Giải SGK Toán 11 CTST Bài 3: Các công thức lượng giác có đáp án

Đặt chiều rộng cổng AH = d.

⇒ OA = OB = $\frac{1}{2}$d.

Xét tam giác OBB’ vuông tại B’, có:

$\sin \widehat{BOB\text{'}}=\frac{BB\text{'}}{OB}=\frac{27}{\frac{d}{2}}=\frac{54}{d}$.

Vì $\stackrel{⏜}{AB}=\stackrel{⏜}{BC}$ nên sđ$\stackrel{⏜}{AC}$ = 2.sđ$\stackrel{⏜}{AB}$ $\Rightarrow \widehat{AOC}=2\widehat{BOB\text{'}}$

Xét tam giác OCC’ vuông tại C’, có:

$\sin \widehat{COC\text{'}}=\frac{CC\text{'}}{OC}\iff CC\text{'}=OC.\sin \widehat{COC\text{'}}=OC.\sin \left(2\widehat{BOB\text{'}}\right)$

Sau bài học này ta sẽ giải quyết tiếp được bài toán như sau:

$\sin \left(2\widehat{BOB\text{'}}\right)=2\sin \widehat{BOB\text{'}}.\cos \widehat{BOB\text{'}}=2.\frac{54}{d}.\sqrt{1-{\left(\frac{54}{d}\right)}^2}=\frac{108}{d}\sqrt{1-{\left(\frac{54}{d}\right)}^2}$.

Ta có: cos(α – β) = x_M.x_N + y_M.y_N = cosα.cosβ + sinα.sinβ.

Ta có: cos(α + β) = cos(α – (– β)) = cosα.cos(–β) + sinα.sin(–β) = cosα.cosβ – sinα.sinβ.

$\sin \frac{\pi }{12}=\sin \left(\frac{\pi }{3}-\frac{\pi }{4}\right)=\sin \frac{\pi }{3}.\text{cos}\frac{\pi }{4}-\text{cos}\frac{\pi }{3}.\text{sin}\frac{\pi }{4}=\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}-\frac{1}{2}.\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$.

Ở ví dụ 1 ta có: $\text{cos}\frac{\pi }{12}=\frac{\sqrt{6}+\sqrt{2}}{4}$

Ta có:

cos2α = cos(α + α) = cosα.cosα – sinα.sinα = cos²α – sin²α = cos²α + sin²α – 2sin²α = 1 – 2sin²α = 2cos²α – 1.

sin2α = sin(α + α) = sinα.cosα + cosα.sinα = 2.sinα.cosα .

$\tan 2\alpha =\tan \left(\alpha +\alpha \right)=\frac{\tan \alpha +\tan \alpha }{1-\tan \alpha .\tan \alpha }=\frac{2\tan \alpha }{1-{\tan }^2\alpha }$.

+) Ta có: $\text{cos}\frac{\pi }{4}=\text{cos}\left(\text{2.}\frac{\pi }{8}\right)=2{\text{cos}}^2\frac{\pi }{8}-1=\frac{\sqrt{2}}{2}$

$\Rightarrow {\text{cos}}^2\frac{\pi }{8}=\frac{\sqrt{2}+2}{4}$

$\Rightarrow \text{cos}\frac{\pi }{8}=\sqrt{\frac{\sqrt{2}+2}{4}}=\frac{\sqrt{2+\sqrt{2}}}{2}$ (vì $0<\frac{\pi }{8}<\frac{\pi }{2}$).

