Câu hỏi của Bảo Ngọc

\(A=4-sin^245^0+2\cdot cos^260^0-3\cdot cot^345^0\)\(=4-\left(\dfrac{\sqrt{2}}{2}\right)^2+2\cdot\left(\dfrac{1}{2}\right)^2-3\cdot1^3\)\(=4-\dfrac{1}{2}+2\cdot\dfrac{1}{4}-3=1-\dfrac{1}{2}+\dfrac{1}{2}=1\)\(B=tan45\cdot cos30\cdot cot30=1\cdot cos30\cdot cot30\)\(=cos30\cdot\dfrac{cos30}{sin30}=\dfrac{cos^230^0}{sin30}=\left(\dfrac{\sqrt{3}}{2}\right)^2:\dfrac{1}{2}\)\(=\dfrac{3}{4}:\dfrac{1}{2}=\dfrac{3}{4}\cdot2=\dfrac{3}{2}\)\(C=sin15+sin75-cos15-cos75+sin30\)\(=\left(sin15-cos75\right)+\left(sin75-cos15\right)+sin30\)\(=\left(sin15-sin15\right)+\left(sin75-sin75\right)+\dfrac{1}{2}\)\(=\dfrac{1}{2}\)Câu trả lời: \(A=4-sin^245^0+2\cdot cos^260^0-3\cdot cot^345^0\)\(=4-\left(\dfrac{\sqrt{2}}{2}\right)^2+2\cdot\left(\dfrac{1}{2}\right)^2-3\cdot1^3\)\(=4-\dfrac{1}{2}+2\cdot\dfrac{1}{4}-3=1-\dfrac{1}{2}+\dfrac{1}{2}=1\)\(B=tan45\cdot cos30\cdot cot30=1\cdot cos30\cdot cot30\)\(=cos30\cdot\dfrac{cos30}{sin30}=\dfrac{cos^230^0}{sin30}=\left(\dfrac{\sqrt{3}}{2}\right)^2:\dfrac{1}{2}\)\(=\dfrac{3}{4}:\dfrac{1}{2}=\dfrac{3}{4}\cdot2=\dfrac{3}{2}\)\(C=sin15+sin75-cos15-cos75+sin30\)\(=\left(sin15-cos75\right)+\left(sin75-cos15\right)+sin30\)\(=\left(sin15-sin15\right)+\left(sin75-sin75\right)+\dfrac{1}{2}\)\(=\dfrac{1}{2}\)