Trigonometri

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Şimdiden teşekkür ederim.

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Tamamdır hallettim.

[stuart clark]

Let $A=\cos(1^{\circ})\cdot \cos(3^{\circ})\cdots \cdots \cdots \cos(89^{\circ})$

So $A=\sin(1^{\circ})\cdot \sin(3^{\circ})\cdots \cdots \cdots \sin(89^{\circ})$

So $B=\sin(2^{\circ})\cdot \sin(4^{\circ})\cdots \cdots \cdots \sin(88^{\circ})$

So $\displaystyle AB=\frac{1}{2^{44}}\bigg[\sin (2^{\circ})\cdot \sin (4^{\circ})\cdots \cdots \sin (178^{\circ}) \bigg]$

So $AB=\frac{1}{2^{44}}\bigg[\sin(2^{\circ})\cdot \sin(4^{\circ})\cdots \cdots \cdots \sin(88^{\circ})\bigg]\frac{1}{\sqrt{2}}=\frac{B}{\sqrt{2}}$

So $\displaystyle B\bigg(A-\frac{1}{\sqrt{2}}\frac{1}{2^{44}}\bigg)=0$

So $\displaystyle A = \frac{1}{\sqrt{2}}\cdot \frac{1}{2^{44}},$ bcz $B\neq 0$