Question

cho góc α thỏa mãn \(\dfrac{\pi}{2}\)<α<π và \(\sin\dfrac{\alpha}{2}\)= \(\dfrac{2}{\sqrt{5}}\) .Tính giá trị biểu thức A= \(\tan\left(\dfrac{\alpha}{2}-\dfrac{\pi}{4}\right)\)

Answer 1

điều kiện : \(\dfrac{\pi}{2}\) < α < \(\pi\) (1)

\(\sin^2\dfrac{\alpha}{2}+\cos^2\dfrac{\alpha}{2}=1\)

⇔ \(\left(\dfrac{2}{\sqrt{5}}\right)^2+\cos^2\dfrac{\alpha}{2}=1\)

⇒ \(\cos\dfrac{\alpha}{2}=\pm\dfrac{1}{\sqrt{5}}\)

Do (1) nên ta có \(\dfrac{\pi}{4}< \dfrac{\alpha}{2}< \dfrac{\pi}{2}\): \(\cos\dfrac{\alpha}{2}>0\) ⇒ \(\cos\dfrac{\alpha}{2}=\dfrac{1}{\sqrt{5}}\) ⇒ \(\tan\dfrac{\alpha}{2}=\dfrac{\sin\dfrac{\alpha}{2}}{\cos\dfrac{\alpha}{2}}=\dfrac{\dfrac{2}{\sqrt{5}}}{\dfrac{1}{\sqrt{5}}}=2\)

Khi đó ta có:

A = \(\dfrac{\tan\dfrac{\alpha}{2}-\tan\dfrac{\pi}{4}}{1+\tan\dfrac{\alpha}{2}.\tan\dfrac{\pi}{4}}\) = \(\dfrac{2-1}{1+2.1}\) =\(\dfrac{1}{3}\)

VẬY..............................