(16x fourth power + 4x square + x) △ x

(16x fourth power + 4x square + x) △ x

(16x fourth power + 4x square + x) △ x
=16x³ + 4x + 1
sin10°cos45°+sin80°cos55°
Original formula = sin10 ° cos45 ° + sin (90 ° - 10 °) cos (90 ° - 45 °)
=sin10°cos45°-sin10°cos45°
=0
The cos55 ° in the title should be cos45 ° right?
cos20°cos(α-20°)-cos70°sin(α-20°)
Cos70 ° = sin20 ° original formula = cos20 ° cos (α - 20 °) - cos20 ° sin (α - 20 °) = cos (20 + α - 20 °) = cos α
How to solve "4 (the square of x) + 16x -- 1 = 4x (x + 4) -- 1"?
4(x^2)+16x-1=4x(x+4)-1
4x^2+16x-1=4x*x+16x-1
4x^2+16x=4x^2+16x
4x^2-4x^2=16x-16x
0=0
So, X can be any number
Is there any solution? Isn't it an identity?
personnel
Find the value of cos20 ° cos40 ° cos80 °
It's the multiplication of three numbers
cos20°cos40°cos80°
=2sin20°cos20°cos40°cos80°/2sin20°
=sin40°cos40°cos80°/2sin20°
=sin80°cos80°/4sin20°
=sin160°/8sin20°
=sin20°/8sin20°
=1/8
namely
zero point nine three nine six
zero point seven six six zero
zero point one seven three six
sin^215º+cos^275º+sin15ºcos75º
Trigonometric identity transformation problem, need complete problem-solving process
The original formula = 3sin ^ 215 & ordm=
=3*(-1/2)*(cos(15+15)-cos(15-15))
=1 / 4 * (6-3 * radical 3)
The value of the square of 3x - 12x + 16 is always greater than zero
The square of 3x - 12x + 16
=3(x^2-4x+4)+4
=3(x-2)^2+4
≥4
The value of cos20 ° cos40 ° cos80 ° is______ .
Cos20 ° cos40 ° cos80 ° = 8sin20 ° cos20 ° cos40 ° cos80 ° 8sin20 ° = sin160 ° 8sin20 ° = 18
Sin & sup2; 10 & ordm; + cos & sup2; 40 & ordm; + sin40 & ordm; cos40 & ordm; evaluation
It may be an example of the old edition of the third year mathematics book
I've got the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum of the sum;
（sin15º=½√（2-√3））
If there is an accurate value, please give the joint method; if not, forget it
It's better to give more similar questions
It's high school content. The title is wrong,
It can be changed to sin & sup2; 10 & ordm; + cos & sup2; 40 & ordm; + sin 10 & ordm; Cos 40 & ordm;,
The conventional solution is the sum difference product of the first two terms,
The unconventional solution of the third term product sum difference is given below
Let x = Sin & sup2; 10 & ordm; + cos & sup2; 40 & ordm; + sin 10 & ordm; Cos 40 & ordm;,
y=cos²10º+sin²40º+cos10ºsin40º,
Then x + y = 2 + sin50 & ordm;,
x-y=-cos20º+cos80º-sin30º
=-cos(50º-30º)+cos(50º+30º)-1/2=-sin50º-1/2,
The sum of the two formulas gives 2x = 3 / 2, x = 3 / 4,
We can also construct △ ABC, a = 10 & ordm;, B = 50 & ordm;, C = 120 & ordm;,
From the cosine theorem, C & sup2; = A & sup2; + B & sup2; + AB,
Then sine and cosine theorem is used to get Sin & sup2; 120 & ordm; = Sin & sup2; 10 & ordm; + Sin & sup2; 50 & ordm; + sin10 & ordm; sin50 & ordm;,
That is, 3 / 4 = Sin & sup2; 10 & ordm; + cos & sup2; 40 & ordm; + sin 10 & ordm; Cos 40 & ordm
It is proved that - 2x2 + 4x - 10 ＜ 0 is constant
It is proved that: - 2x2 + 4x-10 = - 2 (x2-2x) - 10 = - 2 (x2-2x + 1-1) - 10 = - 2 (x-1) 2-8, ∵ 2 (x-1) 2 ≥ 0, ∵ - 2 (x-1) 2 ≤ 0, ∵ 2 (x-1) 2-8 < 0, i.e. - 2x2 + 4x-10 < 0