Giải SBT Toán 11 CTST Bài 2. Giá trị lượng giác của một góc lượng giác có đáp án

a) sin(‒1693°) = ‒sin(1693°)

= ‒sin(4.360° + 180° + 73°)

= sin73°

= cos(90° ‒ 73°) = cos17°.

b) $\text{cos}\frac{1003\pi }{3}=\text{cos}\left(334\pi +\frac{\pi }{3}\right)=\text{cos}\frac{\pi }{3}=\text{sin}\frac{\pi }{6}$

c) tan 885° = tan(180.4 + 165°) = tan165° = tan(180° ‒ 15°) = ‒tan15°.

d) $\cot \left(-\frac{53\pi }{10}\right)=-\cot \left(\frac{53\pi }{10}\right)=-\cot \left(\frac{50\pi }{10}+\frac{3\pi }{10}\right)$

$=-\cot \left(5\pi +\frac{3\pi }{10}\right)=-\cot \left(\frac{3\pi }{10}\right)$

$-\cot \left(\frac{\pi }{2}-\frac{\pi }{5}\right)=-\tan \left(\frac{\pi }{5}\right).$

Vì $\text{sin}\alpha =\frac{3}{5}$ và $\frac{\pi }{2}<\alpha <\pi$ nên $\text{cos}\alpha =-\frac{4}{5},\text{tan}\alpha =-\frac{3}{4}$ và $\text{cot}\alpha =-\frac{4}{3}$.

a) $A=\frac{3\cdot \frac{3}{5}}{2\cdot \left(-\frac{4}{5}\right)-\left(-\frac{3}{4}\right)}=\frac{\frac{9}{5}}{-\frac{8}{5}+\frac{3}{4}}=\frac{\frac{9}{5}}{-\frac{17}{20}}=-\frac{36}{17}.$

b) $B=\frac{{\left(-\frac{4}{3}\right)}^2-\frac{3}{5}}{-\frac{3}{4}+2\cdot \left(-\frac{4}{5}\right)}=\frac{\frac{16}{9}-\frac{3}{5}}{-\frac{3}{4}-\frac{8}{5}}=\frac{\frac{53}{45}}{-\frac{47}{20}}=-\frac{212}{423}.$

a) sin605° = sin(3.180° + 65°) = ‒sin65°.

sin1645° = sin(9.180° + 25°) = ‒sin25° = ‒sin(90° ‒ 65°) = ‒cos65°.

cot25° = cot(90° ‒ 65°) = tan65°.

sin2605° + sin21645° + cot225°

= (‒sin65°)2 + (‒cos65°)2 + (tan65°)2

= 1 + tan265°

$=\frac{1}{{\text{cos}}^2{65}^{\circ }}$

b) sin530° = sin(3.180° ‒ 10°) = sin10°.

sin640° = sin(4.180° ‒ 80°) = ‒sin80° = ‒sin(90° ‒ 10°) = ‒cos10°.

$\frac{\text{sin}530°}{1+\text{sin}640°}=\frac{\text{sin}10°}{1-\text{cos}10°}=\frac{{\text{sin}}^{2}10°}{\text{sin}10°\left(1-\text{cos}10°\right)}$

$=\frac{1-{\text{cos}}^{2}10°}{\text{sin}10°\left(1-\text{cos}10°\right)}=\frac{\left(1+\text{cos}10°\right)\left(1-\text{cos}10°\right)}{\text{sin}10°\left(1-\text{cos}10°\right)}$

$=\frac{1+\text{cos}10°}{\text{sin}10°}=\frac{111}{\text{sin}10°}+\text{cot}10°\text{ .}$