Evaluate cos10 ° cos20 ° cos40 °

Evaluate cos10 ° cos20 ° cos40 °

cos10°cos20°cos40°=2sin10° cos10°cos20°cos40°/2sin10°
=2sin20°cos20°cos40°/4sin10° =2sin40°cos40°/8sin10° =sin80°/8sin10° =tan80°/8
cos10*cos20*cos40 =2sin10cos10cos20cos40/(2sin10)=2sin20cos20cos40/(4sin10)
=2sin40cos40/(8sin10)=sin80/(8sin10)=cos10/(8sin10)=1/8*cot10
Calculation: LG25 + 2lg2 + 2 ^ 1 + log (2) (3) main log (2) (3) how to calculate this
Calculation: LG25 + 2lg2 + 2 ^ 1 + log (2) (3) main log (2) (3) how to calculate this
There's a formula
a^[loga(N)]=N
Original formula = lg5 & sup2; + 2lg2 + 2 ^ 1 × 2 ^ [log2 (3)]
=2lg5+2lg2+2×3
=2(lg5+lg2)+6
=2+6
=8
Given the positive scale function y = (1-2a) x, if the value of Y decreases with the increase of the value of X, then the value range of a is______ .
According to the fact that the value of Y decreases with the increase of the value of X, we know that K ＜ 0, i.e. 1-2a ＜ 0, a ＞ 12
Log base 3 logarithm of 18 minus log base 3 logarithm of 2
log3（18）-log3（2）=log3（18/2）=log3(9)=2
The positive proportional function y = (3-K) x.1. If y increases with the increase of X, the value range of K is obtained. 2. If y decreases with the increase of X, the value range of K is obtained
When 3-K > 0, the image is located in the first and third quadrants, from left to right, y increases with the increase of X (monotonic increase), which is an increasing function;
When 3-K
1. K is less than 3
2. K is greater than 3
First order function.
In the same direction: 3-K > 0
Opposite direction: 3-K
Log 5 is the logarithm of base 35 minus log 5 is the logarithm of base 1 / 50 minus log 5 is the logarithm of base 14 plus log 1 / 2 is the logarithm of base 1
The original formula = log5 (35 / 50 / 14) + 0
=log5(125)
=log5(5³)
=3
In the image with known positive scale function y = (k-1) x, y decreases with the increase of X, then the value range of X is?
1：k＜1,2：k＞1,3：8,4：16
The answer is K
Lgx + lgY = 2lg (x-2y) for log change sign 2x / Y
That is, there is 2lg (x-2y) - lgx - lgY = 0 and X / Y > 0, x-2y > 0, that is, X / Y > 2, that is, LG ((x-2y) ^ 2 / XY) = 0, that is, LG (x / Y - 4 + 4Y / x) = 0, so x / Y - 4 + 4Y / x = 1, and X / y as a whole, z = x / Y > 2, then the above equation is Z-5 + 4 / z = 0, that is, Z ^ 2-5z + 4 = 0, and the solution is Z = 1 (rounding off)
Given the function y = 2mx + 1-3m, what is the value of M, it is a positive proportional function? What is the value of M, it is a linear function? Why is the value of M? Why does the value of function y increase with the increase of X?
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When 1-3m = 0, i.e. M = 1 / 3, it is a positive proportional function
When m ≠ 0, it is a linear function
When 2m > 0, i.e. m > 0, the function value y increases with the increase of X
It is known that: lgx + lgY = 2lg (x-2y), then the value of log [2 ^ (1 / 2)] [(x / y)] is____
∵lgx+lgy=2lg(x-2y)
∴lg(xy)=lg(x-2y)^2
∴xy=(x-2y)^2
xy=(x-2y)^2
xy=x^2-4xy+4y^2
Divide by Y ^ 2 at the same time
(x/y)^2-5(x/y)+4=0
Replace "x / Y" with "t"
t^2-5t+4=0
formula
t^2-5t+(5/2)^2=9/4
t-5/2=±3/2
t1=1 t2=4
∵(x/y)=t
(x / y) = 1 or 4
Then log [2 ^ (1 / 2)] [(x / y)] is 0 or 4
log???
lgxy=lg(x-2y)^2
xy=(x-2y)^2
x^2-5xy+4y^2=0
(x-4y)(x-y)=0
Ψ x = 4Y or x = y
Because x > 0, Y > 0, x-2y > 0
If x = y does not meet the condition, it will be rounded off
∴x=4y
∴x/y=4
The original formula = log (√ 2) 4
=4