Given sin (a + b) = 1, find Tan (2a + b) + tanb =?

Given sin (a + b) = 1, find Tan (2a + b) + tanb =?

sin(A+B)=1
Then a + B = 2K π + π / 2
2A+B=(A+B)+A
B=(A+B)-A
So the original formula = Tan (2k π + π / 2 + a) + Tan (2k π + π / 2-A)
=tan(π/2+A)+tan(π/2-A)
=-cotA+cotA
=0

Sin (a + b) = 1, it is proved that Tan (2a + b) + Tan B = 0

sin(A+B)=1
A+B=2kπ+π/2
2A+2B=4kπ+π
tan(2A+2B)=tan(4kπ+π)=0
tan[(2A+B)+B]=0
So [Tan (2a + b) + tanb] / [1-tan (2a + b) * tanb] = 0
So tan (2a + b) + tanb = 0

Given Tan α = 1 / 2, find Sin & # 178; α + sin α cos α + 2

sin²α+sinαcosα+2
=(sin²α+sinαcosα)/(sin²α+cos²α) +2
=(Tan & # 178; α + Tan α) / (Tan & # 178; α + 1) + 2 (the numerator denominator is also divided by cos & # 178; α)
=(1/4+1/2)/(1/4+1) +2
=(3/4)/(5/4) +2
=3/5+2
=Two and three fifths

We know that θ is an acute angle, and prove that sin θ + cos θ is less than π / 2

sinθ+cosθ=√2sin(θ+π/4)
∵0＜θ＜π/2
∴π/4＜θ+π/4＜3π/4
√2 /2＜sin(θ+π/4)≤1
∴1＜√2sin(θ+π/4)≤√2
That is: 1 ＜ sin θ + cos θ ≤ √ 2

It is known that 2 α + β = 90 degrees, and α is an acute angle. It is proved that sin α = √ (1-sin β) / 2, cos α = √ (1 + sin β) / 2
I haven't studied mathematics for a long time, and now I forget about it,

2 α + β = 90 degrees
2 α = 90 degrees - β
cos2α=sinβ
Half angle formula:
sinα=√[(1-cos2α)/2]=√[(1-sinβ)/2]
cosα=√[(1+cos2α)/2]=√[(1+sinβ)/2]

It is known that α is an acute angle, and it is proved that sin α + cos α > 1
How does = √ 2 (cos45 sin α + sin45 cos α) change to = √ 2Sin (α + 45)

sinα＋cosα
=√2（√2/2sinα＋√2/2cosα）
=√2（cos45 sinα＋sin45 cosα）
=√2sin(α＋45）
0

The proof of sin ^ 3 α + cos ^ 3 α
Wrong, sin ^ 3 α + cos ^ 5 α

You wrote (sin α) ^ 3 + (COS α) ^ 5

If it is an acute angle, sin α - cos α = 1 / 2, find Sin & # 179; α - cos & # 179; α

Use a
Square on both sides
sin²a+cos²a-2sinacosa=1/4
1-2sinacosa=1/4
sinacosa=3/8
So the original formula = (Sina COSA) (Sin & # 178; a + sinacosa + cos & # 178; a)
=1/2*(1+3/8)
=11/16

Knowledge structure framework of mathematical triangle

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Seeking triangle knowledge tree of junior high school mathematics

Triangles are divided into acute triangles, right triangles and obtuse triangles
Divide by side: unequal triangle isosceles triangle equilateral triangle
Similar triangles: triangles with equal corresponding angles and proportionate corresponding sides judge similar triangles: 1. Equal corresponding angles 2. Proportionate corresponding sides 3. Two corresponding sides are proportionate and the included angles of the two sides are equal 4. The triangle formed by a line parallel to a triangle and the other two sides of the triangle is similar to this triangle
Congruent triangles: similar triangles with similarity ratio of 1 are special cases of similar triangles. Congruent triangles are determined as follows: 1. Three corresponding sides are equal; 2. Two angles are equal and any side is equal; 3. Congruent triangles with equal sides are congruent
Pythagorean theorem: the sum of the squares of the two right angles of a right triangle is equal to the square of the hypotenuse
The relationship between the three sides of a triangle: the sum of any two sides must be greater than the third side, and the difference between any two sides must be less than the third side
The sum of the three internal angles of a triangle is 180 degrees
The height of the side opposite the vertex of an isosceles triangle is in line with the bisector of the vertex and the middle line
The two base angles of an isosceles triangle are equal
The height of an equilateral triangle is equal to 3 ^ 0.5/2 times of its side length
The area of a triangle is the base times the height divided by two
Trigonometric function: sine (SIN), cosine (COS), tangent (tan) and cot
Sina = the edge of the pair of angles a divided by the hypotenuse cosa = the adjacent edge of angle a divided by the hypotenuse Tana = the opposite edge of angle a divided by the adjacent edge of angle a COTA = the adjacent edge of angle a divided by the opposite edge of angle a
（sinA)^2+（cosA）^2=1 sinA=tanA*cosA tanA=1/cotA