Question

Cho tan a=2.Hãy tính giá trị của biểu thức sau 

a)A=sin a+cos a/sin a-cos a b)B=sin^2 a+2.sin a cos a-3.cos^3 a

Answer 1

.

Answer 2

\(tan\alpha=2=>cos\alpha=\sqrt{\dfrac{1}{1+tan^2\alpha}}=\sqrt{\dfrac{1}{1+2^2}}=\dfrac{\sqrt{5}}{5}\)

\(=>sin\alpha=\sqrt{1-\left(\dfrac{\sqrt{5}}{5}\right)^2}=\dfrac{2\sqrt{5}}{5}\)

\(a.A=\dfrac{sin\alpha+cos\alpha}{sin\alpha-cos\alpha}=\dfrac{\dfrac{2\sqrt{5}}{5}+\dfrac{\sqrt{5}}{5}}{\dfrac{2\sqrt{5}}{5}-\dfrac{\sqrt{5}}{5}}=3\\ b.B=sin^2\alpha+2sin\alpha\cdot cos\alpha-3cos^3\alpha\\ =\left(\dfrac{2\sqrt{5}}{5}\right)^2+2\cdot\dfrac{2\sqrt{5}}{5}\cdot\dfrac{\sqrt{5}}{5}-3\cdot\left(\dfrac{\sqrt{5}}{5}\right)^3\\ =\dfrac{40-3\sqrt{5}}{25}\)

Answer 3

Ta có :

\(\tan^2a+1=\dfrac{1}{\cos^2a}\)

\(\Rightarrow\cos^2a=\dfrac{1}{\tan^2a+1}=\dfrac{1}{4+1}=\dfrac{1}{5}\)

\(\Rightarrow\cos a=\dfrac{\sqrt[]{5}}{5}\left(\tan a=2>0\right)\)

\(\Rightarrow\sin a=\tan a.\cos a=2.\dfrac{\sqrt[]{5}}{5}=\dfrac{2\sqrt[]{5}}{5}\left(\tan a=\dfrac{\sin a}{\cos a}\right)\)

a) \(A=\dfrac{\sin a+\cos a}{\sin a-\cos a}=\dfrac{\dfrac{2\sqrt[]{5}}{5}+\dfrac{\sqrt[]{5}}{5}}{\dfrac{2\sqrt[]{5}}{5}-\dfrac{\sqrt[]{5}}{5}}=\dfrac{\dfrac{3\sqrt[]{5}}{5}}{\dfrac{\sqrt[]{5}}{5}}=3\)

b) \(B=\sin^2a+2\sin a.\cos a-3\cos^3a\)

\(\Rightarrow B=\left(\dfrac{2\sqrt[]{5}}{5}\right)^2+2.\dfrac{2\sqrt[]{5}}{5}.\dfrac{\sqrt[]{5}}{5}-3.\left(\dfrac{\sqrt[]{5}}{5}\right)^3\)

\(\Rightarrow B=\dfrac{4}{5}+\dfrac{4}{5}-\dfrac{3}{5\sqrt[]{5}}=\dfrac{8\sqrt[]{5}-3}{5\sqrt[]{5}}\)