Tự chứng minh từng cái này rồi suy ra cái đó nhé b.

Ta có: \(sin\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}-sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=sin^2\frac{A}{2}\)

Tương tự ta suy ra: 

\(sin\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}+cos\frac{A}{2}sin\frac{B}{2}cos\frac{C}{2}+cos\frac{A}{2}cos\frac{B}{2}sin\frac{C}{2}=sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}+3sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}\left(1\right)\)

Tiếp theo chứng minh:

\(2sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=\frac{cosA+cosB+cosC-1}{2}\left(2\right)\)

\(sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}=\frac{3}{2}-\frac{cosA+cosB+cosC}{2}\left(3\right)\)

\(tan\frac{A}{2}tan\frac{B}{2}+tan\frac{B}{2}tan\frac{C}{2}+tan\frac{C}{2}tan\frac{A}{2}=1\left(4\right)\)

Từ (1), (2), (3), (4) suy được điều phải chứng minh

ko hiểu ( vì em mới học lớp 6)

trinh le na

cho bạn 4 năm nữa cũng chưa hiểu đâu

Tham khảo:

a) Ta có: \(\widehat {AMB} + \widehat {AMC} = {180^o}\)

\( \Rightarrow \cos \widehat {AMB} =  - \cos \widehat {AMC}\)

Hay \(\cos \widehat {AMB} + \cos \widehat {AMC} = 0\)

b) Áp dụng định lí cos trong tam giác AMB ta có:

 \(\begin{array}{l}A{B^2} = M{A^2} + M{B^2} - 2MA.MB\;\cos \widehat {AMB}\\ \Leftrightarrow M{A^2} + M{B^2} - A{B^2} = 2MA.MB\;\cos \widehat {AMB}\;\;(1)\end{array}\)

Tương tự, Áp dụng định lí cos trong tam giác AMB ta được:

\(\begin{array}{l}A{C^2} = M{A^2} + M{C^2} - 2MA.MC\;\cos \widehat {AMC}\\ \Leftrightarrow M{A^2} + M{C^2} - A{C^2} = 2MA.MC\;\cos \widehat {AMC}\;\;(2)\end{array}\)

c) Từ (1), suy ra \(M{A^2} = A{B^2} - M{B^2} + 2MA.MB\;\cos \widehat {AMB}\;\)

Từ (2), suy ra \(M{A^2} = A{C^2} - M{C^2} + 2MA.MC\;\cos \widehat {AMC}\;\)

Cộng vế với vế ta được:

\(2M{A^2} = \left( {A{B^2} - M{B^2} + 2MA.MB\;\cos \widehat {AMB}} \right)\; + \left( {A{C^2} - M{C^2} + 2MA.MC\;\cos \widehat {AMC}} \right)\;\)

\( \Leftrightarrow 2M{A^2} = A{B^2} + A{C^2} - M{B^2} - M{C^2} + 2MA.MB\;\cos \widehat {AMB} + 2MA.MC\;\cos \widehat {AMC}\)

Mà: \(MB = MC = \frac{{BC}}{2}\) (do AM là trung tuyến)

\( \Rightarrow 2M{A^2} = A{B^2} + A{C^2} - {\left( {\frac{{BC}}{2}} \right)^2} - {\left( {\frac{{BC}}{2}} \right)^2} + 2MA.MB\;\cos \widehat {AMB} + 2MA.MB\;\cos \widehat {AMC}\)

\( \Leftrightarrow 2M{A^2} = A{B^2} + A{C^2} - 2.{\left( {\frac{{BC}}{2}} \right)^2} + 2MA.MB\;\left( {\cos \widehat {AMB} + \;\cos \widehat {AMC}} \right)\)

\( \Leftrightarrow 2M{A^2} = A{B^2} + A{C^2} - {\frac{{BC}}{2}^2}\)

\(\begin{array}{l} \Leftrightarrow M{A^2} = \frac{{A{B^2} + A{C^2} - {{\frac{{BC}}{2}}^2}}}{2}\\ \Leftrightarrow M{A^2} = \frac{{2\left( {A{B^2} + A{C^2}} \right) - B{C^2}}}{4}\end{array}\) (đpcm)

Cách 2:

Theo ý a, ta có: \(\cos \widehat {AMC} =  - \cos \widehat {AMB}\)

Từ đẳng thức (1): suy ra \(\cos \widehat {AMB} = \frac{{M{A^2} + M{B^2} - A{B^2}}}{{2.MA.MB}}\)

 \( \Rightarrow \cos \widehat {AMC} =  - \cos \widehat {AMB} =  - \frac{{M{A^2} + M{B^2} - A{B^2}}}{{2.MA.MB}}\)

Thế \(\cos \widehat {AMC}\)vào biểu thức (2), ta được:

\(M{A^2} + M{C^2} - A{C^2} = 2MA.MC.\left( { - \frac{{M{A^2} + M{B^2} - A{B^2}}}{{2.MA.MB}}} \right)\)

Lại có: \(MB = MC = \frac{{BC}}{2}\) (do AM là trung tuyến)

