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. given cot α = 3, find the value of cos? α + 1 / 2sin2 α? (to calculate the process) 2sin2 α is the denominator
If a is substituted for cosa / Sina = COTA = 3cosa = 3sina, and the identity sin? 2A + cos? A = 1, then sin? A = 1 / 10cos? A = 9 / 10sin2a = 2sinacosa = 2sina (3sina) = 6sin? A = 6 / 101 / 2sin2a, is this Sina in the numerator or in the denominator
How to simplify f (x) = sin ^ 2 θ + SiNx + cos ^ 2 θ + cosx into the form of F (x) = asin (ω x + φ) + B?
f(x)=sin^2θ+sinx+cos^2θ+cosx
=1 + Radix 2 (Radix 2 / 2 * SiNx + Radix 2 / 2 * cosx)
=1 + radical 2 * sin (x + π / 4)
=Radical 2 * sin (x + π / 4) + 1
Where a = radical 2, ω = 1, φ = π / 4, B = 1
(1) Find the maximum and minimum of the function y = 3sinx + 4cosx. (2) you can use a, B to represent the maximum and minimum of the function y = asinx + bcosx
(1) From the auxiliary angle formula, y = 5sin (x + arc tan4 / 3), then the maximum and minimum values are ± 5 (2) from the auxiliary angle formula, y = √ (a + b) × sin (x + arc tanb / a), then the maximum and minimum values are ± √ (a + b)
Simplify a complex function formula Let f (x) = (a + bi) (P + Qi) ^ x + (a-bi) (p-qi) ^ X be reduced to a real function. This result should be a real number
It is convenient to use the exponential form to simplify P + Qi = re ^ I θ, where r = √ (P ^ 2 + Q ^ 2), θ = arctan (Q / P), then p-qi = re ^ (- I θ), let a + bi = CE ^ is, where C = √ (a ^ 2 + B ^ 2), s = arctan (B / a), then a-bi = CE ^ (- is). Therefore, there are: F (x) = CR ^ Xe ^ I (s + θ x) + CR ^ Xe ^ (- I (s + θ x))
(1) 2Sin α + cos α （2）2sinαcosα+cos²+1 The value domain (1) y = cosx / 1-2sinx (2) y = 1-2x - √ 2x + 1 (by means of substitution and monotonicity), please,
(1) (2) the original formula = sin2a + 1 / 2 (1 + cos2a) + 1 = 1 = sin2a + 1 / 2cos2a + 3 / 2 = 5 / 2 (2 / 2 / 5) sin2a + 1 / 2cos2a + 3 / 2 (1 / 2cos2a) + 3 / 2 = 5 / 2 (2 / 2) sin2a + 1 / 2 (1 + cos2a) + 1 = sin2a + 1 / 2cos2a + 3 / 2 = √ 5 / 2 [(2 / 5) sin2a + (1 / √ 5) cos2a] + 3 / 2 = 5 / 2 = 5 / 2Sin (2a + φ) + 3 / 2 (where COS is cos (COS) 3 / 2 = 5 / 2Sin (2a + φ) + 3 / 2 (where COS is cos (where COS is cos (where COS is a + 2) + 3 / 2 (φ = 2 / √ 5
Please simplify this function for me S=32-x[32/(x+2)-1]
S=32-x[32/(x+2)-1]
=32-32x/(x+2)+x
=64/(x+2)+x
How to simplify the function y = sin2x + cos2x? How to find the value range after the simplification?
Y = sin2x + cos2x = root 2Sin (2x + 45 degrees), so the maximum root 2, the minimum negative root 2, period π
Simplification of sin2x + cos2x
sin2x+cos2x
=Root 2 · [cos45 · sin2x + sin45 · cos2x]
=Root 2 · sin [2x + 45]
It is known that X1 is the root of X · log2x = 3 and X2 is the root of X · 2 ^ x = 3. Find the value of X1 + x2
The intersection points of y = log2x and y = 3 / x, y = 2 ^ X and y = 3 / X are a, B, y = 3 / X and y = x at point C, AB and y = x at point D, then a (x1,3 / x1), B (x2,3 / x2), C (3 ^ (1 / 2), 3 ^ (1 / 2)) because y = log2x and y = 2 ^
Given function f (x) = (3-ax)) / (A-1) 1. If a > 0, find the definition field 2. If f (x) is a decreasing function on (0,1], find the range of real number a. It's mainly the second question. The first one can do it,
Help you solve it
Question 1: the definition domain of a function. Every expression has meaning
Therefore, there is 3-ax > = 0, because a > 0, it is easy to get x0, divided into two sections, (0,1) and (1, + ∞)
When a > 1, the denominator is greater than 0, the larger the value of X, the smaller the value of numerator, and the smaller the value of function. At this time, it is a subtraction function, meeting the requirements of the topic!
Let's look at (0,1). In this interval, similarly, it is obviously an increasing function!
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