdot 360^\circ$ for integer $n$ , which means that $x = 4^\circ, 8^\circ, 12^\circ, \ldots, 360^\circ$ satisfy the former equation.

Recall the properties of Chebyshev polynomials, the leading coefficient of polynomial $\cos(90x)$ in terms of $\cos(x)$ is $2^{89}$ and because $90$ is an even number not divisible by $4$ , the constant of this same polynomial is equals to $-1$ . Thus

$\begin{aligned} 1 &=& \cos(90x) \\ &=& 2^{89} \cos^{90}(x) - \ldots -1 \\ 0 &=& 2^{89} \cos^{90}(x) - \ldots - 2 \end{aligned}$

By Vieta's formula, and apply these formulae: $\cos(A) = -\cos(180^\circ -A) = -\cos(A - 180^\circ) = \cos(360^\circ - A)$ , we get

$\begin{aligned} \cos(4^\circ) \cos(8^\circ) \cos(12^\circ) \ldots \cos(360^\circ) &=& -\frac2{2^{89}} \\ \bigg[\cos(4^\circ) \cos(8^\circ) \cos(12^\circ) \dots \cos(88^\circ) \bigg]^4 \cdot \cos(180^\circ) \cos(360^\circ) &=& -\frac1{2^{88}} \\ C^4 \cdot -1 \cdot 1 &=& -\frac1{2^{88}} \\ C &=& \frac1{2^{22}} \\ \end{aligned}$

Hence, putting it all together, the product equals to $2^{-22.5} \cdot 2^{-22} = 2^{-44.5} = \left( \frac12\right)^{44.5}$ . Lastly, apply the log,

$\large \log_{\frac12} \left( \frac12\right)^{44.5} = 44.5 \cancel{\log_{\frac12} \frac12 }= \boxed{44.5}$