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°.

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

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

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

a) Ta có:

sin197° = sin(180° + 17°) = ‒sin17°.

sin73° = sin(90° ‒ 17°) = cos17°.

cos163° = cos(180° ‒ 17°) = ‒cos17°.

Suy ra:

sin 17°sin197° + sin73°cos163°

= sin 17°.(‒sin17°) + cos17°.(‒cos17°)

= ‒(sin²17° + cos²17°) = ‒1.

a) tan³α + cot³α = (tanα + cotα)³ ‒ 3tanαcotα(tanα + cotα)

= (tanα + cotα)³ ‒ 3 (tanα + cotα) (*)

Thay tanα + cotα = 2 vào biểu thức (*) ta có: 2³ ‒ 3.2 = 2.

b) (sinα + cosα)² = sin²α + cos²α + 2 sinαcosα = 1 + 2 sinαcosα.

Do đó $\text{sin}\alpha \text{cos}\alpha =\frac{1}{2}\left[{(\text{sin}\alpha +\text{cos}\alpha )}^2-1\right]=\frac{1}{2}\left[{\left(\frac{1}{4}\right)}^2-1\right]=-\frac{15}{32}$.

c) sin³α + cos³α

= (sinα + cosα)(sin²α ‒ sinαcosα + cos²α)

= (sinα + cosα)(1 ‒ sinαcosα)

Mà $\text{sin}\alpha \text{cos}\alpha =\frac{1}{2}\left[{(\text{sin}\alpha +\text{cos}\alpha )}^2-1\right]=\frac{1}{2}\left[{\left(\frac{1}{2}\right)}^2-1\right]=-\frac{3}{8}$, nên

${\text{sin}}^3\alpha +{\text{cos}}^3\alpha =\frac{1}{2}\cdot \left[1-\left(-\frac{3}{8}\right)\right]=\frac{1}{2}\cdot \frac{11}{8}=\frac{11}{16}$.

Vì tanx xác định nên cosx ≠ 0. Chia tử và mẫu của phân thức cho luỹ thừa thích hợp của cosx để biểu diễn biểu thức theo tanx.

a) $\frac{3\text{sin}x-4\text{cos}x}{5\text{sin}x+2\text{cos}x}=\frac{3\cdot \frac{\text{sin}x}{\text{cos}x}-4}{5\cdot \frac{\text{sin}x}{\text{cos}x}+2}=\frac{3\text{tan}x-4}{5\text{tan}x+2}=\frac{3.2-4}{5.2+2}=\frac{1}{6}$.

b) $\frac{{\text{sin}}^{3}x+2{\text{cos}}^{3}x}{2\text{sin}x+3\text{cos}x}=\frac{\frac{{\text{sin}}^{3}x}{{\text{cos}}^{3}x}+2}{\left(2\frac{\text{sin}x}{\text{cos}x}+3\right)\cdot \frac{1}{{\text{cos}}^{2}x}}=\frac{{\text{tan}}^{3}x+2}{\left(2\text{tan}x+3\right)\left({\text{tan}}^{2}x+1\right)}$

$=\frac{2^3+2}{\left(2\cdot 2+3\right)\left(2^2+1\right)}=\frac{2}{7}$

Thay t = 31 vào công thức trên ta có:

$d\left(31\right)=4\sin \left[\frac{2\pi }{365}\left(31-80\right)\right]+12\approx \mathrm{9,01}$ (giờ)

Vậy thành phố X vào  ngày 31 tháng 1 có 9,01 giờ có Mặt Trời chiếu sáng.