Chương 6: CUNG VÀ GÓC LƯỢNG GIÁC. CÔNG THỨC LƯỢNG GIÁC - Hỏi đáp

Sử dụng công thức cộng là xong thôi:

\(\frac{1}{2}\left(cos2x+cos\left(2a-2x\right)\right)\)

\(=\frac{1}{2}.2cos\left(\frac{2x+2a-2x}{2}\right).cos\left(\frac{2x-2a+2x}{2}\right)\)

\(=cosa.cos\left(2x-a\right)\)

Đề bài sai

\(5sin\left(a+b-b\right)=3sin\left(a+b+b\right)\)

\(\Leftrightarrow5sin\left(a+b\right)cosb-5cos\left(a+b\right)sinb=3sin\left(a+b\right)cosb+3cos\left(a+b\right)sinb\)

\(\Leftrightarrow2sin\left(a+b\right)cosb=8cos\left(a+b\right)sinb\)

\(\Rightarrow\frac{sin\left(a+b\right)}{cos\left(a+b\right)}=\frac{4sinb}{cosb}\Rightarrow tan\left(a+b\right)=4tanb\)

2.

\(2sin\frac{A}{2}cos\frac{A}{2}=\frac{2sin\frac{B+C}{2}cos\frac{B-C}{2}}{2cos\frac{B+C}{2}cos\frac{B-C}{2}}=\frac{cos\frac{A}{2}}{sin\frac{A}{2}}\)

\(\Leftrightarrow2sin^2\frac{A}{2}=1\Leftrightarrow1-2sin^2\frac{A}{2}=0\)

\(\Leftrightarrow cosA=0\Rightarrow A=90^0\)

\(A=\frac{sinx}{cosx}+\frac{cosx}{sinx}+\frac{sin3x}{cos3x}+\frac{cos3x}{sin3x}\)

\(=\frac{sin^2x+cos^2x}{sinx.cosx}+\frac{sin^23x+cos^23x}{sin3x.cos3x}=\frac{2}{2sinx.cosx}+\frac{2}{2sin3x.cos3x}\)

\(=\frac{2}{sin2x}+\frac{2}{sin6x}=\frac{2\left(sin2x+sin6x\right)}{sin2x.sin6x}=\frac{4sin4x.cos2x}{sin2x.sin6x}\)

\(=\frac{8sin2x.cos^22x}{sin2x.sin6x}=\frac{8cos^22x}{sin6x}\)

\(B=\frac{sin30}{cos30}+\frac{sin60}{cos60}+\frac{sin40}{cos40}+\frac{sin50}{cos50}=\frac{sin30.cos60+cos30.sin60}{cos30.cos60}+\frac{sin40.cos50+sin50.cos40}{cos40.cos50}\)

\(=\frac{sin90}{cos30.cos60}+\frac{sin90}{cos40.cos50}=\frac{1}{\frac{1}{2}.\frac{\sqrt{3}}{2}}+\frac{1}{\frac{1}{2}cos90+\frac{1}{2}cos10}\)

\(=\frac{4\sqrt{3}}{3}+\frac{2}{cos10}=\frac{4\sqrt{3}\left(cos10+\frac{\sqrt{3}}{2}\right)}{3cos10}=\frac{4\sqrt{3}\left(cos10+cos30\right)}{3cos10}\)

\(=\frac{8\sqrt{3}cos20.cos10}{3cos10}=\frac{8\sqrt{3}}{3}cos20\)

\(A=\frac{1}{2}-\frac{1}{2}cos\left(2a-2b\right)+\frac{1}{2}-\frac{1}{2}cos2b+2sin\left(a-b\right)sinb.cosa\)

\(=1-\frac{1}{2}\left[cos\left(2a-2b\right)+cos2b\right]+2sin\left(a-b\right)sinb.cosa\)

\(=1-cosa.cos\left(a-2b\right)+2sin\left(a-b\right).sinb.cosa\)

\(=1-cosa\left[cos\left(a-2b\right)-2sin\left(a-b\right)sinb\right]\)

\(=1-cosa\left[cos\left(a-2b\right)+cosa-cos\left(a-2b\right)\right]\)

\(=1-cosa^2=sin^2a\)

Hoàn toàn tương tự:

\(B=1+cos\left(2a+b\right).cosb-2cosa.cosb.cos\left(a+b\right)\)

\(=1+cosb\left[cos\left(2a+b\right)-2cosa.cos\left(a+b\right)\right]\)

