Question

Rút gọn các biểu thức :

a) \(\dfrac{\sin2\alpha+\sin\alpha}{1+\cos2\alpha+\cos\alpha}\)

b) \(\dfrac{4\sin^2\alpha}{1-\cos^2\dfrac{\alpha}{2}}\)

c) \(\dfrac{1+\cos\alpha-\sin\alpha}{1-\cos\alpha-\sin\alpha}\)

d) \(\dfrac{1+\sin\alpha-2\sin^2\left(45^0-\dfrac{\alpha}{2}\right)}{4\cos\dfrac{\alpha}{2}}\)

Answer 1

a) \(\dfrac{\sin2\text{a}+\cos a}{1+\cos2\text{a}+\cos a}=2\tan a\)

Answer 2

a) \(\dfrac{sin2\alpha+sin\alpha}{1+cos2\alpha+cos\alpha}=\dfrac{2sin\alpha cos\alpha+sin\alpha}{2cos^2\alpha+cos\alpha}\)\(=\dfrac{sin\alpha\left(2cos\alpha+1\right)}{cos\alpha\left(2cos\alpha+1\right)}=\dfrac{sin\alpha}{cos\alpha}=tan\alpha\).

Answer 3

b) \(\dfrac{4sin^2\alpha}{1-cos^2\dfrac{\alpha}{2}}=\dfrac{4sin^2\alpha}{sin^2\dfrac{\alpha}{2}}=\dfrac{4.sin^2\dfrac{\alpha}{2}.cos^2\dfrac{\alpha}{2}}{sin^2\dfrac{\alpha}{2}}=4sin^2\dfrac{\alpha}{2}\).

Answer 4

c) \(\dfrac{1+cos\alpha-sin\alpha}{1-cos\alpha-sin\alpha}=\dfrac{2cos^2\dfrac{\alpha}{2}-2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}}{2sin^2\dfrac{\alpha}{2}-2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}}\)
\(=\dfrac{2cos\dfrac{\alpha}{2}\left(cos\dfrac{\alpha}{2}-sin\dfrac{\alpha}{2}\right)}{2sin\dfrac{\alpha}{2}\left(cos\dfrac{\alpha}{2}-sin\dfrac{\alpha}{2}\right)}\)
\(=\dfrac{cos\dfrac{\alpha}{2}}{sin\dfrac{\alpha}{2}}=cot\dfrac{\alpha}{2}\).

Answer 5

d) \(\dfrac{1+sin\alpha-2sin^2\left(45^o-\dfrac{\alpha}{2}\right)}{4cos\dfrac{\alpha}{2}}\)\(=\dfrac{\left(1-2sin^2\left(45^o-\dfrac{\alpha}{2}\right)\right)+sin\alpha}{4cos\dfrac{\alpha}{2}}\)\(=\dfrac{cos\left(90^o-\alpha\right)+sin\alpha}{4cos\dfrac{\alpha}{2}}=\dfrac{sin\alpha+sin\alpha}{4cos\dfrac{\alpha}{2}}\)\(=\dfrac{2sin\alpha}{4cos\dfrac{\alpha}{2}}=\dfrac{1}{2sin\dfrac{\alpha}{2}}.\)