Question

[Thầy Đức Anh]

Tính giá trị biểu thức (không sử dụng máy tính cầm tay)

a) \(M=\dfrac{\sin48^{\circ}}{\cos42^{\circ}}-\cos60^{\circ}+\tan27^{\circ}.\tan63^{\circ}\)

b) \(N=\cot27^{\circ}.\cot60^{\circ}.\cot63^{\circ}+\sin^244+\sin^246\)

Answer 1

[Xyz OLM]

Ta có \(M=\dfrac{sin48^{\text{o}}}{cos42^{\text{o}}}-cos60^{\text{o}}+tan27^{\text{o}}.tan63^{\text{o}}\)

\(=\dfrac{sin48^{\text{o}}}{sin48^o}-cos60^{\text{o}}+tan27^o.cot27^o=1-cos60^{\text{o}}+1\)

\(=2-cos60^{\text{o}}=2-\dfrac{1}{2}=\dfrac{3}{2}\)

N = \(cot27^{\text{o}}.cot60^{\text{o}}.cot63^{\text{o}}+sin^244^{\text{o}}+sin^246^{\text{o}}\)

\(=cot27^{\text{o}}.tan27^{\text{o}}.cot60^{\text{o}}+sin^244^{\text{o}}+cos^244^{\text{o}}=cot60^{\text{o}}+1=\dfrac{1}{\sqrt{3}}+1\)

\(=\dfrac{\sqrt{3}+3}{3}\)

Answer 2

[Vũ Thị Minh Hà]

Ta có $M=\genfrac{}{}{0ex}{}{sin48^{\text{o}}}{cos42^{\text{o}}}-cos60^{\text{o}}+tan27^{\text{o}}.tan63^{\text{o}}$

$=\genfrac{}{}{0ex}{}{sin48^{\text{o}}}{sin48^o}-cos60^{\text{o}}+tan27^o.cot27^o=1-cos60^{\text{o}}+1$

$=2-cos60^{\text{o}}=2-\genfrac{}{}{0ex}{}{1}{2}=\genfrac{}{}{0ex}{}{3}{2}$

N = $cot27^{\text{o}}.cot60^{\text{o}}.cot63^{\text{o}}+si{n}^{2}44^{\text{o}}+si{n}^{2}46^{\text{o}}$

$=cot27^{\text{o}}.tan27^{\text{o}}.cot60^{\text{o}}+si{n}^{2}44^{\text{o}}+co{s}^{2}44^{\text{o}}=cot60^{\text{o}}+1=\genfrac{}{}{0ex}{}{1}{\sqrt{3}}+1$

$=\genfrac{}{}{0ex}{}{\sqrt{3}+3}{3}$