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

Do x thuộc cung phần tư thứ \(IV\) \(\Rightarrow\left\{{}\begin{matrix}sinx< 0\\cosx>0\end{matrix}\right.\) \(\Rightarrow sinx-cosx< 0\)

\(sinx+cosx=m\Rightarrow\left(sinx+cosx\right)^2=m^2\)

\(\Rightarrow1+2sinx.cosx=m^2\Rightarrow2sinx.cosx=m^2-1\)

Đặt \(P=sinx-cosx< 0\Rightarrow P^2=\left(sinx-cosx\right)^2=1-2sinx.cosx\)

\(\Rightarrow P^2=1-\left(m^2-1\right)=2-m^2\Rightarrow P=-\sqrt{2-m^2}\) (do \(P< 0\))

Câu 3:

Ta có \(sin^2\frac{A}{2}=\frac{1-cosA}{2}=\frac{1-\frac{b^2+c^2-a^2}{2bc}}{2}=\frac{a^2-b^2-c^2+2bc}{4bc}=\frac{a^2-\left(b-c\right)^2}{4bc}\)

\(=\frac{\left(a+b-c\right)\left(a+c-b\right)}{4bc}=\frac{\left(p-c\right)\left(p-b\right)}{bc}\Rightarrow sin\frac{A}{2}=\sqrt{\frac{\left(p-b\right)\left(p-c\right)}{bc}}\)

Tương tự ta có \(sin\frac{B}{2}=\sqrt{\frac{\left(p-a\right)\left(p-c\right)}{ac}}\) ; \(sin\frac{C}{2}=\sqrt{\frac{\left(p-a\right)\left(p-b\right)}{ab}}\)

\(\Rightarrow4Rsin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=4\left(\frac{abc}{4S}\right)\sqrt{\frac{\left(p-a\right)^2\left(p-b\right)^2\left(p-c\right)^2}{a^2b^2c^2}}\)

\(=\frac{abc.\left(p-a\right)\left(p-b\right)\left(p-c\right)}{S.abc}=\frac{\left(p-a\right)\left(p-b\right)\left(p-c\right)}{S}=\frac{\left(p-a\right)\left(p-b\right)\left(p-c\right)}{\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}}=\sqrt{\frac{\left(p-a\right)\left(p-b\right)\left(p-c\right)}{p}}=r\)

Câu 1:

\(a.sin\left(B-C\right)=a.sinBcosC-a.cosB.sinC\)

\(bsin\left(C-A\right)=bsinC.cosA-bcosC.sinA\)

\(csin\left(A-B\right)=csinAcosB-csinB.cosA\)

Cộng lại:

\(VT=cosA\left(bsinC-c.sinB\right)+cosB\left(c.sinA-a.sinC\right)+cosC\left(a.sinB-bsinA\right)\)

\(=cosA\left(\frac{b.c}{2R}-\frac{bc}{2R}\right)+cosB\left(\frac{ac}{2R}-\frac{ac}{2R}\right)+cosC\left(\frac{ab}{2R}-\frac{ab}{2R}\right)=0\)

Câu 2:

\(sin^2A+sin^2B+sin^2C=\frac{1}{2}-\frac{1}{2}cos2A+\frac{1}{2}-\frac{1}{2}cos2B+1-cos^2C\)

\(=2-\frac{1}{2}\left(cos2A+cos2B\right)-cosC.cosC\)

\(=2-cos\left(A+B\right)cos\left(A-B\right)+cosC.cos\left(A+B\right)\)

\(=2+cosC.cos\left(A-B\right)+cosC.cos\left(A+B\right)\)

\(=2+cosC\left[cos\left(A-B\right)+cos\left(A+B\right)\right]\)

\(=2+2cosA.cosB.cosC\)

\(A=sina+\sqrt{3}cosa=2\left(\frac{1}{2}sina+\frac{\sqrt{3}}{2}cosa\right)\)

\(2=\left(sina.cos\frac{\pi}{3}+cosa.sin\frac{\pi}{3}\right)=2sin\left(a+\frac{\pi}{3}\right)\ge-2\)

\(\Rightarrow A_{min}=-2\) khi \(sin\left(a+\frac{\pi}{3}\right)=-1\Rightarrow a=-\frac{5\pi}{6}+k2\pi\)

\(VT=\frac{tanx}{sinx}-\frac{sinx}{cotx}\\ =\frac{sinx}{cosx}.\frac{1}{sinx}-\frac{sinx.sinx}{cosx}\\ =\frac{1-sin^2x}{cosx}\\ =cosx\\ =VP\)

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

\(=\frac{sinx.cos^2x}{cosx.sinx}=cosx\)

\(\frac{sin2x-cosx}{2sinx-1}+sinx=\frac{2sinx.cosx-cosx}{2sinx-1}+sinx\)

\(=\frac{cosx\left(2sinx-1\right)}{2sinx-1}+sinx=cosx+sinx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)

\(A=\frac{1}{2}+\frac{1}{2}cos\left(2a+2b\right)+\frac{1}{2}+\frac{1}{2}cos\left(2a-2b\right)-cos2a.cos2b\)

\(=1+\frac{1}{2}\left[cos\left(2a+2b\right)+cos\left(2a-2b\right)\right]-cos2a.cos2b\)

