Giải bài tập 1.55 trang 28 SBT toán 11 tập 1 kết nối

a) $\frac{sin(45^{o}+\alpha)-cos(45^{o}+\alpha)}{ sin(45^{o}+\alpha)+cos(45^{o}+\alpha)}$

$=\frac{(sin45^{o}cos\alpha+cos45^{o}sin\alpha)-(cos45^{o}cos\alpha-sin45^{o}sin\alpha)}{(sin45^{o}cos\alpha+cos45^{o}sin\alpha)+(cos45^{o}cos\alpha -sin45^{o}sin\alpha)}$

$=\frac{(\frac{\sqrt{2}}{2}cos\alpha + \frac{\sqrt{2}}{2}sin\alpha)-(\frac{\sqrt{2}}{2}cos\alpha -\frac{\sqrt{2}}{2}sin\alpha)}{ (\frac{\sqrt{2}}{2}cos\alpha+ \frac{\sqrt{2}}{2}sin\alpha)+(\frac{\sqrt{2}}{2}cos\alpha -\frac{\sqrt{2}}{2}sin\alpha)}$

$=\frac{\sqrt{2}sin\alpha}{\sqrt{2}cos\alpha}$

$=tan\alpha$

b) $\frac{sin2\alpha + sin\alpha}{1+cos2\alpha + cos\alpha}$

$=\frac{2sin\alpha cos\alpha +sin\alpha}{1+(2cos^{2}\alpha-1)+cos\alpha}$

$=\frac{2sin\alpha(cos\alpha+1)}{2cos\alpha(cos\alpha+1)}$

$=\frac{sin\alpha}{cos\alpha}$

$=tan\alpha$

c)  $\frac{1+cos\alpha -sin\alpha}{1-cos\alpha-sin\alpha}$

$=\frac{1+(2cos^{2}\frac{\alpha}{2}-1)-2sin\frac{\alpha}{2}cos\frac{\alpha}{2}}{1-(1-2sin^{2}\frac{\alpha}{2})-2sin\frac{\alpha}{2}cos\frac{\alpha}{2}}{1-(1-2sin^{2}\frac{\alpha}{2})-2sin\frac{\alpha}{2}cos\frac{\alpha}{2}}$

$=\frac{2cos^{2}\frac{\alpha}{2}-2sin\frac{\alpha}{2}cos\frac{\alpha}{2}}{2sin^{2}\frac{\alpha}{2}-2sin\frac{\alpha}{2}cos\frac{\alpha}{2}}$

$=\frac{2cos\frac{\alpha}{2}(cos\frac{\alpha}{2}-sin\frac{\alpha}{2})}{2sin\frac{\alpha}{2}(sin\frac{\alpha}{2}-cos\frac{\alpha}{2})}$

$=\frac{cos\frac{\alpha}{2}}{-sin\frac{\alpha}{2}}$

$=-cot\frac{\alpha}{2}$

d) $\frac{sin\alpha + sin3\alpha + sin5\alpha}{cos\alpha + cos3\alpha + cos5\alpha}$

$=\frac{(sin5\alpha + sin\alpha)+sin3\alpha}{(cos5\alpha + cos\alpha)+cos3\alpha}$

$=\frac{2sin\frac{5\alpha+\alpha}{2}cos\frac{5\alpha-\alpha}{2}+sin3\alpha}{2cos\frac{5\alpha+\alpha}{2}cos\frac{5\alpha-\alpha}{2}+cos3\alpha}$

$=\frac{2sin3\alpha cos2\alpha+sin3\alpha}{2cos3\alpha cos2\alpha +cos3\alpha}$

$=\frac{sin3\alpha (2cos2\alpha + 1)}{cos3\alpha(2cos2\alpha + 1)}$

$=\frac{sin3\alpha}{cos3\alpha}$

$=tan3\alpha$