`A= sinx. sin(60^o - x) . sin (60^o +x)`

`= sinx . 1/2(cos2x - cos120^o)`

`=sinx . 1/2 cos 2x + 1/4 sinx`

\(A=4sinx.sin\left(60^0-x\right).sin\left(60^0+x\right)\)

\(=2.sinx.\left(cos2x-cos120^0\right)\)

\(=2sinx\left(cos2x+\dfrac{1}{2}\right)\)

\(=2sinx.cos2x+sinx\)

a) \(A=2sin30^o+3cos45^o-sin60^0\)

\(\Leftrightarrow A=2.\dfrac{1}{2}+3.\dfrac{\sqrt[]{2}}{2}-\dfrac{\sqrt[]{3}}{2}\)

\(\Leftrightarrow A=1+\dfrac{3\sqrt[]{2}}{2}-\dfrac{\sqrt[]{3}}{2}\)

\(\Leftrightarrow A=1+\dfrac{\sqrt[]{3}\left(\sqrt[]{6}-1\right)}{2}\)

b) \(B=3cos30^o+3sin45^o-cos45^o\)

\(\Leftrightarrow B=3\dfrac{\sqrt[]{3}}{2}+3\dfrac{\sqrt[]{2}}{2}-\dfrac{\sqrt[]{2}}{2}\)

\(\Leftrightarrow B=\dfrac{3\sqrt[]{3}}{2}+\dfrac{2\sqrt[]{2}}{2}\)

\(\Leftrightarrow B=\dfrac{3\sqrt[]{3}}{2}+\sqrt[]{2}\)

 = \(\sqrt{3}\)

k mk nha

chị ơi cho em chịu câu này ạ

2028⁰F

ko bik đúng hay sai nữa

60 độ C = 140 độ F

d

D

\(A=sin42^0-cos48^0=cos\left(90^0-42^0\right)-cos48^0=cos48^0-cos48^0=0\)

\(B=cot56^0-tan34^0=tan\left(90^0-56^0\right)-tan34^0=tan34^0-tan34^0=0\)

\(C=sin30^0-cot50^0-cos60^0+tan40^0\)

\(=cos\left(90^0-30^0\right)-tan\left(90^0-50^0\right)-cos60^0+tan40^0\)

\(=cos60^0-tan40^0-cos60^0+tan40^0=0\)

\(A=\sin42^0-\cos48^0=\sin42^0-\sin42^0=0\)

\(B=\cot56^0-\tan34^0=\tan34^0-\tan34^0=0\)

= 60\(^0\)

60\(^0\)