Chương 6: CUNG VÀ GÓC LƯỢNG GIÁC. CÔNG THỨC LƯỢNG GIÁC

`sin \alpha +cos \alpha=3/5`

`<=>1+2sin \alpha cos \alpha=9/25`

`<=>sin \alpha cos \alpha=-8/25`

Có: `C=sin (180^o -\alpha).cos (180^o -\alpha)`

         `=sin \alpha .(-cos \alpha)`

         `=-sin \alpha.cos \alpha=-(-8/25)=8/25`

Mình sửa đề chút nha: 

\(C=sin\left(180-\alpha\right)cos\left(180-\alpha\right)\)

\(C=sin180^0-sin\alpha+cos180^0-cos\alpha\)

\(C=0+\left(-1\right)-\left(sin\alpha+cos\alpha\right)\)

\(C=-1-\dfrac{3}{5}=-\dfrac{8}{5}\)

Xét ΔABC có \(cosC=\dfrac{CA^2+CB^2-AB^2}{2\cdot CA\cdot CB}\)

\(\Leftrightarrow\dfrac{6^2+6.3^2-AB^2}{2\cdot6\cdot6.3}=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow AB\simeq4.72\left(cm\right)\)

Xét ΔABC có \(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=0.44\)

hay góc A=64 độ

=>góc B=71 độ

=sin 5x-sin x-4sin x

=2*cos3x*sin 2x-4*sinx

=4*cos3x*sinx*cosx-4sinx

=4sinx(cos3x*cosx-1)

\(A=1+\sin\left(\dfrac{x}{2}\right)\cdot\sin x-\cos\left(\dfrac{x}{2}\right)\cdot\sin^2x-2\cdot\dfrac{1+\cos\left(\dfrac{\Pi}{2}-x\right)}{2}\)

\(=1+\sin\left(\dfrac{x}{2}\right)\cdot\sin x-\cos\left(\dfrac{x}{2}\right)\cdot\sin^2x-1-\sin x\)

\(=\sin x\left[\sin\left(\dfrac{x}{2}\right)-\cos\left(\dfrac{x}{2}\right)\cdot\sin x-1\right]\)

\(=1+\sin x-\sin7x-\cos5x-2\cdot\dfrac{1+\cos3x}{2}\)

\(=1+\sin\left(\dfrac{x-7x}{2}\right)\cdot\cos\left(\dfrac{x+7x}{2}\right)-\cos5x-1-\cos3x\)

\(=\sin\left(-3x\right)\cdot\cos4x-\left(\cos5x+\cos3x\right)\)

\(=-\sin3x\cdot\cos4x-2\left[\cos4x\cdot\cos x\right]\)

\(=-\cos4x\left(\sin3x+2\cos x\right)\)

A=4cos^3x-8cos^2x+5cosx-1

=4cos^3x-4cos^2x-4cos^2x+4cosx+cosx-1

=(cosx-1)(4cos^2x-4cosx+1)

=(cosx-1)(2cosx-1)^2