§3. Công thức lượng giác

BC/sinA=2R

=>2R=28,5:1/2=57

=>d=57

a.

\(S=\dfrac{1}{2}bc.sinA=125\sqrt{3}\)

\(a=\sqrt{b^2+c^2-2bc.cosA}=5\sqrt{21}\)

\(\Rightarrow h_a=\dfrac{2S}{a}=\dfrac{50}{\sqrt{7}}\)

\(p=\dfrac{a+b+c}{2}=\dfrac{20+25+5\sqrt{21}}{2}\Rightarrow r=\dfrac{S}{p}=...\)

\(R=\dfrac{abc}{4S}=\dfrac{20.25.5\sqrt{21}}{4.125\sqrt{3}}=...\)

b.

\(p=\dfrac{a+b+c}{2}=21\)

\(S=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}=84\)

\(h_b=\dfrac{2S}{b}=12\)

\(R=\dfrac{abc}{4S}=\dfrac{65}{8}\)

\(r=\dfrac{S}{p}=4\)

\(P=\dfrac{\dfrac{sina}{cos^3a}+\dfrac{cos^3a}{cos^3a}}{\dfrac{sina}{cos^3a}+\dfrac{2cosa}{cos^3a}}=\dfrac{tana.\dfrac{1}{cos^2a}+1}{tana.\dfrac{1}{cos^2a}+\dfrac{2}{cos^2a}}=\dfrac{tana\left(1+tan^2a\right)+1}{tana.\left(1+tan^2a\right)+2\left(1+tan^2a\right)}\)

Tới đây thay số và bấm máy

\(\widehat{A}=180^o-56^o13'-71^o=52^o47'\)

áp dụng định lý sin

\(\dfrac{BC}{sinA}=\dfrac{AB}{sinC}\\ =>AB=\dfrac{BC.sinC}{sinA}=\dfrac{16,8\times sin71^o}{sin52^o47'}=19,94\left(cm\right)\)

Đặt x=a/2

=>tan x=cot x-2cot 2x

=>\(tanx-cotx=-2\cdot\dfrac{cos2x}{sin2x}\)

=>\(\dfrac{sinx}{cosx}-\dfrac{cosx}{sinx}=\dfrac{-2\cdot cos2x}{sin2x}\)

=>\(\dfrac{sin^2x-cos^2x}{sinx\cdot cosx}=\dfrac{-cos2x}{sinx\cdot cosx}\)

=>sin^2x-cos^2x=-cos2x(luôn đúng)

2: =2*sin2x*cosx+sin2x

=2*sin2x(cosx+1)

3: =2*cos2x*cosx+cos2x

=2*cos2x(cosx+1)

`1) cos x + sin 2x - cos 3x`

`= -2sin 2x . (-sin x) + sin 2x`

`= sin 2x ( 2 sin x + 1 )`

Cấu 2 hình như sai đề bạn ạ phải là `sin 3x + sin x` chứ :v

\(A=cosx+cos\left(n+y\right)+cos\left(x+2y\right)+...+cos\left(x+ny\right)=\left(n+1\right)cosn\)

\(\dfrac{sinx+sin3x+sin5x+...+sin\left(2n-1\right)x}{cosx+cos3x+cos5x+...+cos\left(2n-1\right)x}\) 

                                            \(=tan\left(nx\right)\)

\(sinx+sin2x+sin3x+...+sinnx\)

               \(=\dfrac{sin\dfrac{nx}{2}sin\dfrac{\left(n+1\right)x}{2}}{sin\dfrac{x}{2}}\)

\(cosx+cos2x+cos3x+cosnx\)

              \(=\dfrac{sin\dfrac{nx}{2}cos\dfrac{\left(n+1\right)x}{2}}{sin\dfrac{x}{2}}\)