a) $cos\alpha=0,8$
Ta có: $sin^2\alpha+cos^2\alpha=1$
→ $sin^2\alpha+(0,8)^2=1$
→ $sin^2\alpha=\dfrac{9}{25}$
→ \(\left[ \begin{array}{l}sin\alpha=\dfrac{3}{5}\\sin\alpha=-\dfrac{3}{5}\end{array} \right.\)
+) Với giá trị $sin\alpha=\dfrac{3}{5}$
→ $tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{3}{5}÷0,8=\dfrac{3}{4}$
→ $cot\alpha=\dfrac{1}{tan\alpha}=1÷\dfrac{3}{4}=\dfrac{4}{3}$
+) Với giá trị $sin\alpha=-\dfrac{3}{5}$
→ $tan\alpha=\dfrac{sin\alpha}{cos\alpha}=-\dfrac{3}{5}÷0,8=-\dfrac{3}{4}$
→ $cot\alpha=\dfrac{1}{tan\alpha}=1÷(-\dfrac{3}{4})=\dfrac{4}{3}$
b) $tan\alpha=\dfrac{3}{4}$
→ $cot\alpha=\dfrac{4}{3}$
Ta có: $1+tan^2\alpha=\dfrac{1}{cos^2\alpha}$
⇒ $1+(\dfrac{3}{4})^2=\dfrac{1}{cos^2\alpha}$
⇔ $cos^2\alpha=\dfrac{16}{25}$
⇔ \(\left[ \begin{array}{l}cos\alpha=\dfrac{4}{5}\\cos\alpha=-\dfrac{4}{5}\end{array} \right.\)
+) Với giá trị $cos\alpha=\dfrac{4}{5}$
→ $sin\alpha=tan\alpha×cos\alpha=\dfrac{3}{4}×\dfrac{4}{5}=\dfrac{3}{5}$
+) Với giá trị $cos\alpha=-\dfrac{4}{5}$
→ $sin\alpha=tan\alpha×cos\alpha=\dfrac{3}{4}×\dfrac{4}{5}=-\dfrac{3}{5}$
c) $cot\alpha=\dfrac{\sqrt{7}}{3}$
→ $tan\alpha=\dfrac{3\sqrt{7}}{7}$
Ta có: $1+cot^2\alpha=\dfrac{1}{sin^2\alpha}$
⇒ $1+(\dfrac{\sqrt{7}}{3})^2=\dfrac{1}{sin^2\alpha}$
⇔ $sin^2\alpha=\dfrac{9}{16}$
⇔ \(\left[ \begin{array}{l}sin\alpha=\dfrac{3}{4}\\sin\alpha=-\dfrac{3}{4}\end{array} \right.\)
+) Với giá trị $sin\alpha=\dfrac{3}{4}$
→ $cos\alpha=cot\alpha×sin\alpha=\dfrac{\sqrt{7}}{3}×\dfrac{3}{4}=\dfrac{\sqrt{7}}{4}$
+) Với giá trị $sin\alpha=-\dfrac{3}{4}$
→ $cos\alpha=cot\alpha×sin\alpha=\dfrac{\sqrt{7}}{3}×(-\dfrac{3}{4})=-\dfrac{\sqrt{7}}{4}$
