Bài tập nâng cao · Bài 15
Bài tập nâng cao Giá trị lượng giác của góc lượng giác · Bài 15
a) Cho $\cos\alpha=\frac{2}{3}$. Tính $A=\frac{\tan\alpha+3\cot\alpha}{\tan\alpha+\cot\alpha}$.
b) Cho $\tan\alpha=3$. Tính $B=\frac{\sin\alpha-\cos\alpha}{\sin^{3}\alpha+3\cos^{3}\alpha+2\sin\alpha}$
c) Cho $\cot\alpha=\sqrt{5}$. Tính $C=\sin^{2}\alpha-\sin\alpha\cos\alpha+\cos^{2}\alpha$
b) Cho $\tan\alpha=3$. Tính $B=\frac{\sin\alpha-\cos\alpha}{\sin^{3}\alpha+3\cos^{3}\alpha+2\sin\alpha}$
c) Cho $\cot\alpha=\sqrt{5}$. Tính $C=\sin^{2}\alpha-\sin\alpha\cos\alpha+\cos^{2}\alpha$
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Lời giải
a) Ta có $A=\frac{\tan\alpha+3\frac{1}{\tan\alpha}}{\tan\alpha+\frac{1}{\tan\alpha}}=\frac{\tan^{2}\alpha+3}{\tan^{2}\alpha+1}=\frac{\frac{1}{\cos^{2}\alpha}+2}{\frac{1}{\cos^{2}\alpha}}=1+2\cos^{2}\alpha$
Suy ra $A=1+2.\frac{4}{9}=\frac{17}{9}$
b) $B=\frac{\frac{\sin\alpha}{\cos^{3}\alpha}-\frac{\cos\alpha}{\cos^{3}\alpha}}{\frac{\sin^{3}\alpha}{\cos^{3}\alpha}+\frac{3\cos^{3}\alpha}{\cos^{3}\alpha}+\frac{2\sin\alpha}{\cos^{3}\alpha}}=\frac{\tan\alpha\left( \tan^{2}\alpha+1 \right)-\left( \tan^{2}\alpha+1 \right)}{\tan^{3}\alpha+3+2\tan\alpha\left( \tan^{2}\alpha+1 \right)}$
Suy ra $B=\frac{3\left( 9+1 \right)-\left( 9+1 \right)}{27+3+2.3\left( 9+1 \right)}=\frac{2}{9}$
c) Ta có $C=\sin^{2}\alpha.\frac{\sin^{2}\alpha-\sin\alpha\cos\alpha+\cos^{2}\alpha}{\sin^{2}\alpha}=\sin^{2}\alpha\left( 1-\frac{\cos\alpha}{\sin\alpha}+\frac{\cos^{2}\alpha}{\sin^{2}\alpha} \right)$
$=\frac{1}{1+\cot^{2}\alpha}\left( 1-\cot\alpha+\cot^{2}\alpha \right)=\frac{1}{1+\left( \sqrt{5} \right)^{2}}\left( 1-\sqrt{5}+5 \right)=\frac{6-\sqrt{5}}{6}$
Suy ra $A=1+2.\frac{4}{9}=\frac{17}{9}$
b) $B=\frac{\frac{\sin\alpha}{\cos^{3}\alpha}-\frac{\cos\alpha}{\cos^{3}\alpha}}{\frac{\sin^{3}\alpha}{\cos^{3}\alpha}+\frac{3\cos^{3}\alpha}{\cos^{3}\alpha}+\frac{2\sin\alpha}{\cos^{3}\alpha}}=\frac{\tan\alpha\left( \tan^{2}\alpha+1 \right)-\left( \tan^{2}\alpha+1 \right)}{\tan^{3}\alpha+3+2\tan\alpha\left( \tan^{2}\alpha+1 \right)}$
Suy ra $B=\frac{3\left( 9+1 \right)-\left( 9+1 \right)}{27+3+2.3\left( 9+1 \right)}=\frac{2}{9}$
c) Ta có $C=\sin^{2}\alpha.\frac{\sin^{2}\alpha-\sin\alpha\cos\alpha+\cos^{2}\alpha}{\sin^{2}\alpha}=\sin^{2}\alpha\left( 1-\frac{\cos\alpha}{\sin\alpha}+\frac{\cos^{2}\alpha}{\sin^{2}\alpha} \right)$
$=\frac{1}{1+\cot^{2}\alpha}\left( 1-\cot\alpha+\cot^{2}\alpha \right)=\frac{1}{1+\left( \sqrt{5} \right)^{2}}\left( 1-\sqrt{5}+5 \right)=\frac{6-\sqrt{5}}{6}$