\(=1+cos2a.cos2b-cos2a.cos2b\)

\(=1\)

Theo Viet ta có \(\left\{{}\begin{matrix}tana+tanb=p\\tana.tanb=q\end{matrix}\right.\)

\(\Rightarrow tan\left(a+b\right)=\frac{tana+tanb}{1-tana.tanb}=\frac{p}{1-q}\)

\(A=cos^2\left(a+b\right)\left[1+p.tan\left(a+b\right)+q.tan^2\left(a+b\right)\right]\)

\(A=\frac{1}{1+tan^2\left(a+b\right)}\left[1+\frac{p^2}{1-q}+\frac{q.p^2}{\left(1-q\right)^2}\right]\)

\(A=\frac{\left(1-q\right)^2}{p^2+\left(1-q\right)^2}\left(1+\frac{p^2}{\left(1-q^2\right)}\right)\)

\(A=\frac{\left(1-q^2\right)}{p^2+\left(1-q\right)^2}.\left(\frac{p^2+\left(1-q\right)^2}{\left(1-q\right)^2}\right)=1\)

\(=cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+sin\left(\frac{\pi}{2}-x-\frac{\pi}{6}\right)sin\left(\frac{\pi}{2}-x-\frac{3\pi}{4}\right)\)

\(=cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+sin\left(\frac{\pi}{3}-x\right)sin\left(-x-\frac{\pi}{4}\right)\)

\(=cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+sin\left(x-\frac{\pi}{3}\right)sin\left(x+\frac{\pi}{4}\right)\)

\(=cos\left(x-\frac{\pi}{3}-x-\frac{\pi}{4}\right)=cos\left(-\frac{7\pi}{12}\right)=cos\frac{7\pi}{12}=\frac{\sqrt{2}-\sqrt{6}}{4}\)

\(T=a\left(2a+1\right)+b\left(b+1\right)+\frac{2a+3b}{ab}\)

\(T=\left(2a^2+b^2\right)+a+b+\frac{2}{b}+\frac{3}{a}\)

Xét: \(2a^2+b^2=\frac{4a^2}{2}+b^2\ge\frac{\left(2a+b\right)^2}{2}\ge\frac{9}{2}\)

\(a+b+\frac{2}{b}+\frac{3}{a}=a+\frac{1}{a}+\frac{2}{a}+b+\frac{1}{b}+\frac{1}{b}\)

\(\ge4+\frac{2}{a}+\frac{1}{b}=\frac{4}{2a}+\frac{1}{b}\ge4+\frac{25}{3}=\frac{37}{3}\)

Cộng theo vế tìm được min T

Câu 2: đường tròn tâm \(O\left(1;2\right)\) ; \(R=2\)

Do \(M\in d\Rightarrow M\left(a;a+7\right)\)

\(OM^2=\left(a-1\right)^2+\left(a+5\right)^2=2a^2+8a+26\)

\(\Rightarrow MA^2=MB^2=OM^2-R^2=\left(a-1\right)^2+\left(a+5\right)^2-4=2a^2+8a+22\)

Ta có \(\Delta OAM=\Delta OBM\Rightarrow S_{OAMB}=2S_{OAM}=OA.AM=R.AM\)

Mặt khác do \(OM\perp AB\) (tính chất đường tròn)

\(\Rightarrow S_{OAMB}=AB.OM\)

\(\Rightarrow AB.OM=R.AM\Rightarrow AB^2=\frac{R^2.AM^2}{OM^2}=\frac{4\left(2a^2+8a+22\right)}{2a^2+8a+26}=\frac{4\left(a^2+4a+11\right)}{a^2+4a+13}\)

\(\Rightarrow AB^2=4-\frac{8}{a^2+4a+13}=4-\frac{8}{\left(a+2\right)^2+9}\ge4-\frac{8}{9}=\frac{28}{9}\)

\(\Rightarrow AB_{min}=\frac{2\sqrt{7}}{3}\) khi \(a=-2\Rightarrow b=5\Rightarrow a+b=3\)

Câu 1:

\(P=4sin\frac{A+B}{2}cos\frac{A-B}{2}-2\left(2cos^2\frac{C}{2}-1\right)\)

\(P=4cos\frac{C}{2}cos\frac{A-B}{2}-4cos^2\frac{C}{2}+2\)

\(\Leftrightarrow4cos^2\frac{C}{2}-4cos\frac{C}{2}.cos\frac{A-B}{2}+P-2=0\)

Đặt \(x=cos\frac{C}{2}\)

\(\Rightarrow4x^2-4cos\frac{A-B}{2}.x+P-2=0\) (1)

Do góc C luôn tồn tại \(\Rightarrow\) phương trình (1) luôn có ít nhất 1 nghiệm

\(\Delta'=4cos^2\frac{A-B}{2}-4\left(P-2\right)\ge0\)

\(\Leftrightarrow4cos^2\frac{A-B}{2}+8\ge4P\Rightarrow P\le cos^2\frac{A-B}{2}+2\le3\)

\(\Rightarrow P_{max}=3\) khi \(\left\{{}\begin{matrix}A=B\\cos\frac{C}{2}=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}A=B=30^0\\C=120^0\end{matrix}\right.\)