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:

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

Vì $\overset{\frown}{AB}=\overset{\frown}{BC}$ nên sđ$\overset{\frown}{AC}$ = 2.sđ$\overset{\frown}{AB}$ $\Rightarrow \widehat{AOC}=2\widehat{BOB'}$

Xét tam giác OCC’ vuông tại C’, có:

$\sin \widehat{COC'}=\frac{CC'}{OC}\Leftrightarrow CC'=OC.\sin \widehat{COC'}=OC.\sin (2\widehat{BOB'})$

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

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

Vậy khoảng cách này từ điểm C đến AH là $\frac{108}{d}\sqrt{1-{(\frac{54}{d})}^{2}}$.

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.

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

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

Suy ra$\tan \frac{π}{12}=\frac{\sin \frac{π}{12}}{\text{cos}\frac{π}{12}}=\frac{\frac{\sqrt{6}-\sqrt{2}}{4}}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}=2-\sqrt{3}$

Ta có:

cos2α = cos(α + α) = cosα.cosα – sinα.sinα = cos$^{2}$α – sin$^{2}$α = cos$^{2}$α + sin$^{2}$α – 2sin$^{2}$α = 1 – 2sin$^{2}$α = 2cos$^{2}$α – 1.

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

$\tan 2α=\tan (α+α)=\frac{\tan α+\tan α}{1-\tan α.\tan α}=\frac{2\tan α}{1-{\tan }^{2}α}$.

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

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

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

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

$\Leftrightarrow 1-{\tan }^{2}\frac{π}{8}=2\tan \frac{π}{8}$

$\Leftrightarrow {\tan }^{2}\frac{π}{8}+2\tan \frac{π}{8}-1=0$

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

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:

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

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

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

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

Ta có:

$\cos a.\cos b=\cos \frac{α+β}{2}.\cos \frac{α-β}{2}=\frac{1}{2}[\cos (\frac{α+β}{2}-\frac{α-β}{2})+\cos (\frac{α+β}{2}+\frac{α-β}{2})]$

$=\frac{1}{2}(\cos β+\cos α)$.

$\sin a.\sin b=\sin \frac{α+β}{2}.\sin \frac{α-β}{2}=\frac{1}{2}[\cos (\frac{α+β}{2}-\frac{α-β}{2})-\cos (\frac{α+β}{2}+\frac{α-β}{2})]$

$=\frac{1}{2}(\cos β-\cos α)$.

$\sin a.\text{cos}b=\sin \frac{α+β}{2}.\text{cos}\frac{α-β}{2}=\frac{1}{2}[\sin (\frac{α+β}{2}-\frac{α-β}{2})+\sin (\frac{α+β}{2}+\frac{α-β}{2})]$

$=\frac{1}{2}(\sin β+\sin α)$.

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

Xét tam giác OBB’ vuông tại B’, có:

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

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

Vì $\overset{\frown}{AB}=\overset{\frown}{BC}$ nên sđ$\overset{\frown}{AC}$ = 2.sđ$\overset{\frown}{AB}$ $\Rightarrow \widehat{AOC}=2\widehat{BOB'}$

Xét tam giác OCC’ vuông tại C’, có:

$\sin \widehat{COC'}=\frac{CC'}{OC}\Leftrightarrow CC'=OC.\sin \widehat{COC'}=OC.\sin (2\widehat{BOB'})$

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

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

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

a) Ta có:

+) $\text{cos}\frac{5π}{12}=\text{cos}(\frac{π}{4}+\frac{π}{6})=\text{cos}\frac{π}{4}.\text{cos}\frac{π}{6}-\text{sin}\frac{π}{4}.\sin \frac{π}{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}$+.

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

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

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

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

b) Ta có:

– 555° = $\frac{π.(-555^{\circ})}{180^{\circ}}=-\frac{37π}{12}=-(3π+\frac{π}{12})$ rad.

