Area formula and deformation of grade 5 in Beijing Normal University It must be in the fifth grade of Beijing Normal University

Area formula and deformation of grade 5 in Beijing Normal University It must be in the fifth grade of Beijing Normal University

Area of square = side length times side length
S = the square of a
The area of a rectangle = length times width
S=AB
Area of trapezoid = (top bottom + bottom) times height divided by two
S = (a + b) H divided by 2

Kneel down to seek primary school north Normal University Edition fifth grade second volume mathematics unknown unit one and known unit one formula! Just the equation.

Let's be more specific

What are the area formulas in the first volume of the fifth grade mathematics of Beijing Normal University

Rectangle = length * width
Square = side length * side length
Triangle = (bottom * height) / 2
Trapezoid = (top bottom + bottom) * height / 2

It's about the sine formula of sum and difference of two angles, 1. Proposition a: 3sinb = sin (2a + b) is the condition of proposition B: Tan (a + b) = 2tana 2. Because of the printing reasons, the content on the horizontal line of the following question can't be recognized. If the conclusion is known, please fill in one condition of the original question on the horizontal line, known as: A, B, are acute angles, and Sina CoSb= -0.5,_______________ Then sin (a-b) = 59 / 72 3. Known: A, B belong to 180 to 270 degrees, Sina = - radical 5 / 5. CoSb = - root 10 / 10, find the value of a-b I have worked out the third question. What I don't understand is that I use sin and COS respectively. One is 45 degrees and the other is - 45 degrees,

One
3sinB=sin2AcosB+cos2AsinB
3sinB=2sinAcosAcosB+(cosA^2-(1-cosA^2))sinB
3sinB=2sinAcosAcosB+(2cosA^2-1)sinB
2sinA/cosA=(tanA+tanB)/1-tanAtanB
Expand, let those items have the same position as the top, and then take the value to make it true
Two
Sina + CoSb = 1 (you can give this as long as it is between 2 and - 2)
Sina-cosb = - 0.5, work out the formula to the end, and then you can get a relationship. You can sort this relationship down and write it in the air
Three
SINB and cosa can be calculated
sinA^2+cosA^2=1
The sine and cosine of the third quadrant are all negative
sin(A-B)=sinAcosB-cosAsinB=
And then take the positive and negative according to this

What is the value of 3 / 20 of sine minus 1 / 20 of cosine plus 64 times of 20 degree square of sine

YGE, do you want the result of the following equation? We can use trigonometric function formula to decompose and evaluate
3/SIN20^2-1/COS20^2+64SIN20^2=32
Answer 32

Why is the square of a sine equal to one minus half of the cosine angle?

Because:
cos(a+b)=coacosb-sinasinb
Let a = b = D
cos2d=(cosd)^2-(sind)^2=(cosd)^2-[1-(cosd)^2]=2(cosd)^2-1
So (cosd) ^ 2 = (cos2d + 1) / 2
Substituting D / 2 for D, cosd / 2 = ± √ [(1 + cosd) / 2]
But cos2d = (cosd) ^ 2 - (sind) ^ 2 = [1 - (sind) ^ 2] - (sind) ^ 2 = 1-2 (sind) ^ 2
So (sind) ^ 2 = (1-cos2d) / 2
In the same way, sind / 2 = ± √ [(1-cosd) / 2]

In a triangle, the square of the sine of a multiplied by the square of the cosine of B, and then the square of the cosine of a multiplied by the square of the sine of B is equal to the square of the sine of C, so as to find the shape of the triangle

right triangle
The left side of the equation is reduced to sin (a + b) * sin (a-b)
The square of sinc on the right side of the equation is equal to the square of sin (a + b)
If sin (a + b) is deleted from both sides, sin (a + b) = sin (a-b) is obtained
Obviously, a + B is not equal to A-B, so (a + b) + (a-b) = 180 degrees
So a = 90 degrees

(2 times cosine 10 degrees minus sine 20 degrees) divided by sine 70 degrees is what?

Molecular: 2cos10-sin20
=2cos(30-20)-sin20
=2cos30*cos20+2sin30*sin20-sin20
=√3cos20
Denominator: sin70 = cos20
So the original formula = √ 3

What are the sine and cosine of 20 and 40 degrees

sin20=0.342,cos20=0.940
sin40=2*sin20*cos20=0.643
cos40=(cos20)^2-(sin20)^2=0.766

How long is the hypotenuse of a right triangle? What is the long side 120mm and the short side 75mm? The calculation formula and method are required!

Not even Pythagorean theorem?
The square root of the sum of squares of the long side and the short side
Bevel = (120 ^ 2 + 75 ^ 2) ^ (1 / 2)
=141.509717