Bài tự luận · Bài 9
Dạng 1. Tính giá trị của các biểu thức chứa giá trị lượng giác
Bài tự luận Công thức lượng giác · Bài 9
(CTST11) Cho $\cos\alpha=\frac{11}{61}$ và $-\frac{\pi}{2}<\alpha<0$, tính giá trị của các biểu thức sau:
a) $\sin\left( \frac{\pi}{6}-\alpha \right)$
b) $\cot\left( \alpha+\frac{\pi}{4} \right)$
c) $\cos\left( 2\alpha+\frac{\pi}{3} \right)$
d) $\tan\left( \frac{3\pi}{4}-2\alpha \right)$
a) $\sin\left( \frac{\pi}{6}-\alpha \right)$
b) $\cot\left( \alpha+\frac{\pi}{4} \right)$
c) $\cos\left( 2\alpha+\frac{\pi}{3} \right)$
d) $\tan\left( \frac{3\pi}{4}-2\alpha \right)$
Xem lời giải
Lời giải
a) Vì $-\frac{\pi}{2}<\alpha<0$ nên $\sin\alpha<0$.
Do đó, $\sin\alpha=-\sqrt{1-\cos^{2}\alpha}=-\sqrt{1-\left( \frac{11}{61} \right)^{2}}=-\frac{60}{61}$.
Suy ra
$\sin\left( \frac{\pi}{6}-\alpha \right)=\sin\frac{\pi}{6}\cos\alpha-\cos\frac{\pi}{6}\sin\alpha=\frac{1}{2}\cdot \frac{11}{61}-\frac{\sqrt{3}}{2}\cdot \left( -\frac{60}{61} \right)=\frac{11+60\sqrt{3}}{122}.$
b) Ta có $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{-\frac{60}{61}}{\frac{11}{61}}=-\frac{60}{11}$.
Do đó $\cot\left( \alpha+\frac{\pi}{4} \right)=\frac{1}{\tan\left( \alpha+\frac{\pi}{4} \right)}=\frac{1-\tan\alpha\tan\frac{\pi}{4}}{\tan\alpha+\tan\frac{\pi}{4}}=\frac{1-\left( -\frac{60}{11} \right)\cdot 1}{\left( -\frac{60}{11} \right)+1}=-\frac{71}{49}$.
c) Ta có $\cos2\alpha=2\cos^{2}\alpha-1=2\cdot \left( \frac{11}{61} \right)^{2}-1=-\frac{3479}{3721}$;
$\sin2\alpha=2\sin\alpha\cos\alpha=2\cdot \left( -\frac{60}{61} \right)\cdot \frac{11}{61}=-\frac{1320}{3721}$. Suy ra $\cos\left( 2\alpha+\frac{\pi}{3} \right)=\cos2\alpha\cos\frac{\pi}{3}-\sin2\alpha\sin\frac{\pi}{3}=-\frac{3479}{3721}\cdot \frac{1}{2}-\left( -\frac{1320}{3721} \right)\cdot \frac{\sqrt{3}}{2}$
$=\frac{-3479+1320\sqrt{3}}{7442}.$
d) Ta có $\tan2\alpha=\frac{\sin2\alpha}{\cos2\alpha}=\frac{-\frac{1320}{3721}}{-\frac{3479}{3721}}=\frac{1320}{3479}$.
Suy ra $\tan\left( \frac{3\pi}{4}-2\alpha \right)=\frac{\tan\frac{3\pi}{4}-\tan2\alpha}{1+\tan\frac{3\pi}{4}\tan2\alpha}=\frac{-1-\frac{1320}{3479}}{1+(-1)\cdot \frac{1320}{3479}}=-\frac{4799}{2159}$.
Do đó, $\sin\alpha=-\sqrt{1-\cos^{2}\alpha}=-\sqrt{1-\left( \frac{11}{61} \right)^{2}}=-\frac{60}{61}$.
Suy ra
$\sin\left( \frac{\pi}{6}-\alpha \right)=\sin\frac{\pi}{6}\cos\alpha-\cos\frac{\pi}{6}\sin\alpha=\frac{1}{2}\cdot \frac{11}{61}-\frac{\sqrt{3}}{2}\cdot \left( -\frac{60}{61} \right)=\frac{11+60\sqrt{3}}{122}.$
b) Ta có $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{-\frac{60}{61}}{\frac{11}{61}}=-\frac{60}{11}$.
Do đó $\cot\left( \alpha+\frac{\pi}{4} \right)=\frac{1}{\tan\left( \alpha+\frac{\pi}{4} \right)}=\frac{1-\tan\alpha\tan\frac{\pi}{4}}{\tan\alpha+\tan\frac{\pi}{4}}=\frac{1-\left( -\frac{60}{11} \right)\cdot 1}{\left( -\frac{60}{11} \right)+1}=-\frac{71}{49}$.
c) Ta có $\cos2\alpha=2\cos^{2}\alpha-1=2\cdot \left( \frac{11}{61} \right)^{2}-1=-\frac{3479}{3721}$;
$\sin2\alpha=2\sin\alpha\cos\alpha=2\cdot \left( -\frac{60}{61} \right)\cdot \frac{11}{61}=-\frac{1320}{3721}$. Suy ra $\cos\left( 2\alpha+\frac{\pi}{3} \right)=\cos2\alpha\cos\frac{\pi}{3}-\sin2\alpha\sin\frac{\pi}{3}=-\frac{3479}{3721}\cdot \frac{1}{2}-\left( -\frac{1320}{3721} \right)\cdot \frac{\sqrt{3}}{2}$
$=\frac{-3479+1320\sqrt{3}}{7442}.$
d) Ta có $\tan2\alpha=\frac{\sin2\alpha}{\cos2\alpha}=\frac{-\frac{1320}{3721}}{-\frac{3479}{3721}}=\frac{1320}{3479}$.
Suy ra $\tan\left( \frac{3\pi}{4}-2\alpha \right)=\frac{\tan\frac{3\pi}{4}-\tan2\alpha}{1+\tan\frac{3\pi}{4}\tan2\alpha}=\frac{-1-\frac{1320}{3479}}{1+(-1)\cdot \frac{1320}{3479}}=-\frac{4799}{2159}$.