How do you solve this math problem? Help! Thank you!!

Solution: 0cosa = 1 7, sina = 4 3 7

sin2a=2sinacosa=8√3/49cos2a=2(cosa)^2-1=-47/49tan2a=sin2a/cos2a=-8√3/47cos(a-b)=13/14

sin(a-b)=3√3/14

sinb=sin[a-(a-b)]

sinacos(a-b)-cosasin(a-b)0If you have any questions, please feel free to ask!

1) cos = 1 7, because 0 2 , so sin = (1-cos) = [1-(1 7) ]= 4 3 7

So tan = sin cos = 4 3 according to the tangent formula of the double angle:

tan2 = (2tan ) (1 - tan )2)cos( -=13 14, because - 2 2, so sin( -= 1-(cos ( = 1-(13 14) ]= 3 3 14

According to the cosine formula of the difference between the two angles:

cos[α = cosαcos(α-sinαsin(α-cosβ = (1/7) *13/14) +4 √ 3 / 7) *3√3 /14)

Get =60 degrees.

The dividend and the divisor are expanded to 100 times at the same time, the quotient is constant, and the remainder is also expanded to 100 times, that is, the original quotient is 13, the remainder is 4, and the dividend is y and the divisor is x, then.

y=x*13+4, x+y+13+4=413, x=28, y=368

These four numbers are all given to you, which result do you want?

If the dividend and the divisor are enlarged by 100 times at the same time, the quotient is 13 and the number of remaining countries is 400

Then the original quotient is 13, and the remainder is 4

If the divisor is x, then the dividend = 13x+4

The column equation is: 13x+4+x+13+4=41314x=392

x=2813x+4=368

Dividend = 368, divisor = 28

Nine-sixteens three-fourths multiplied by three-quarters.

1+2+3+..2011=(1+2011)/2=1006

f(x)=2011x+1/x+2/x+3/x+..2010/x+1006

f(-x)=-2011x-1/x-2/x-3/x-..2010/x+1006

Let g(x)=f(x)-1006, then g(x)=2011x+1 x+2 x+3 x+.2010/x

g(-x)=-2011x-1/x-2/x-3/x-..2010/x

Then g(-x)=-g(x), so g(x)=f(x)-1006 is an odd function, then the center of symmetry of g(x) is (0,0), and from f(x)=g(x)+1006, we can see that f(x) is 1006 units of upward translation of the image obtained by g(x), so the center of symmetry of f(x) is (0,1006).

Solution: pm hm, proof as follows:

From p+ pne= p+ pmq=90°, pne= pmq and hqn= mqp=90°, qn=qm, so qnh qmp, i.e. pm=nh in rt qnh, hn qn, and hm qm, thus pm hm

Just finished typing didn't save the crash amount Agree with the answer upstairs.

Solution: Let the area be s, the straight corner side is a, and the other right angle side is 8-as=1 2 a (8-a).

1/2(a²-8a)

1/2(a-4)²+8

This is a quadratic function.

When a = 4, s has a maximum value of 8

At this point, the right triangle is an isosceles right triangle.

Solution: Let the right-angled sides of the triangle be x and y respectively

x+y=8 and s =1 2xy

From the mean theorem: x+y 2 xy

8 ≥2√xy ∴xy≤16

If and only if x=y, i.e., x=4, y=4, xy has a maximum value at which time xy=16

s△=1/2xy=1/2×16=8

Answer: When the two right-angled sides are 4 each, the area of the right-angled triangle is the largest, and the maximum value is 8

It is known that x+y=8Find s=x*y 2;

x+y)^2>=(4xy)=8s;The equal sign holds if and only if x=y=4.

s=8^2/8=8;

4 on both sidesThe maximum area is 8