Gọi tam giác vuông đề bài cho là ΔABC vuông tại A, có \(\hat{B}=\alpha\)

Xét ΔABC vuông tại A có

\(\sin\alpha=\sin B=\frac{AC}{BC};cos\alpha=cosB=\frac{AB}{BC}\)

\(\tan\alpha=\tan B=\frac{AC}{AB};\cot\alpha=\cot B=\frac{AB}{AC}\)

\(\frac{\sin\alpha}{cos\alpha}=\frac{AC}{BC}:\frac{AB}{BC}=\frac{AC}{AB}=tan\alpha\)

\(\frac{cosa}{\sin a}=\frac{cosB}{\sin B}=\frac{AB}{BC}:\frac{AC}{BC}=\frac{AB}{AC}=\cot\alpha\)

\(\tan a\cdot\cot\alpha=\frac{\sin\alpha}{cos\alpha}\cdot\frac{cos\alpha}{\sin\alpha}=1\)

\(\sin^2\alpha+cos^2\alpha=\left(\frac{AB}{BC}\right)^2+\left(\frac{AC}{BC}\right)^2=\frac{AB^2+AC^2}{BC^2}=\frac{BC^2}{BC^2}=1\)

sin3x=sin(2x+x)=sin2xcoxx+cox2xsinx 
=2sinxcox^2 x+(1-2sin^2 x)sinx 
=2sinxcox^2 x+ sinx-2sin^3 x 
=sinx(2cos^2 x +1) - 2sin^3 x 
=sinx(2-2sin^2 x +1) - 2sin^3 x 
=3sinx - 4 sin^3 x. 
cos3x=cox(2x+x)=cos2xcosx-sin2xsinx 
=(2cos^2 x-1)cosx-2sin^2 xcosx 
=2cos^3 x-cosx-(2-cos^2 x)cosx 
=2cos^3 x -cosx-2coxx+2cos^3 x 
=4cos^3 x - 3cosx. 

=> tan 3a= sin3a/cos3a rồi ra

1/

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

\(=\frac{2\left(\frac{1-cos2x}{2}\right)}{sin2x}=\frac{1-cos2x}{sin2x}\)

2/

\(\frac{sin\left(60-x\right)cos\left(30-x\right)+cos\left(60-x\right)sin\left(30-x\right)}{sin4x}=\frac{sin\left(60-x+30-x\right)}{sin4x}=\frac{sin\left(90-2x\right)}{2sin2x.cos2x}\)

\(=\frac{cos2x}{2sin2x.cos2x}=\frac{1}{2sin2x}\)

3/

\(4cos\left(60+a\right)cos\left(60-a\right)+2sin^2a\)

\(=2\left(cos\left(60+a+60-a\right)+cos\left(60+a-60+a\right)\right)+2sin^2a\)

\(=2cos120+2cos2a+2\left(\frac{1-cos2a}{2}\right)\)

\(=-1+2cos2a+1-cos2a=cos2a\)

\(cos2a=\frac14\)

=>\(2\cdot cos^2a-1=\frac14\)

=>\(2\cdot cos^2a=\frac54\)

=>\(cos^2a=\frac58\)

\(\sin^2a+cos^2a=1\)

=>\(\sin^2a=1-\frac58=\frac38\)

\(P=3\cdot\sin^2a-4\cdot cos^2a\)

\(=3\cdot\frac38-4\cdot\frac58=\frac{9-20}{8}=-\frac{11}{8}\)

\(C=sin^4a\left(3-2sin^2a\right)+cos^4a\left(3-2cos^2a\right)\)

\(=sin^4a\left(1+2cos^2a\right)+cos^4a\left(1+2sin^2a\right)\)

\(=sin^4a+cos^4a+2sin^2a.cos^2a\left(sin^2a+cos^2a\right)\)

\(=sin^4a+cos^4a+2sin^2a.cos^2a=\left(sin^2a+cos^2a\right)^2=1\)

a: \(a^4=b^4+c^4\)

=>\(a^4>c^4\) và \(a^4>b^4\)

=>a>c và a>b

=>a là cạnh lớn nhất

=>\(\hat{BAC}\) là góc lớn nhất

\(a^4=b^4+c^4\)

=>\(a^4 = (b^2+c^2)^2 - 2b^2c^2 < (b^2+c^2)^2 \Rightarrow a^2 < b^2 + c^2\)

=>\(b^2+c^2-a^2>0\)

=>\(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\) >0

=>\(\hat{A}<90^0\)

mà \(\hat{BAC}\) đã là góc lớn nhất trong ΔABC

nên \(\hat{ABC}<90^0;\hat{ACB}<90^0\)

b: Xét ΔABC có \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\)

=>\(\sin A=\frac{a}{2R}\) ; \(\sin B = \frac{b}{2R}, \quad \sin C = \frac{c}{2R}\)

Xét ΔABC có \(\cos B = \frac{a^2 + c^2 - b^2}{2ac}, \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab}\)

\(\tan B \cdot \tan C = \frac{\sin B \cdot \sin C}{\cos B \cdot \cos C}\)

\(=\frac{\frac{b}{2R} \cdot\frac{c}{2R}}{\frac{a^2 + c^2 - b^2}{2ac} \cdot\frac{a^2 + b^2 - c^2}{2ab}}\)

\(=\frac{\frac{bc}{4R^2}}{\frac{(a^2 + c^2 - b^2)(a^2 + b^2 - c^2)}{4a^2bc}}=\frac{a^2 b^2 c^2}{R^2 (a^2 + c^2 - b^2)(a^2 + b^2 - c^2)}\)

\(=\frac{a^2b^2c^2}{R^2\cdot\left\lbrack(a^2+c^2-b^2)(a^2+b^2-c^2)\right\rbrack}\)

\(=\frac{a^2b^2c^2}{R^2\left\lbrace\left[a^2+(c^2-b^2)\right]\left[a^2-(c^2-b^2)\right]\right\rbrace}\)

\(=\frac{a^2b^2c^2}{R^2\cdot\left\lbrack a^4-\left(c^2-b^2\right)^2\right\rbrack}=\frac{a^2b^2c^2}{R^2\cdot\left\lbrack a^4-(c^4-2b^2c^2+b^4)\right\rbrack}\)

\(=\frac{a^2b^2c^2}{R^2\cdot\left\lbrack a^4-(b^4+c^4)+2b^2c^2\right\rbrack}=\frac{a^2b^2c^2}{R^2\cdot2b^2c^2}=\frac{a^2}{2R^2}=2\cdot\sin^2A\)

MN K BT?