+) $\tan \frac{\pi }{4}=\tan \left(2.\frac{\pi }{8}\right)=\frac{2\tan \frac{\pi }{8}}{1-{\tan }^2\frac{\pi }{8}}=1$

$\iff 1-{\tan }^2\frac{\pi }{8}=2\tan \frac{\pi }{8}$

$\iff {\tan }^2\frac{\pi }{8}+2\tan \frac{\pi }{8}-1=0$

$\iff \tan \frac{\pi }{8}=-1+\sqrt{2}$(vì $0<\frac{\pi }{8}<\frac{\pi }{2}$).

a) Ta có: cos(α – β) = cosα.cosβ + sinα.sinβ; cos(α + β) = cosα.cosβ – sinα.sinβ

Khi đó:

cos(α – β) + cos(α + β) = cosα.cosβ + sinα.sinβ + cosα.cosβ – sinα.sinβ

= 2cosα.cosβ.

cos(α – β) – cos(α + β) = cosα.cosβ + sinα.sinβ – cosα.cosβ + sinα.sinβ

= 2sinα.sinβ .

b) Ta có: sin(α – β) = sinα.cosβ + cosα.sinβ; sin(α + β) = sinα.cosβ – cosα.sinβ

Khi đó:

sin(α – β) + sin(α + β) = sinα.cosβ + cosα.sinβ + sinα.cosβ – cosα.sinβ = 2sinα.cosβ.

sin(α – β) – sin(α + β) = sinα.cosβ + cosα.sinβ – sinα.cosβ + cosα.sinβ = 2cosα.sinβ.

Ta có:

$\sin \frac{\pi }{24}\text{cos}\frac{5\pi }{24}=\frac{1}{2}\left[\sin \left(\frac{\pi }{24}-\frac{5\pi }{24}\right)+\sin \left(\frac{\pi }{24}+\frac{5\pi }{24}\right)\right]=\frac{1}{2}\left[\sin \left(-\frac{\pi }{6}\right)+\sin \left(\frac{\pi }{4}\right)\right]$

$=\frac{1}{2}\left[-\frac{1}{2}+\frac{\sqrt{2}}{2}\right]=\frac{\sqrt{2}-1}{4}$

$\sin \frac{7\pi }{8}\text{sin}\frac{5\pi }{8}=\frac{1}{2}\left[\text{cos}\left(\frac{7\pi }{8}-\frac{5\pi }{8}\right)-\text{cos}\left(\frac{7\pi }{8}+\frac{5\pi }{8}\right)\right]=\frac{1}{2}\left(\text{cos}\frac{\pi }{4}-\text{cos}\frac{3\pi }{2}\right)$

$=\frac{1}{2}\left(\frac{\sqrt{2}}{2}+0\right)=\frac{\sqrt{2}}{4}$.

Ta có:

$\cos a.\cos b=\cos \frac{\alpha +\beta }{2}.\cos \frac{\alpha -\beta }{2}=\frac{1}{2}\left[\cos \left(\frac{\alpha +\beta }{2}-\frac{\alpha -\beta }{2}\right)+\cos \left(\frac{\alpha +\beta }{2}+\frac{\alpha -\beta }{2}\right)\right]$

$=\frac{1}{2}\left(\cos \beta +\cos \alpha \right)$.

$\sin a.\sin b=\sin \frac{\alpha +\beta }{2}.\sin \frac{\alpha -\beta }{2}=\frac{1}{2}\left[\cos \left(\frac{\alpha +\beta }{2}-\frac{\alpha -\beta }{2}\right)-\cos \left(\frac{\alpha +\beta }{2}+\frac{\alpha -\beta }{2}\right)\right]$

$=\frac{1}{2}\left(\cos \beta -\cos \alpha \right)$.

$\sin a.\text{cos}b=\sin \frac{\alpha +\beta }{2}.\text{cos}\frac{\alpha -\beta }{2}=\frac{1}{2}\left[\sin \left(\frac{\alpha +\beta }{2}-\frac{\alpha -\beta }{2}\right)+\sin \left(\frac{\alpha +\beta }{2}+\frac{\alpha -\beta }{2}\right)\right]$

$=\frac{1}{2}\left(\sin \beta +\sin \alpha \right)$.

Ta có:  OA = OB = $\frac{120}{2}=60$ cm.

Xét tam giác OBB’ vuông tại B’, có:

$\sin \widehat{BOB\text{'}}=\frac{BB\text{'}}{OB}=\frac{27}{60}=\frac{9}{20}$.

