2026年通成学典课时作业本九年级数学全一册浙教版第59页答案
$\sin 30° = \_\_\_\_\_\_, \cos 30° = \_\_\_\_\_\_, \tan 30° = \_\_\_\_\_\_; \sin 60° = \_\_\_\_\_\_, \cos 60° = \_\_\_\_\_\_, \tan 60° = \_\_\_\_\_\_; \sin 45° = \_\_\_\_\_\_, \cos 45° = \_\_\_\_\_\_, \tan 45° = \_\_\_\_\_\_.$

答案

$\sin 30°=\dfrac{1}{2},\cos 30°=\dfrac{\sqrt{3}}{2},\tan 30°=\dfrac{\sqrt{3}}{3};\sin 60°=\dfrac{\sqrt{3}}{2},\cos 60°=\dfrac{1}{2},\tan 60°=\sqrt{3};\sin 45°=\dfrac{\sqrt{2}}{2},\cos 45°=\dfrac{\sqrt{2}}{2},\tan 45°=1$
1 [2025 天津]$\tan 45° - \sqrt{2}\cos 45°$的值为(
A


A.0
B.1
C.$1-\dfrac{\sqrt{2}}{2}$
D.$1-\sqrt{2}$

答案

1.A
2 已知α为锐角,且$\cosα=\frac{\sqrt{3}}{2}$,则α等于(
A


A.$30°$
B.$45°$
C.$60°$
D.$90°$

答案

2.A
3 在$\mathrm{Rt}△ ABC$中,$∠ C=90°$,$∠ B=30°$,$AB=8$,则$BC$的长为(
D


A.$\dfrac{4\sqrt{3}}{3}$
B.$4$
C.$8\sqrt{3}$
D.$4\sqrt{3}$

答案

3.D
4 在$△ ABC$中,若$∠ A = 105°$,$∠ B = 45°$,则$\tan C$的值为
$\dfrac{\sqrt{3}}{3}$

答案

4.$\dfrac{\sqrt{3}}{3}$
5 计算:
(1) $\dfrac{2\cos^2 30° - \sin 30°}{\tan^2 60° - 4\cos 45°}$;
(2) $2\cos 45° · \sin 45° - 2\sin 30° · \tan 45° + \sqrt{6}\tan 60°$。

答案

5.(1)原式$=\dfrac{2×(\dfrac{\sqrt{3}}{2})^2 - \dfrac{1}{2}}{(\sqrt{3})^2 - 4×\dfrac{\sqrt{2}}{2}} = \dfrac{2×\dfrac{3}{4} - \dfrac{1}{2}}{3-2\sqrt{2}} = \dfrac{1}{3-2\sqrt{2}} = 3+2\sqrt{2}$
(2)原式$=2×\dfrac{\sqrt{2}}{2}×\dfrac{\sqrt{2}}{2} - 2×\dfrac{1}{2}×1 + \sqrt{6}×\sqrt{3} = 1-1+3\sqrt{2}=3\sqrt{2}$
6 若α为锐角,且$\cos(α - 20°)=\frac{\sqrt{2}}{2}$,则α等于(
D


A.$25°$
B.$30°$
C.$45°$
D.$65°$

答案

6.D
7 若α为锐角,且$\tan(α + 15°) = \sqrt{3}$,则$\tanα$的值为
1

答案

7.1
8 设α为锐角,若$\cos^2α=\dfrac{3}{4}$,则α=
$30°$
;若$\tan(α - 10°)=\dfrac{\sqrt{3}}{3}$,则α=
$40°$

答案

8.$30°\ \ 40°$
9(1)填空:
① $\frac{\sin 30°}{\cos 30°}=$
$\dfrac{\sqrt{3}}{3}$
;$\tan 30°=$
$\dfrac{\sqrt{3}}{3}$

② $\frac{\sin 60°}{\cos 60°}=$
$\sqrt{3}$
;$\tan 60°=$
$\sqrt{3}$
.
(2)观察(1)的结果,猜想$\frac{\sin α}{\cos α}$与$\tan α$的大小关系.
(3)如图,在$Rt△ ABC$中,$∠ C=90°$,$∠ A=α$,$△ ABC$的三边长分别为$a,b,c$.试利用三角函数的定义证明(2)中的猜想.

答案

9.(1)① $\dfrac{\sqrt{3}}{3}\ \ \dfrac{\sqrt{3}}{3}$ ② $\sqrt{3}\ \ \sqrt{3}$
(2)$\dfrac{\sin α}{\cos α}=\tan α$
(3)因为$\sin α=\dfrac{a}{c},\cos α=\dfrac{b}{c}$,所以$\dfrac{\sin α}{\cos α}=\dfrac{a}{c}÷\dfrac{b}{c}=\dfrac{a}{b}$,而$\tan α=\dfrac{a}{b}$,所以$\dfrac{\sin α}{\cos α}=\tan α$