二倍角公式 · 历年高考数学真题与解析

本页汇总 高考真题检索 的「二倍角公式」高考数学真题共 60 道,覆盖 2008–2026 年,最常出题型为 单选题;含完整答案与解析。

60questions
Past questions included
2008–2026
Years covered
区分题为主
Overall difficulty
Single-choice
Most common question type
Practice This Topic Do 5 reliable past questions first, then see the answers and explanations after you submit
常用解题方法化归与转化函数与方程整体代换
常见易错点符号错误范围错误漏解
Core competencies应用

Past Question List

2026 Shanxi Gaokao Single choice Differentiating Question 7 2026_山西选择性考试

7. 已知α为第二象限角,且 $3 \sin 2 \alpha \cos \alpha = 8 \sin \alpha \cos 2 \alpha$ ,则 $\frac { 1 + \sin \alpha } { 2 - \cos \alpha } =$

A. $\frac 3 4$
B. $\frac { 3 } { 5 }$
C. $\frac { 1 } { 2 }$
D. $\frac { 1 5 } { 1 2 }$
2025 National Gaokao Single choice Hardest Question 8 2025_新课标II Paper

8. 已知 $0<\alpha<\pi, \cos \frac{\alpha}{2}=\frac{\sqrt{5}}{5}$ ,则 $\sin \left(\alpha-\frac{\pi}{4}\right)=$

A. $\frac{\sqrt{2}}{10}$
B. $\frac{\sqrt{2}}{5}$
C. $\frac{3 \sqrt{2}}{10}$
D. $\frac{7 \sqrt{2}}{10}$ 二、 多选题:本题共3小题,每小题6分,共18分。在每小题给出的四个选项中,有多项符合题目要求。全部选对的得 6 分,部分选对的得部分分,有选错的得 0 分。
2025 National Gaokao Single choice Differentiating Question 8 2025_新课标 II Paper

8. $0 < \alpha < \pi$ $\cos { \frac { \alpha } { 2 } } = { \frac { \sqrt { 5 } } { 5 } }$ $\sin ( \alpha - { \frac { \pi } { 4 } } ) =$ )

A. $\frac { \sqrt { 2 } } { 1 0 }$
B. $\frac { \sqrt { 2 } } { 5 }$
C. $\frac { 3 { \sqrt { 2 } } } { 1 0 }$
D. $\frac { 7 { \sqrt { 2 } } } { 1 0 }$
2023 ?? Gaokao Free response Hardest Question 22 2023_全国甲卷 (2023·文)

22.已知点 $P(2,1)$ ,直线 $l:\left\{\begin{array}{c}x=2+t \cos \alpha, \\ y=1+t \sin \alpha\end{array}\right.$( $t$ 为参数),$\alpha$ 为 $l$ 的倾斜角,$l$ 与 $x$ 轴正半轴、 $y$ 轴正半轴分别交于 $A, B$ ,且 $|P A| \cdot|P B|=4$ .
(1)求 $\alpha$ ;
(2)以坐标原点为极点,$x$ 轴正半轴为极轴建立极坐标系,求 $l$ 的极坐标方程.

2023 ?? Gaokao Single choice Differentiating Question 6 2023_新课标 I 卷 (2023)

6.过点 $(0,-2)$ 与圆 $x^{2}+y^{2}-4 x-1=0$ 相切的两条直线的夹角为 $\alpha$ ,则 $\sin \alpha=$()

A. 1
B. $\frac{\sqrt{15}}{4}$
C. $\frac{\sqrt{10}}{4}$
D. $\frac{\sqrt{6}}{4}$
2022 Beijing Gaokao Single choice Differentiating Question 5 2022_北京卷 (2022)

5.已知函数 $f(x)=\cos ^{2} x-\sin ^{2} x$ ,则( )

A. $f(x)$ 在 $\left(-\frac{\pi}{2},-\frac{\pi}{6}\right)$ 上单调递减
B. $f(x)$ 在 $\left(-\frac{\pi}{4}, \frac{\pi}{12}\right)$ 上单调递增
C. $f(x)$ 在 $\left(0, \frac{\pi}{3}\right)$ 上单调递减
D. $f(x)$ 在 $\left(\frac{\pi}{4}, \frac{7 \pi}{12}\right)$ 上单调递增
2021 National Gaokao Single choice Hardest Question 11 2021_全国甲卷 (2021·文)

11.若 $\alpha \in\left(0, \frac{\pi}{2}\right), \tan 2 \alpha=\frac{\cos \alpha}{2-\sin \alpha}$ ,则 $\tan \alpha=$