\(=1+cosb\left[cos\left(2a+b\right)-cos\left(2a+b\right)-cosb\right]\)

\(=1-cos^2b=sin^2b\)

Lời giải:

$A=\frac{2\cos \frac{2x+y}{2}\sin \frac{x}{2}}{2\sin \frac{2x+y}{2}.\cos \frac{x}{2}}-\frac{2\cos \frac{2x+y}{2}\cos \frac{x}{2}}{-2\sin \frac{2x+y}{2}\sin \frac{x}{2}}$

$=\tan \frac{x}{2}.\cot \frac{2x+y}{2}+\cot \frac{x}{2}.\cot \frac{2x+y}{2}=\cot \frac{2x+y}{2}(\tan \frac{x}{2}+\cot \frac{x}{2})$

Đề bài sai

Ví dụ: cho \(x=0\) thì vế phải bằng 0, vế trái bằng 5

Hai vế ko hề bằng nhau

Lời giải:

$8\sin ^2x\sin (x+60)\sin (x-60)=-4\sin ^2x[-2\sin (x+60)\sin (x-60)]$

$=-4\sin ^2x(\cos 2x-\cos 120^0)$

$=-4\sin ^2x(\cos 2x+\frac{1}{2})=-2(1-\cos 2x)(\cos 2x+\frac{1}{2})$

$=(\cos 2x-1)(2\cos 2x+1)=2\cos ^22x-1-\cos 2x$

$=\cos ^22x-\sin ^22x-\cos 2x=\cos 4x-\cos 2x$ (đpcm)

\(A=sin^21^o+c\text{os}^22^o+sin^23^o+c\text{os}^24^o+...+sin^2179^o+c\text{os}^2180^o\)

\(=sin^21^o+c\text{os}^22^o+sin^23^o+c\text{os}^24^o+...+c\text{os}^290^o-sin^289^o-c\text{os}^288^o-...-sin^21^o-c\text{os}^20^o\)

\(=c\text{os}^290^o-c\text{os}^20^o\)

\(=-1\)

Chúc bn học tốt

A=sin²40+cos²10+2sin40cos10-cos²40-sin²10-2sin10cos40+cos(90+50)

A=(sin²40-cos²40)+(cos²10-sin²10)+2(sin40cos10-cos40sin10)-sin50

A=(sin40-cos40)(sin40+cos40)-(sin10-cos10)(sin10+cos10)+1-sin50

A=\(\sqrt{2}\) sin(40-\(\frac{\pi}{4}\))\(\sqrt{2}\) cos(40-\(\frac{\pi}{4}\))-\(\sqrt{2}\)sin(10-\(\frac{\pi}{4}\))\(\sqrt{2}\) cos(10-\(\frac{\pi}{4}\))+1-sin50

A=-2sin5cos5+2sin35cos35+1-sin50

A= - sin10+sin70+1-sin50

A= 2cos40sin30-sin(90-40)+1

A=cos40-cos40+1 =1

\(sin5x-2sinx\left(cos4x+cos2x\right)=sin5x-2.2sinx.cosx.cos3x\)

\(=sin5x-2sin2x.cos3x\)

\(=sin5x-\left(sin5x+sin\left(-x\right)\right)\)

\(=-sin\left(-x\right)=sinx\)

\(=\frac{1+cosx}{sinx}\left(1-\frac{1+cos^2x-2cosx}{sin^2x}\right)=\frac{1+cosx}{sinx}\left(\frac{sin^2x-1-cos^2x+2cosx}{sin^2x}\right)\)

\(=\frac{1+cosx}{sinx}\left(\frac{-cos^2x-cos^2x+2cosx}{sin^2x}\right)=\frac{\left(1+cosx\right)2cosx\left(1-cosx\right)}{sinx.sin^2x}\)

\(=\frac{2cosx\left(1-cos^2x\right)}{sinx.sin^2x}=\frac{2cosx.sin^2x}{sinx.sin^2x}=\frac{2cosx}{sinx}=2cotx\)

Lời giải:

a)

\(\sin x\cos x=\frac{(\sin x+\cos x)^2-\sin^2x-\cos^2x}{2}=\frac{m^2-1}{2}\)

b) Áp dụng kết quả phần a.

\(\sin^4 x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\left (\frac{m^2-1}{2}\right)^2=\frac{2m^2-m^4+1}{2}\)