\(\begin{array}{l} \Rightarrow M{A^2} + {\left( {\frac{{BC}}{2}} \right)^2} - A{C^2} = 2MA.MB.\left( { - \frac{{M{A^2} + M{B^2} - A{B^2}}}{{2.MA.MB}}} \right)\\ \Leftrightarrow M{A^2} + {\left( {\frac{{BC}}{2}} \right)^2} - A{C^2} =  - \left( {M{A^2} + M{B^2} - A{B^2}} \right)\\ \Leftrightarrow M{A^2} + {\left( {\frac{{BC}}{2}} \right)^2} - A{C^2} + M{A^2} + {\left( {\frac{{BC}}{2}} \right)^2} - A{B^2} = 0\\ \Leftrightarrow 2M{A^2} - A{B^2} - A{C^2} + {\frac{{BC}}{2}^2} = 0\\ \Leftrightarrow 2M{A^2} = A{B^2} + A{C^2} - {\frac{{BC}}{2}^2}\\ \Leftrightarrow M{A^2} = \frac{{A{B^2} + A{C^2} - {{\frac{{BC}}{2}}^2}}}{2}\\ \Leftrightarrow M{A^2} = \frac{{2\left( {A{B^2} + A{C^2}} \right) - B{C^2}}}{4}\end{array}\)

\(A+B+C=180^0\Rightarrow A+B=180^0-C\)

\(\Rightarrow sin\left(A+B\right)=sin\left(180^0-C\right)=sinC\)

\(cos\left(A+B\right)=cos\left(180^0-C\right)=-cosC\)

\(tan\left(A+B\right)=tan\left(180^0-C\right)=-tanC\)

b/ \(\frac{A+B+C}{2}=90^0\Rightarrow\frac{A+B}{2}=90^0-\frac{C}{2}\)

\(\Rightarrow sin\frac{A+B}{2}=sin\left(90^0-\frac{C}{2}\right)=cos\frac{C}{2}\)

\(cos\frac{A+B}{2}=cos\left(90^0-\frac{C}{2}\right)=sin\frac{C}{2}\)

\(tan\frac{A+B}{2}=tan\left(90-\frac{C}{2}\right)=cot\frac{C}{2}\)

c/ \(A+B=180^0-C\Rightarrow tan\left(A+B\right)=-tanC\)

\(\Leftrightarrow\frac{tanA+tanB}{1-tanA.tanB}=-tanC\)

\(\Leftrightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)

\(\Leftrightarrow tanA+tanB+tanC=tanA.tanB.tanC\)

d/ \(sinA+sinB+sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)

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

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

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

\(=4cos\frac{C}{2}.cos\frac{A}{2}.cos\frac{B}{2}\)

e/

\(cosA+cosB+cosC=2cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2sin^2\frac{C}{2}\)

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

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

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

\(=1+4sin\frac{C}{2}.sin\frac{A}{2}sin\frac{B}{2}\)

f/

\(sin2A+sin2B+sin2C=2sin\left(A+B\right).cos\left(A-B\right)+2sinC.cosC\)

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

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

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

\(=4sinC.sinA.sinB\)

g/

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

\(=1+\frac{1}{2}\left(cos2A+cos2B\right)+cos^2C\)

\(=1+cos\left(A+B\right).cos\left(A-B\right)+cos^2C\)

\(=1-cosC.cos\left(A-B\right)+cos^2C\)

\(=1-cosC\left(cos\left(A-B\right)-cosC\right)\)

\(=1-cosC\left[cos\left(A-B\right)+cos\left(A+B\right)\right]\)

\(=1-2cosC.cosA.cosB\)

Xét các vec tơ đơn vị \(\frac{\overrightarrow{AB}}{AB};\frac{\overrightarrow{BC}}{BC};\frac{\overrightarrow{CA}}{CA}\) trên các cạnh AB, BC, CA của tam giác ABC

Có \(0\le\left(\frac{\overrightarrow{AB}}{AB};\frac{\overrightarrow{BC}}{BC};\frac{\overrightarrow{CA}}{CA}\right)^2=3-2\left(\cos A+\cos B+\cos C\right)\)

Suy ra \(\cos A+\cos B+\cos C\le\frac{3}{2}\) => Điều cần chứng minh

1) \(sin\left(A+2B+C\right)=sin\left(\pi-B+2B\right)\)

=\(sin\left(\pi+B\right)=sin\left(-B\right)=-sinB\)

2) \(sinBsinC-cosBcosC=-cos\left(B+C\right)\)

\(=-cos\left(\pi-A\right)=cosA\)

4) bạn ơi +2 vào vế phải mới đúng nhé

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

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

=\(-cos^2C-cos\left(A-B\right)cos\left(A+B\right)+2\)

\(=-cos^2C-\frac{1}{2}\left(cos2A+cos2B\right)+2\)

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

\(=-cos^2C-cos^2A+\frac{1}{2}-cos^2C+\frac{1}{2}+2\)

= sin²C - 1 + sin²A - 1 + sin²C - 1 + 3

= sin²A + sin²B + sin²C

Xét tam giác ABC, ta có:

\(\widehat A + \widehat B + \widehat C = {180^o} \Rightarrow \frac{{\widehat A}}{2} + \frac{{\widehat B + \widehat C}}{2} = {90^o}\)

Do đó \(\frac{{\widehat A}}{2}\) và \(\frac{{\widehat B + \widehat C}}{2}\) là hai góc phụ nhau.

a) Ta có: \(\sin \frac{A}{2} = \cos \left( {{{90}^o} - \frac{A}{2}} \right) = \cos \frac{{B + C}}{2}\)

b) Ta có: \(\tan \frac{{B + C}}{2} = \cot \left( {{{90}^o} - \frac{{B + C}}{2}} \right) = \cot \frac{A}{2}\)