$\Rightarrow \text{cos}\widehat{BOB\text{'}}=\sqrt{1-{\left(\frac{9}{20}\right)}^2}=\frac{\sqrt{319}}{20}$

Vì $\stackrel{⏜}{AB}=\stackrel{⏜}{BC}$ nên sđ$\stackrel{⏜}{AC}$ = 2.sđ$\stackrel{⏜}{AB}$ $\Rightarrow \widehat{AOC}=2\widehat{BOB\text{'}}$

Xét tam giác OCC’ vuông tại C’, có:

$\sin \widehat{COC\text{'}}=\frac{CC\text{'}}{OC}\iff CC\text{'}=OC.\sin \widehat{COC\text{'}}=OC.\sin \left(2\widehat{BOB\text{'}}\right)$

Sau bài học này ta sẽ giải quyết tiếp được bài toán như sau:

$\sin \left(2\widehat{BOB\text{'}}\right)=2\sin \widehat{BOB\text{'}}.\cos \widehat{BOB\text{'}}=2.\frac{9}{20}.\frac{\sqrt{319}}{20}=\frac{9\sqrt{319}}{200}$.

Vậy khoảng cách này từ điểm C đến AH là $60.\frac{9\sqrt{319}}{200}\approx 48,2\left(cm\right)$.

a) Ta có:

+) $\text{cos}\frac{5\pi }{12}=\text{cos}\left(\frac{\pi }{4}+\frac{\pi }{6}\right)=\text{cos}\frac{\pi }{4}.\text{cos}\frac{\pi }{6}-\text{sin}\frac{\pi }{4}.\sin \frac{\pi }{6}=\frac{\sqrt{2}}{2}.\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}.\frac{1}{2}$$=\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}=\frac{\sqrt{6}-\sqrt{2}}{4}$+.

+) +$\sin \frac{5\pi }{12}=\sin \left(\frac{\pi }{4}+\frac{\pi }{6}\right)=\sin \frac{\pi }{4}.\text{cos}\frac{\pi }{6}\text{+\hspace{0.17em}cos}\frac{\pi }{4}.\sin \frac{\pi }{6}=\frac{\sqrt{2}}{2}.\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}.\frac{1}{2}$

$=\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}=\frac{\sqrt{6}+\sqrt{2}}{4}$.

+) $\tan \frac{5\pi }{12}=\frac{\sin \frac{5\pi }{12}}{\text{cos}\frac{5\pi }{12}}=\frac{\frac{\sqrt{6}-\sqrt{2}}{4}}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}=2-\sqrt{3}$.

+) $\cot \frac{5\pi }{12}=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}$.

b) Ta có:

– 555° = $\frac{\pi .\left(-555°\right)}{180°}=-\frac{37\pi }{12}=-\left(3\pi +\frac{\pi }{12}\right)$ rad.

Khi đó:

+) $\text{cos}\left(-555°\right)=\text{cos}\left(3\pi +\frac{\pi }{12}\right)=-\text{cos}\left(\frac{\pi }{12}\right)=-\text{cos}\left(\frac{\pi }{3}-\frac{\pi }{4}\right)$

$=-\left(\text{cos}\frac{\pi }{3}\text{cos}\frac{\pi }{4}+\sin \frac{\pi }{3}\sin \frac{\pi }{4}\right)=-\left(\frac{1}{2}.\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}\right)=-\frac{\sqrt{2}+\sqrt{6}}{4}$.

+) $\sin \left(-555°\right)=\sin \left(3\pi +\frac{\pi }{12}\right)=\sin \left(\frac{\pi }{12}\right)=\sin \left(\frac{\pi }{3}-\frac{\pi }{4}\right)$

$=\sin \frac{\pi }{3}\text{cos}\frac{\pi }{4}-\text{cos}\frac{\pi }{3}\sin \frac{\pi }{4}=\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}-\frac{1}{2}.\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$.

