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In △ ABC, if sin2a ≤ sin2b + sin2c sinbsinc, then the value range of a is () A. （0，π 6] B. [π 6，π） C. （0，π 3] D. [π 3，π）
According to the sine theorem, a = 2rsina, B = 2rsinb, C = 2rsinc
∵sin2A≤sin2B+sin2C-sinBsinC，
∴a2≤b2+c2-bc
∴cosA=b2+c2−a2
2bc≥1
Two
∴A≤π
Three
∵A＞0
The value range of a is (0, π)
3]
Therefore, C is selected
In △ ABC, the opposite sides of ∠ a, ∠ B, and ∠ C are ABC respectively. When a, B, C satisfy what relationship, △ ABC is an acute or obtuse triangle
If the sum of squares of the lengths of two sides of a triangle is the square of the length of the third side, the triangle is an obtuse angle triangle as long as there is one case like this. When the sum of squares of the lengths of any two sides is the square of the third side, it is an acute angle triangle, I hope I can understand
Cosine theorem: in △ ABC, given (a + B + C) (a + B-C) = 3AB, find the size of the angle Detailed process
Step one: why is a + B bracketed?
The three sides of a right triangle are 3M, 4m and 5m respectively. What is the height on the longest side of this right triangle?
The three sides of a right triangle are 3M, 4m and 5m respectively. What is the height on the longest side of this right triangle?
Hypotenuse * height on hypotenuse = right angle side * right angle side
Height on hypotenuse = right angle side * right angle side △ bevel side
3*4÷5=12/5
If - A is one-third of A-7, it is equal to one-third of the sharp angle
Choose B
Sin α =3/5, α is acute angle
Then Tan α = 3 / 4
tan（α- π/4）
=（tanα-tan π/4）/（1+tanαtan π/4）
=（3/4-1）/（1+3/4）
=-1/7
If cos (a + b) cos (a-b) = 1 / 3, then (COSA) ^ 2 - (SINB) ^ 2 =? The answer is one third
∵cos(a+b)cos(a-b)=1/3,
∴cos(a+b)cos(a-b)=(cos2a+cos2b)/2 ∴cos2a+cos2b=2/3.
(cosa)^2-(sinb)^2=（1+cos2a-1+cos2b）/2
=(cos2a+cos2b)/2
=1/3
(attached: in the second step, cos α cos β = [cos (α + β) + cos (α - β)] / 2 in the sum difference formula of integration
The number of zeros of the function f (x) = xcos2x on the interval [0, 2 π] is () A. 2 B. 3 C. 4 D. 5
0
0
Select c
After calculation, f (- 2) = - 3.864
The approximate interval of the positive zeros of the function f (x) = - x ^ 2-3x + 5?
X = (root 29-3) / 2
(1,1.5)
The number of zeros of the function f (x) = 2x + x3-2 in the interval (0, 1) is______ One
∵f（x）=2x+x3-2，
ν f ′ (x) = 2xln2 + 3x2 > 0 is always true on (0, 1),
The function f (x) =2x+x3-2 increases monotonically in the interval (0, 1),
∵ f (0) = - 1 < 0, and f (1) = 1 > 0,
∴f（0）f（1）＜0，
The function f (x) = 2x + x3-2 has a unique zero point in the interval (0, 1),
So the answer is: 1
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