A question on the Friends of Mathematics in the third year of high school

cosθsinx-sin(x-θ)tanθ-2)sinx-sinθcosθsinx-sinxcosθ+cosxsinθ+(tanθ-2)sinx-sinθ

cosxsin +(tan -2)sinx-sin root number [(tan -2) 2+sin 2]*sin(x+a)-sin

When sin(x+a)=-1, the above equation obtains the minimum value.

Under the root number -[(tan -2) 2+sin 2]-sin =0(tan -2) 2+sin 2]=-sin We can see that sin <0 is squared on both sides (tan -2) 2=0

tanθ=2

ctgθ=1/2

1 (sin ) 2=1+(ctg ) 2=5 4sin =(-2 5) 5

Tips: Organize the terms of sinx and cosx, and the terms with parameter constants, use the asinx+bcosx method to merge the terms of sinx and cosx, find their minimum values, only contain the parameters, make them equal to 0, get the trigonometric equation with , and solve sin

The detailed rent is as follows, click to enlarge:

af∩β=m。。This should be M.

Let cd = n, it can be proved that the inner quadrilateral bmen is a parallelogram.

BM CF NE, BN AD ME, two sides are parallel, the third side is parallel to the intersection of the two planes).

The angle between AD and CF is fixed, that is, the size of the angle NBM is fixed, so only the length of BM and ME affects the area of the triangle.

It can be assumed that AD is perpendicular to CF, and the area of the triangle is the area of the parallelogram BMEF, and when it is maximum, it is when BM*ME is maximum.

g:(g+h)＝bm:cf=am:af=1-mf:af=1-me:ad

g/(g+h)=bm/cf=(ad-me)/ad

bm*me=[g*cf/(g+h)]*ad*[1-(g/(g+h))]

gh/(g+h)²]cf*ad

When the question changes to how much g h. gh (g+h) maximum.

Let g h=x and gh (g+h) =1 (x+1 x+2).

The problem changes to find the minimum when x is equal to x+1 x.

You made a mistake in copying the question, isn't the f-point on the plane? The condition behind you says af = f, so it means that point f is on . . .

Let's take a second look at the question

In the analysis, BM CF is analyzed, and the angle of CM AD BMC and AD CF are the same or complementary in size, which is denoted as M

BM=(G (G+H))CF cm=[H (G+H)]AD, so the area of the triangle is 1 2 sinm BM cm=1 2sinm(g (g+H))cf[h (g+H)]AD

It is sufficient to find the maximum value of (g (g+h))[h (g+h)] :(g (g+h))[h (g+h)]=gh (g 2+2gh+h 2)=1 [g h+h g+2]<=1 [2+2]=1 4 if and only if g h=h g, so the maximum area of the triangle is 1 2 *1 4*sinm=1 8sinm when g h=1

Note: The logarithm of base b is loga(b).

by 1-log6(3).

log6(6)-log6(3)

log6(6/3)

log6(2).

log6(18) again

log6(3×6)

log6(3)+log6(6)

1+log6(3).

Thus log6(2) log6(18).

log6(2)×(1+log6(3).

log6(4)

log6(2²)

2log6(2).

log6(2)×log6(18)}÷log6(4)=÷[2log6(2)]

1.Composed. If we know log18(9)=a, we get 2log18(3)=a

Thus log18(3)=a2.

and log18(2) = log18(18 9) = log18(18)-log18(9).

1-log18(9)

1-a.log18(5)=b.

The formula for changing the bottom has :

log36(15)

log18(15)/log18(36)

log18(3)+log18(5)]/[2log18(6)]=[log18(3)+log18(5)]/[2log18(2)+2log18(3)]

(a/2)+b]/[2(1-a)+2a/2]=[(a+2b)/2]/(2-a)

a+2b)/[2(2-a)]=

Alas, what gave me the courage to open this question, fifteen years after graduating from high school, and I couldn't even forget what these operators meant!! hidden, afraid of embarrassment

When summing, this is to see what is the common ratio of your medium ratio series, if the common ratio is 2 is 2SN, if the common ratio is 4 is 4SN and then make the difference.