+) $\tan \left(-555°\right)=\frac{\sin \left(-555°\right)}{\text{cos}\left(-555°\right)}=\frac{\frac{\sqrt{6}-\sqrt{2}}{4}}{-\frac{\sqrt{6}+\sqrt{2}}{4}}=-2+\sqrt{3}$.

+)  $\cot \left(-555°\right)=\frac{1}{-2+\sqrt{3}}=-2-\sqrt{3}$.

Ta có:  $\text{cos}\alpha =-\sqrt{1-{\left(-\frac{5}{13}\right)}^2}=-\frac{12}{13}$ (vì  $\pi <\alpha <\frac{3\pi }{2}$).

Ta lại có:

 $\sin \left(\alpha +\frac{\pi }{6}\right)=\left(-\frac{5}{13}\right).\frac{\sqrt{3}}{2}+\left(-\frac{12}{13}\right).\frac{1}{2}=\frac{-12+5\sqrt{3}}{26}$;

 $\text{cos}\left(\frac{\pi }{4}-\alpha \right)=\text{cos}\frac{\pi }{4}\text{cos}\alpha +\sin \frac{\pi }{4}\sin \alpha =\frac{\sqrt{2}}{2}.\left(-\frac{12}{13}\right)+\frac{\sqrt{2}}{2}.\left(-\frac{5}{13}\right)=\frac{-17\sqrt{2}}{26}$.

a) Ta có: $\text{cos}\alpha =\sqrt{1-{\left(\frac{\sqrt{3}}{3}\right)}^2}=\frac{\sqrt{6}}{3}$ (vì $0<\alpha <\frac{\pi }{2}$).

Khi đó:

$\sin 2\alpha =2.\sin \alpha .\text{cos}\alpha =2.\frac{\sqrt{3}}{3}.\frac{\sqrt{6}}{3}=\frac{2\sqrt{2}}{3}$;

$\text{cos}2\alpha =2.{\text{cos}}^2\alpha -1=2.{\left(\frac{\sqrt{6}}{3}\right)}^2-1=\frac{1}{3}$;

$\tan 2\alpha =\frac{\sin 2\alpha }{\text{cos\hspace{0.17em}}2\alpha }=\frac{\frac{2\sqrt{2}}{3}}{\frac{1}{3}}=2\sqrt{2}$;

$\cot 2\alpha =\frac{1}{\tan 2\alpha }=\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}$.

b) Ta có:  $\text{cos}\frac{\alpha }{2}=-\sqrt{1-{\left(\frac{3}{4}\right)}^2}=-\frac{\sqrt{7}}{4}$ (vì  $\pi <\alpha <2\pi \iff \frac{\pi }{2}<\frac{\alpha }{2}<\pi$).

Khi đó:

 $\sin \alpha =2.\sin \frac{\alpha }{2}.\text{cos}\frac{\alpha }{2}=2.\frac{3}{4}.\left(-\frac{\sqrt{7}}{4}\right)=-\frac{3\sqrt{7}}{8}$;

 $\text{cos}\alpha =2.{\text{cos}}^2\frac{\alpha }{2}-1=2.{\left(-\frac{\sqrt{7}}{4}\right)}^2-1=-\frac{1}{8}$;

 $\sin 2\alpha =2.\sin \alpha .\text{cos}\alpha =2.\left(-\frac{3\sqrt{7}}{8}\right).\left(-\frac{1}{8}\right)=\frac{3\sqrt{7}}{32}$;

 $\text{cos}2\alpha =2.{\text{cos}}^2\alpha -1=2.{\left(-\frac{1}{8}\right)}^2-1=-\frac{31}{32}$;

 $\tan 2\alpha =\frac{\sin 2\alpha }{\text{cos\hspace{0.17em}}2\alpha }=\frac{\frac{3\sqrt{7}}{8}}{-\frac{31}{32}}=-\frac{12\sqrt{7}}{31}$;