Khi đó:

+) $\text{cos}(-555^{\circ})=\text{cos}(3π+\frac{π}{12})=-\text{cos}(\frac{π}{12})=-\text{cos}(\frac{π}{3}-\frac{π}{4})$

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

+) $\sin (-555^{\circ})=\sin (3π+\frac{π}{12})=\sin (\frac{π}{12})=\sin (\frac{π}{3}-\frac{π}{4})$

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

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

+) $\cot (-555^{\circ})=\frac{1}{-2+\sqrt{3}}=-2-\sqrt{3}$.

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

Ta lại có:

$\sin (α+\frac{π}{6})=(-\frac{5}{13}).\frac{\sqrt{3}}{2}+(-\frac{12}{13}).\frac{1}{2}=\frac{-12+5\sqrt{3}}{26}$;

$\text{cos}(\frac{π}{4}-α)=\text{cos}\frac{π}{4}\text{cos}α+\sin \frac{π}{4}\sin α=\frac{\sqrt{2}}{2}.(-\frac{12}{13})+\frac{\sqrt{2}}{2}.(-\frac{5}{13})=\frac{-17\sqrt{2}}{26}$.

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

Khi đó:

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

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

$\tan 2α=\frac{\sin 2α}{\text{cos}2α}=\frac{\frac{2\sqrt{2}}{3}}{\frac{1}{3}}=2\sqrt{2}$;

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

b) Ta có: $\text{cos}\frac{α}{2}=-\sqrt{1-{(\frac{3}{4})}^{2}}=-\frac{\sqrt{7}}{4}$ (vì $π<α<2π\Leftrightarrow \frac{π}{2}<\frac{α}{2}<π$).

Khi đó:

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

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

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

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

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

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

a) $\sqrt{2}\sin (α+\frac{π}{4})-\text{cos}α$

$=\sqrt{2}.(\sin α.\text{cos}\frac{π}{4}+\text{cos}α.\sin \frac{π}{4})-\text{cos}α$

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

$=\sin α+\text{cos}α-\text{cos}α$

$=\sin α$.

b) ${(\cos α+\sin α)}^{2}-\sin 2α$

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

=1

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

$\Leftrightarrow {\text{cos}}^{2}α=\frac{7}{10}$

$\Leftrightarrow \text{cos}α=\frac{\sqrt{70}}{10}$ (vì $-\frac{π}{2}<α<0$).

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

$\Leftrightarrow {\sin }^{2}α=\frac{3}{10}$

$\Leftrightarrow \sin α=-\frac{\sqrt{30}}{100}$(vì $-\frac{π}{2}<α<0$).

Khi đó:

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

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

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

Ta có $\frac{π}{2}<α<\frac{3π}{4}\Leftrightarrow π<2α<\frac{3π}{2}$

$\Rightarrow \text{cos}2α=-\sqrt{1-{(-\frac{4}{9})}^{2}}=-\frac{\sqrt{65}}{9}$

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

$\Leftrightarrow {\text{cos}}^{2}α=\frac{9-\sqrt{65}}{18}$

$\Leftrightarrow \text{cos}α=\sqrt{\frac{9-\sqrt{65}}{18}}$ (vì $\frac{π}{2}<α<\frac{3π}{4}$).

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

$\Leftrightarrow {\sin }^{2}α=\frac{\sqrt{65}+1}{18}$

$\Leftrightarrow \sin α=\frac{\sqrt{65}+1}{18}$(vì $\frac{π}{2}<α<\frac{3π}{4}$).

Khi đó:

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

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

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: $\widehat{BAD}=\widehat{BAC}+\widehat{CAD}$

$\Rightarrow \tan \widehat{BAD}=\tan (\widehat{BAC}+\widehat{CAD})=\tan (\widehat{BAC}+30^{\circ})=\frac{\tan \widehat{BAC}+\tan 30^{\circ}}{1-\tan \widehat{BAC}.\tan 30^{\circ}}$

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

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 $α=\frac{π}{2}$ 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 = $\frac{30}{31}$

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

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

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

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

$=\sin \frac{2π}{3}.\text{cos}α-\text{cos}\frac{2π}{3}.\sin α$

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

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

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

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

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

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

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

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