High school math problem known tan28 a, then sin2008 ? with problem solving process .

So: sin2008°

sin(5×360°+180°+28°)

sin(180°+28°)

sin28°

Since: tan28°=a

Then: (sin28° cos28°) = a i.e., sin 28°=a (1-sin 28°) yields: sin 28°=a (1+a).

Since: sin28°>0, tan28°=a>0, so: sin28°=a (1+a).

So: sin2008° = -a (1+a).

That 2008 = 11 * 180 + 28 = 360 * 5 + 180 + 28 so sin2008° = sin(360°*5+180°+28°) = -sin28°

tan28°=a, so sin28°=a 1+a 2 under the root number, so sin2008°=-a 1+a 2 under the root number

1/a+4/b+9/c

a+b+c)/a+4(a+b+c)/b+9(a+b+c)/c

1+b a+c a + 4+4a brother high b+4c b + 9 +9a c+9b c

14+ b/a+4a/b +c/a+9a/c +4c/b+9b/c

When talking about c:b:a=3:2:1, i.e., a=1 6, b=1 3, c=1 2.

The equal sign holds, 10, and substituting a+b+c = 1 into the numerator, gets.

a+b+c)/a + 4(a+b+c)/b + 9(a+b+c)/c =

1 + 4 + 9 + b/a + 4a/b) +c/a + 9a/c) +4c/b + 9b/c)

Let's consider the last three items, first look at b a + 4a b, let b a = t, t 2 - 4t + 4 = t-2) 2 > envy ruler = 0

t^2...0,

Anisotropy: Any two elements in a set are different objects.

by the nature of the set of =

a≠1, otherwise it does not meet the mutuality.

When ab=1, b=1 a, then 1 a=a2 gives a=1, and when it does not meet the mutuality a 2=1, a=-1, then b=ab=-b, and gives b=0 i.e. a=-1, b=0

a^2+1≥2√(a^2*1)=2a

The same goes for b 2+1 2b

c^2+1≥2c

So the original inequality holds.

Mean inequality: a+b2 ab).

Because A is a true subset of B.

For the empty set, it does not exist, and it is lost.

Not an empty set. then a 4 In summary, a 4

Because a is a true subset of b, a≠ba 4

It's good to think about the number line in your head, a>4