The number of male students in a class accounts for 2 / 3 of the number of female students, the number of male students accounts for a few of the number of the whole class, and the number of female students accounts for a few of the number of the whole class

The number of male students in a class accounts for 2 / 3 of the number of female students, the number of male students accounts for a few of the number of the whole class, and the number of female students accounts for a few of the number of the whole class

The number of boys in a class accounts for 2 / 3 of the number of girls, boys account for 2 / 5 of the total number of the class, girls account for 3 / 5 of the total number of the class
PS: to answer this question makes me feel cheated
The calculation process of sin110 ° sin40 ° + cos70 ° cos40 ° is described in detail,
Sin110 ° sin40 ° cos70 ° cos40 ° (plus) = sin (180-110 °) sin40 ° cos70 ° cos40 °
=Sin70 ° sin40 ° cos70 ° cos40 ° (plus) = cos (70 ° - 40 °) = cos30 ° = √ 3 / 2
Sin110 is sin70, not yet? Question: which formula?
There are 23 boys and 22 girls in class 51. The number of girls is a fraction of the number of boys and the number of boys is a fraction of the number of girls
The number of female students is 22 / 23 of the number of male students; the number of male students is 23 / 22 of the number of female students
The value of sin40 ° cos110 ° + cos40 ° sin110 °
Seeking process
sin40°cos110°+cos40°sin110°
=sin（40°+110°）
=sin（150°）
=sin（180°-150°）
=sin30°
=1/2
The number of male students is 2 / 5 more than that of female students. How many times is the number of male students more than that of female students? How many times is the number of female students less than that of male students?
Primary school solution
Suppose the number of girls is 10, and the number of boys is 10 times 5 / 2 plus 10, which is 14
So 14 boys are seven fifths of 10 girls
There are four less girls than boys. It depends on who you are
Let a = (cos40 ° sin40 ° and B = (sin110 ° cos110 °), then the vector ab=___ (expressed by the simplest result)
ab=sin110°cos40°+cos110°sin40°=sin150°=1/2
The number of male students in Grade 6 is 80% of that of female students. Then, what is the number of male students in Grade 6
Regarding female students as the whole "1"
80% for boys
1 + 80% = 180% in the whole class
80% ÷ 180 % = 4/9
Algebra problem cos70'cos40'tsin70'sin40
-- or you'll lose face
cos70'cos40'+sin70'sin40
=cos(70°-40°)
=cos30°
=√3/2
=Cos (70-40) = cos 30 = √ 3 / 2.
The number of male students is 80% of that of female students. How many parts of the class are female students?
Five out of nine,
Let the female be x, then the male be 0.8x
The total score of the class is 1.8x
The ratio of X to 1.8x is five ninths
Find the trigonometric ratio Tan (5pai / 12) sin 165 degree, simplify cos 20 degree cos (A-20 degree) - cos 70 degree sin (A-20 degree)
Given sin square O + sin o = 1, find the value of COS square O + cos fourth power o
1）tan(5π/12)=tan(π/4+π/6)=[tan(π/4)+tan(π/6)]/[1-tan(π/4)tan(π/6)]
=(1+√3/3)/(1-√3/3)=2+√3 ；
sin165°=sin(120°+45°)=sin120°cos45°+cos120°sin45°=√3/2*√2/2-1/2*√2/2=(√6-√2)/4,
So tan (5 π / 12) / sin 165 ° = 4 (2 + √ 3) / (√ 6 - √ 2) = 1 + √ 3
2）cos20°cos(a-20°)-cos70°sin(a-20°)
=sin70°cos(a-20°)-cos70°sin(a-20°)
=sin[70-(a-20°)]=sin(90°-a)=cosa ；
3) Because (sin θ) ^ 2 + sin θ = 1, (sin θ) ^ 2 + (COS θ) ^ 2 = 1,
So sin θ = (COS θ) ^ 2,
So (COS θ) ^ 2 + (COS θ) ^ 4 = sin θ + (sin θ) ^ 2 = 1