A. $\frac{\sqrt{15}}{15}$
B. $\frac{\sqrt{5}}{5}$
C. $\frac{\sqrt{5}}{3}$
D. $\frac{\sqrt{15}}{3}$
2020 National Gaokao Single choice Differentiating Question 9 2020_新课标 I 卷 (2020·理)

9.已知 $\alpha \in(0, \pi)$ ,且 $3 \cos 2 \alpha-8 \cos \alpha=5$ ,则 $\sin \alpha=$( )

A. $\frac{\sqrt{5}}{3}$
B. $\frac{2}{3}$
C. $\frac{1}{3}$
D. $\frac{\sqrt{5}}{9}$
2019 National Gaokao Single choice Differentiating Question 10 2019_新课标 II 卷 (2019·理)

10.已知 $a \in\left(0, \frac{\pi}{2}\right), 2 \sin 2 \alpha=\cos 2 \alpha+1$ ,则 $\sin \alpha=$

A. $\frac{1}{5}$
B. $\frac{\sqrt{5}}{5}$
C. $\frac{\sqrt{3}}{3}$
D. $\frac{2 \sqrt{5}}{5}$
2019 National Gaokao Single choice Differentiating Question 11 2019_新课标 II 卷 (2019·文)

11.已知 $a \in\left(0, \frac{\pi}{2}\right), 2 \sin 2 \alpha=\cos 2 \alpha+1$ ,则 $\sin \alpha=$

A. $\frac{1}{5}$
B. $\frac{\sqrt{5}}{5}$
C. $\frac{\sqrt{3}}{3}$
D. $\frac{2 \sqrt{5}}{5}$
2018 National Gaokao Single choice Differentiating Question 11 2018_新课标 I 卷 (2018·文)

11.(5分)已知角 $\alpha$ 的顶点为坐标原点,始边与 x 轴的非负半轴重合,终边上有两点 $A(1, a), B(2, b)$ ,且 $\cos 2 \alpha=\frac{2}{3}$ ,则 $|a-b|=$( )

A. $\frac{1}{5}$
B. $\frac{\sqrt{5}}{5}$
C. $\frac{2 \sqrt{5}}{5}$
D. 1
2018 Jiangsu Gaokao Free response Differentiating Question 16 2018_江苏卷 (2018)

16.(本小题满分 14 分)
已知 $\alpha, \beta$ 为锐角, $\tan \alpha=\frac{4}{3}, \cos (\alpha+\beta)=-\frac{\sqrt{5}}{5}$ .
(1)求 $\cos 2 \alpha$ 的值;
(2)求 $\tan (\alpha-\beta)$ 的值.

2018 Shanghai Gaokao Free response Differentiating Question 17 2018_上海卷 (2018)

17.(本题满分 14 分,第 1 小题满分 6 分,第 2 小题满分 8 分)

已知 $y=\cos x$ .
(1)若 $f(\alpha)=\frac{1}{3}$ ,且 $\alpha \in[0, \pi]$ ,求 $f\left(\alpha-\frac{\pi}{3}\right)$ 的值;
(2)求函数 $y=f(2 x)-2 f(x)$ 的最小值.

2018 National Gaokao Single choice Differentiating Question 6 2018_新课标 III 卷 (2018·文)

6.(5分)函数 $\mathrm{f}(\mathrm{x})=\frac{\tan \mathrm{x}}{1+\tan ^{2} \mathrm{x}}$ 的最小正周期为( )

A. $\frac{\pi}{4}$
B. $\frac{\pi}{2}$
C. $\pi$
D. $2 \pi$
2018 National Gaokao Single choice Differentiating Question 6 2018_新课标 II 卷 (2018·理)

6.(5分)在 $\triangle \mathrm{ABC}$ 中, $\cos \frac{\mathrm{C}}{2}=\frac{\sqrt{5}}{5}, \mathrm{BC}=1, \mathrm{AC}=5$ ,则 $\mathrm{AB}=$( )

A. $4 \sqrt{2}$
B. $\sqrt{30}$
C. $\sqrt{29}$
D. $2 \sqrt{5}$
2017 Zhejiang Gaokao Fill in the blank Differentiating Question 10 2017_浙江卷 (2017·理)

10.(6 分)(2016 • 浙江)已知 $2 \cos ^{2} \mathrm{x}+\sin 2 \mathrm{x}=\mathrm{A} \sin (\omega \mathrm{x}+\phi)+\mathrm{b}(\mathrm{A}>0)$ ,则 $\mathrm{A}=$ $\_\_\_\_$ , $\mathrm{b}=$ $\_\_\_\_$。

2016 Zhejiang Gaokao Fill in the blank Differentiating Question 10 2016_浙江卷 (2016·理)

10.(6 分)(2016 • 浙江)已知 $2 \cos ^{2} \mathrm{x}+\sin 2 \mathrm{x}=\mathrm{A} \sin (\omega \mathrm{x}+\phi)+\mathrm{b}(\mathrm{A}>0)$ ,则 $\mathrm{A}=$ $\_\_\_\_$ , $\mathrm{b}=$ $\_\_\_\_$ .