 $\cot 2\alpha =\frac{1}{\tan 2\alpha }=-\frac{31}{12\sqrt{7}}$.

a) $\sqrt{2}\sin \left(\alpha +\frac{\pi }{4}\right)-\text{cos}\alpha$

$=\sqrt{2}.\left(\sin \alpha .\text{cos}\frac{\pi }{4}+\text{cos}\alpha .\sin \frac{\pi }{4}\right)-\text{cos}\alpha$

$=\sqrt{2}.\left(\frac{\sqrt{2}}{2}\sin \alpha +\frac{\sqrt{2}}{2}\text{cos}\alpha \right)-\text{cos}\alpha$

 $=\sin \alpha +\text{cos}\alpha -\text{cos}\alpha$

$=\sin \alpha$.

b) ${\left(\cos \alpha +\sin \alpha \right)}^2-\sin 2\alpha$

$={\cos }^2\alpha +{\sin }^2\alpha +2\sin \alpha .\cos \alpha -2\sin \alpha .\text{cos}\alpha$

=1

a) Ta có: $\text{cos}2\alpha =2{\text{cos}}^2\alpha -1=\frac{2}{5}$ 

$\iff {\text{cos}}^2\alpha =\frac{7}{10}$

$\iff \text{cos}\alpha =\frac{\sqrt{70}}{10}$ (vì $-\frac{\pi }{2}<\alpha <0$).

Mặt khác $\text{cos}2\alpha =1-2{\sin }^2\alpha =\frac{2}{5}$

$\iff {\sin }^2\alpha =\frac{3}{10}$

$\iff \sin \alpha =-\frac{\sqrt{30}}{100}$(vì $-\frac{\pi }{2}<\alpha <0$).

Khi đó:

$\tan \alpha =\frac{\sin \alpha }{\text{cos}\alpha }=\frac{-\frac{\sqrt{30}}{100}}{\frac{\sqrt{70}}{100}}=-\frac{\sqrt{3}}{\sqrt{7}}$.

$\cot \alpha =\frac{1}{\tan \alpha }=-\frac{\sqrt{7}}{\sqrt{3}}$.

b) $\sin 2\alpha =-\frac{4}{9}$ và $\frac{\pi }{2}<\alpha <\frac{3\pi }{4}$.

Ta có $\frac{\pi }{2}<\alpha <\frac{3\pi }{4}\iff \pi <2\alpha <\frac{3\pi }{2}$

$\Rightarrow \text{cos\hspace{0.17em}}2\alpha =-\sqrt{1-{\left(-\frac{4}{9}\right)}^2}=-\frac{\sqrt{65}}{9}$

Ta có: $\text{cos}2\alpha =2{\text{cos}}^2\alpha -1=-\frac{\sqrt{65}}{9}$ 

$\iff {\text{cos}}^2\alpha =\frac{9-\sqrt{65}}{18}$

$\iff \text{cos}\alpha =\sqrt{\frac{9-\sqrt{65}}{18}}$ (vì $\frac{\pi }{2}<\alpha <\frac{3\pi }{4}$).

Mặt khác $\text{cos}2\alpha =1-2{\sin }^2\alpha =-\frac{\sqrt{65}}{9}$

$\iff {\sin }^2\alpha =\frac{\sqrt{65}+1}{18}$

$\iff \sin \alpha =\frac{\sqrt{65}+1}{18}$(vì $\frac{\pi }{2}<\alpha <\frac{3\pi }{4}$).

Khi đó:

$\tan \alpha =\frac{\sin \alpha }{\text{cos}\alpha }=\frac{\frac{\sqrt{65}+1}{18}}{\frac{1-\sqrt{65}}{18}}=\frac{\sqrt{65}+1}{1-\sqrt{65}}$.

$\cot \alpha =\frac{1}{\tan \alpha }=\frac{1-\sqrt{65}}{\sqrt{65}+1}$.

