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

\(\dfrac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}=\dfrac{1+cosx+2cos^2x-1+4cos^3x-3cosx}{2cos^2x+cosx-1}\)

\(=\dfrac{4cos^3x+2cos^2x-2cosx}{2cos^2x+cosx-1}=\dfrac{2cosx\left(2cos^2x+cosx-1\right)}{2cos^2x+cosx-1}=2cosx\)

Ta có: \(tan\alpha=2\Leftrightarrow\dfrac{sin\alpha}{cos\alpha}=2\Leftrightarrow sin\alpha=2cos\alpha\)

A = \(\dfrac{16cos^2\alpha+6cos^2\alpha}{20cos^2\alpha-2cos^2\alpha}=\dfrac{22cos^2\alpha}{18cos^2\alpha}=\dfrac{11}{9}\)

\(\frac{\left(sin^2x+tan^2x+1\right)\left(cos^2x-cot^2x+1\right)}{\left(cos^2x+cot^2x+1\right)\left(sin^2x+tan^2x-1\right)}=\frac{\left(sin^2x+\frac{1}{cos^2x}\right)\left(cos^2x-cot^2x+1\right)}{\left(cos^2x+\frac{1}{sin^2x}\right)\left(sin^2x+tan^2x-1\right)}\)

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

thanks ban nha

\(\dfrac{1+cosB}{1-cosB}=\dfrac{2a+c}{2a-c}\)

=>\(\dfrac{1+cosB}{1-cosB}=\dfrac{2sinA+sinC}{2sinA-sinC}\)

=>4*sinA*cosB=2*sinC

=>\(4\cdot\dfrac{a}{2R}\cdot\dfrac{a^2+c^2-b^2}{2ac}=2\cdot\dfrac{c}{2\cdot R}\)

=>a=b

=>ΔBAC cân tại C

\(VT=\frac{cos2a.tan2a}{cos2a\left(tan4a-tan2a\right)}=\frac{sin2a}{\frac{cos2a.sin4a}{cos4a}-sin2a}=\frac{cos4a.sin2a}{sin4a.cos2a-sin2a.cos4a}\)

\(=\frac{cos4a.sin2a}{sin\left(4a-2a\right)}=\frac{cos4a.sin2a}{sin2a}=cos4a\)

Cung tròn số đo 1 có chiều dài \(1.2=2\) cm

\(P=\frac{sin^23a.cos^2a-sin^2a.cos^23a}{\left(sina.cosa\right)^2}=\frac{\left(sin3a.cosa-cos3a.sina\right)\left(sin3a.cosa+cos3a.sina\right)}{\frac{1}{4}sin^22a}\)

\(=\frac{4sin\left(3a-a\right).sin\left(3a+a\right)}{sin^22a}=\frac{4sin2a.sin4a}{sin^22a}\)

\(=\frac{8sin^22a.cos2a}{sin^22a}=8cos2a\)

cung nhỏ nhất có số đo dương là cung lượng giác 165 độ

a: \(D=sin^210^0+\dfrac{1}{2}\cdot cos120-\dfrac{1}{2}\cdot cos20\)

\(=\dfrac{1-cos20}{2}-\dfrac{1}{2}\cdot cos20+\dfrac{1}{2}\cdot\dfrac{\sqrt{3}}{2}\)

\(=\dfrac{1}{2}-cos20+\dfrac{\sqrt{3}}{4}\)

\(=\dfrac{\sqrt{3}+2}{4}-cos20\)

b: \(=\dfrac{1-cos100}{2}+\dfrac{1-cos140}{2}-\dfrac{1}{2}\cdot\left(cos120+cos20\right)\)

\(=\dfrac{2-cos100-cos140-cos20-cos120}{2}\)

c: \(=\dfrac{1}{2}cos14x+\dfrac{1}{2}\cdot cos8x-\dfrac{1}{2}\cdot cos26x-\dfrac{1}{2}\cdot cos8x\)

\(=\dfrac{1}{2}\cdot cos14x-\dfrac{1}{2}\cdot cos26x\)

d: \(=\dfrac{-1}{2}\cdot cos4x+\dfrac{1}{2}\cdot cos2x-\dfrac{1}{2}\cdot cos12x+\dfrac{1}{2}\cdot cos4x\)

\(=\dfrac{1}{2}\cdot cos2x-\dfrac{1}{2}\cdot cos12x\)