2016 Zhejiang Gaokao Free response Differentiating Question 11 2016_浙江卷 (2016·文)

11.(6 分)(2016 •浙江)已知 $2 \cos ^{2} \mathrm{x}+\sin 2 \mathrm{x}=\mathrm{Asin}(\omega \mathrm{x}+\phi)+\mathrm{b}(\mathrm{A}>0)$ ,则 $\mathrm{A}=-\sqrt{2} \_, \mathrm{b}=$ 1 。

2016 Zhejiang Gaokao Free response Differentiating Question 16 2016_浙江卷 (2016·理)

16.(14分)(2016•浙江)在 $\triangle \mathrm{ABC}$ 中,内角 A,B,C 所对的边分别为 a,b,c,已知 $\mathrm{b}+\mathrm{c}=2 \mathrm{a} \cos \mathrm{B}$ .
( I )证明: $\mathrm{A}=2 \mathrm{~B}$
(II)若 $\triangle \mathrm{ABC}$ 的面积 $\mathrm{S}=\frac{\mathrm{a}^{2}}{4}$ ,求角 A 的大小.

2016 National Gaokao Single choice Differentiating Question 9 2016_新课标 II 卷 (2016·理)

9.(5分)若 $\cos \left(\frac{\pi}{4}-\alpha\right)=\frac{3}{5}$ ,则 $\sin 2 \alpha=$( )

A. $\frac{7}{25}$
B. $\frac{1}{5}$
C. $-\frac{1}{5}$
D. $-\frac{7}{25}$
2014 National Gaokao Fill in the blank Differentiating Question 16 2014_大纲版 (2014·理)

16.(5分)若函数 $\mathrm{f}(\mathrm{x})=\cos 2 \mathrm{x}+\mathrm{a} \sin \mathrm{x}$ 在区间 $\left(\frac{\pi}{6}, \frac{\pi}{2}\right)$ 是减函数,则 a 的取值范围是 $\_\_\_\_$ ( $-\infty$ ,2] .

2014 National Gaokao Free response Differentiating Question 17 2014_退役省自主命题 (2014·文)

17.(本小题满分 12 分)已知函数 $f(x)=\sin \left(3 x+\frac{\pi}{4}\right)$ .
(1)求 $f(x)$ 的单调递增区间;
(2)若 $\alpha$ 是第二象限角,$f\left(\frac{\alpha}{3}\right)=\frac{4}{5} \cos \left(\alpha+\frac{\pi}{4}\right) \cos 2 \alpha$ ,求 $\cos \alpha-\sin \alpha$ 的值.

2014 National Gaokao Single choice Differentiating Question 6 2014_新课标 I 卷 (2014·理)

6.(5分)如图,圆 O 的半径为1, A 是圆上的定点, P 是圆上的动点,角 x 的始边为射线 OA ,终边为射线 OP ,过点 P 作直线 OA 的垂线,垂足为 M ,将点 M 到直线 $O P$ 的距离表示为 $x$ 的函数 $f(x)$ ,则 $y=f(x)$ 在 $[0, \pi]$ 的图象大致为(

A.
B.
C.
D.
2014 National Gaokao Single choice Differentiating Question 6 2014_退役省自主命题 (2014·理)

6.若函数 $f(x), g(x)$ 满足 $\int_{-1}^{1} f(x) g(x) d x=0$,则称 $f(x), g(x)$ 在区间 $[-1,1]$ 上的一组正交函数,给出三组函数:①$f(x)=\sin \frac{1}{2} x, g(x)=\cos \frac{1}{2} x$;②$f(x)=x+1, g(x)=x-1$;③$f(x)=x, g(x)=x^{2}$.其中为区间 $[-1,1]$ 的正交函数的组数是

A. 0
B. 1
C. 2
D. 3
2013 National Gaokao Free response Differentiating Question 16 2013_退役省自主命题 (2013·理)

16.(12分)(2013 • 广东)已知函数 $\mathrm{f}(\mathrm{x})=\sqrt{2} \cos \left(\mathrm{x}-\frac{\pi}{12}\right), ~ \mathrm{x} \in \mathrm{R}$ 。
(1)求 $f\left(-\frac{\pi}{6}\right)$ 的值;
(2)若 $\cos \theta=\frac{3}{5}, \quad \theta \in\left(\frac{3 \pi}{2}, 2 \pi\right)$ ,求 $f\left(2 \theta+\frac{\pi}{3}\right)$ .