Xét tam giác ABC, có:

A + B + C = 180° ⇒ A = 180° – (B + C)

sinA = sin(180° – (B + C)) = sin(B + C) = sinB.cosC + sinC.cosB.

Xét tam giác ABC vuông tại B có:

$\tan \widehat{BAC}=\frac{3}{4}$.

Ta lại có: $\widehat{BAD}=\widehat{BAC}+\widehat{CAD}$

$\Rightarrow \tan \widehat{BAD}=\tan \left(\widehat{BAC}+\widehat{CAD}\right)=\tan \left(\widehat{BAC}+30°\right)=\frac{\tan \widehat{BAC}+\tan 30°}{1-\tan \widehat{BAC}.\tan 30°}$

$=\frac{\frac{3}{4}+\frac{\sqrt{3}}{3}}{1-\frac{3}{4}.\frac{\sqrt{3}}{3}}=\frac{48+25\sqrt{3}}{39}\approx 2,34$.

Xét tam giác ABD vuông tại B có:

$\tan \widehat{BAD}=\frac{BD}{AB}\Rightarrow BD=\tan \widehat{BAD}.AB=2,34.4\approx 9,36$.

⇒ CD = BD – BC ≈ 9,36 – 3 = 6,36.

Tại $\alpha =\frac{\pi }{2}$ thì H trùng I, M trùng O nên MH = OI do đó OM = IH.

Xét tam giác AHI vuông tại H có: IH = cosα.IA = 8cosα.

a) Tính sinα và cosα

Từ điểm M kẻ MH vuông góc với Ox, MK vuông góc với Oy.

Ta có: MH = 60 – 30 = 30 m.  

Khi đó hoành độ điểm M là 30.

Mặt khác hoành độ điểm M là: x_M = 31.cosα.

⇒ cosα = $\frac{30}{31}$

⇒ $\sin \alpha =-\sqrt{1-{\left(\frac{30}{31}\right)}^2}=-\frac{\sqrt{61}}{31}$.

b) Vì các cánh quạt tạo thành 3 góc bằng nhau nên $\widehat{MOP}=\widehat{NOP}=\widehat{MON}=120°$

$\Rightarrow \widehat{AOP}=\widehat{MOP}-\widehat{MOA}$

$\Rightarrow \sin \widehat{AOP}=\sin \left(\widehat{MOP}-\widehat{MOA}\right)=\sin \widehat{MOP}.\text{cos}\widehat{MOA}-\text{cos}\widehat{MOP}.\sin \widehat{MOA}$

$=\sin \frac{2\pi }{3}.\text{cos}\alpha -\text{cos}\frac{2\pi }{3}.\sin \alpha$

 $=\frac{\sqrt{3}}{2}.\frac{30}{31}+\frac{1}{2}.\frac{\sqrt{61}}{31}\approx 0,96$.

 $\Rightarrow \sin \left(OA,OP\right)=\sin \widehat{AOP}\approx 0,96$.

Vì vậy chiều cao của điểm P so với mặt đất khoảng: 31.sinα + 60 = 89,76 m.

Ta có:  $\text{cos}\widehat{AOP}\approx \sqrt{1-0,{96}^2}=0,28$.

Ta có:  $\widehat{AON}=\widehat{AOP}+\widehat{PON}$

 $\Rightarrow \sin \widehat{AON}=\sin \left(\widehat{AOP}+\widehat{PON}\right)=\sin \widehat{AOP}.\text{cos}\widehat{PON}\text{+cos}\widehat{AOP}.\sin \widehat{PON}$

 $=0,96.\text{cos}\frac{2\pi }{3}-0,28.\sin \frac{2\pi }{3}$

 $=0,96.\left(-\frac{1}{2}\right)+0,28.\frac{\sqrt{3}}{2}\approx -0,23$.

 $\Rightarrow \sin \left(OA,ON\right)=\sin \widehat{AON}\approx -0,23$.

Vì vậy chiều cao của điểm N so với mặt đất khoảng: 31.sinα + 60 = 89,76 m.