find the exact value of the expression given below. \ncos(-105°)\ncos(-105°)=□\n(simplify your answer…

Explanation:

Step1: Use the even - function property of cosine

Since (\cos(-\alpha)=\cos\alpha), then (\cos(- 105^{\circ})=\cos(105^{\circ})).

Step2: Rewrite (105^{\circ}) as a sum of two special angles

(105^{\circ}=60^{\circ}+45^{\circ}). Then, by the cosine of a sum formula (\cos(A + B)=\cos A\cos B-\sin A\sin B), where (A = 60^{\circ}) and (B=45^{\circ}). We know that (\cos60^{\circ}=\frac{1}{2}), (\cos45^{\circ}=\frac{\sqrt{2}}{2}), (\sin60^{\circ}=\frac{\sqrt{3}}{2}), and (\sin45^{\circ}=\frac{\sqrt{2}}{2}).

Step3: Substitute the values into the formula

[ \begin{align*} \cos(60^{\circ}+45^{\circ})&=\cos60^{\circ}\cos45^{\circ}-\sin60^{\circ}\sin45^{\circ}\ &=\frac{1}{2}\times\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}\ &=\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4}\ &=\frac{\sqrt{2}-\sqrt{6}}{4} \end{align*} ]

Answer:

(\frac{\sqrt{2}-\sqrt{6}}{4})