2013 National Gaokao Free response Differentiating Question 18 2013_退役省自主命题 (2013·文)

18.(本小题满分 12 分)
在 $\triangle A B C$ 中,角 $A, B, C$ 对应的边分别是 $a, b, c$.已知 $\cos 2 A-3 \cos (B+C)=1$.
(I)求角 $A$ 的大小;
(II)若 $\triangle A B C$ 的面积 $S=5 \sqrt{3}, \quad b=5$,求 $\sin B \sin C$ 的值.

2012 Jiangsu Gaokao Free response Differentiating Question 11 2012_江苏卷 (2012)

11.设 $\alpha$ 为锐角,若 $\cos \left(\alpha+\frac{\pi}{6}\right)=\frac{4}{5}$ ,则 $\sin \left(2 \alpha+\frac{\pi}{12}\right)$ 的值为

2012 National Gaokao Single choice Differentiating Question 4 2012_大纲版 (2012·文)

4.(5分)已知 $\alpha$ 为第二象限角, $\sin \alpha=\frac{3}{5}$ ,则 $\sin 2 \alpha=$( )

A. $-\frac{24}{25}$
B. $-\frac{12}{25}$
C. $\frac{12}{25}$
D. $\frac{24}{25}$
2012 National Gaokao Single choice Differentiating Question 7 2012_退役省自主命题 (2012·理)

(7)若 $\theta \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right], \sin 2 \theta=\frac{3 \sqrt{7}}{8}$ ,则 $\sin \theta=$

A. $\frac{3}{5}$
B. $\frac{4}{5}$
C. $\frac{\sqrt{7}}{4}$
D. $\frac{3}{4}$
2012 National Gaokao Single choice Differentiating Question 7 2012_大纲版 (2012·理)

7.(5分)已知 $\alpha$ 为第二象限角, $\sin \alpha+\cos \alpha=\frac{\sqrt{3}}{3}$ ,则 $\cos 2 \alpha=$()

A. $-\frac{\sqrt{5}}{3}$
B. $-\frac{\sqrt{5}}{9}$
C. $\frac{\sqrt{5}}{9}$
D. $\frac{\sqrt{5}}{3}$
2011 National Gaokao Fill in the blank Differentiating Question 14 2011_大纲版 (2011·理)

14.(5分)已知 $\alpha \in\left(\frac{\pi}{2}, \pi\right), \sin \alpha=\frac{\sqrt{5}}{5}$ ,则 $\tan 2 \alpha=$ $\_\_\_\_$ $-\frac{4}{3}$ .

2011 National Gaokao Single choice Differentiating Question 5 2011_老新课标卷 (2011·理)

5.(5分)已知角 $\theta$ 的顶点与原点重合,始边与 x 轴的正半轴重合,终边在直线

$y=2 x$ 上,则 $\cos 2 \theta=$

A. $-\frac{4}{5}$
B. $-\frac{3}{5}$
C. $\frac{3}{5}$
D. $\frac{4}{5}$
2011 National Gaokao Single choice Differentiating Question 7 2011_老新课标卷 (2011·文)

7.(5分)已知角 $\theta$ 的顶点与原点重合,始边与 $x$ 轴的正半轴重合,终边在直线 $y =2 x$ 上,则 $\cos 2 \theta=$( )

A. $-\frac{4}{5}$
B. $-\frac{3}{5}$
C. $\frac{3}{5}$
D. $\frac{4}{5}$
2011 National Gaokao Single choice Differentiating Question 7 2011_退役省自主命题 (2011·理)

7.(5分)(2011•辽宁)设 $\sin \left(\frac{\pi}{4}+\theta\right)=\frac{1}{3}$ ,则 $\sin 2 \theta=~(\quad)$

A. $-\frac{7}{9}$
B. $-\frac{1}{9}$
C. $\frac{1}{9}$
D. $\frac{7}{9}$
2010 National Gaokao Single choice Differentiating Question 3 2010_旧全国 II 卷 (2010·文)

3.(5分)已知 $\sin \alpha=\frac{2}{3}$ ,则 $\cos (\pi-2 \alpha)=$( )

A. $-\frac{\sqrt{5}}{3}$
B. $-\frac{1}{9}$
C. $\frac{1}{9}$
D. $\frac{\sqrt{5}}{3}$

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