Given TaNx = 2010, then [1-sin (9 π / 2-2x)] / sin (9 π - 2x)= The answer is 2010. But I figure it's 1 / 2010

Given TaNx = 2010, then [1-sin (9 π / 2-2x)] / sin (9 π - 2x)= The answer is 2010. But I figure it's 1 / 2010

The original formula = (1-cos2x) / sin2x
=[1-(1-2sin²x)]/2sinxcosx
=2sin²x/2sinxcosx
=sinx/cosx
=tanx
=2010

Given TaNx = - 1 / 3, find 1 / (2sinxcosx + cos ^ 2x)

1/(2sinxcosx+cos^2x)
=(cos^2x+sin^2x)/(2sinxcosx+cos^2x)
Then divide it by cos ^ 2x at the same time. The following calculation will be done by yourself

Given TaNx = 2, find the value of 1sinacosa + cos square a

because
tana=2
therefore
sina/cosa=2
sina=2cosa
also
sin^2a+cos^2a=1
therefore
5cos^2a=1
cos^2a=1/5
2sinacosa + cos square a = 1 / (4cos ^ 2A + cos ^ 2a) = 1 / 1 = 1

2tanx divided by (1 + (TaNx squared)) = 3 / 5 to find sin ((Π / 4) + x)

Sin ((Π / 4) + x) = radical (2) / 2 * (SiNx + cosx) 2 * TaNx / (1 + (TaNx * TaNx)) = 3 / 5 because 1 + (TaNx * TaNx) = ((cosx Square) + (SiNx Square)) / (cosx square) = 1 / (cosx Square), then the above formula can be converted into 2 * TaNx * (cosx Square) = 2 * SiNx * cosx = 3 / 5 (SiNx + cosx) ^ 2 = 1 + 2sinx *

Given 2tanx / 1 + TaNx = 3 / 5, the value process of sin (π / 4 + x) is obtained

Sin (π / 4 + x) = (√ 2 / 2sinx + √ 2 / 2cosx) = 1 / 2 (SiNx + cosx) = (SiNx + 2sinxcosx + cosx) / 2 = (SiNx + 2sinxcosx + cosx) / (2sinx + 2cosx) = (TaNx + 2tanx + 1) / (2tanx + 1) because 2tanx / (1 + TaNx) = 3 / 5 = = > TaNx = 3 or 1 / 3, TaNx substitution is obtained

Mathematical problem -- given - 3 times π ≤ x ≤ 4 parts π, f (x) = the square of TaNx + 2tanx + 2, find the maximum value of F (x) and the corresponding x value

Because f (x) = the square of TaNx + 2tanx + 2 = the square of (TaNx + 1) + 1
And because - 3 parts of π ≤ x ≤ 4 parts of π
Therefore, when x = 4 / π, there is a maximum value of 5
When x = - 4 parts π, there is a minimum value of 1
This is the solution, you look at it more, come on, study hard

F (x) = TaNx squared + 2tanx + 2 find the maximum value of F (x) and the corresponding X

F (x) = TaNx squared + 2tanx + 2 = (TaNx + 1) ^ 2 + 1 the value range of TaNx is R. therefore, when TaNx + 1 = 0, f (x) has a minimum value, and its minimum value is 1. In this case, TaNx = - 1 x = k π - π / 4 (k is an integer)

The known (2x + Tan ^ 2) is (2x + 2) / (2 x)

sin2x=2sinx*cosx=2sinx/cosx*cos^2 x=2tanx/(1/cos^2x)=2tanx/[(sin^2x+cos^2x)/cos^2x]=2tanx/(1+tan^2x) 2tanx/（1+tan^2x） =3/5sin2x=3/5Cos2A=CosA^2-SinA^2=1-2SinA^2sin^2(π/4 +x)=[1-cos2(π/4 +x)]/2=[1-...

TaNx = 2, find the value of sin ^ 2x + sinxcosx-2cos ^ 2x If we change it into a homogeneous fraction, why can we only change the denominator into sin ^ 2x + cos ^ 2x, and then divide it by cos? Can't the original formula be directly divided by cos ^ 2x?

Yes, but the trouble is sin? X + sinxcosx-2cos? X = (sin? 2x + sinxcosx-2cos? X) * cos? X / cos? X (note that the molecule also multiplies cos? X) = (tan? X + tanx-2) * cos? X = 4cos? X and cos? X (1 + tan?)

It is known that SiNx cosx = 0, X belongs to the (0,2 th) school. (1) find the value of TaNx and X, (2) find the value of sinx-2cosx of 2sinx + cosx

SiNx cosx = 0, then SiNx = cosx = TaNx = 1, because SiNx * SiNx + cosx * cosx = 1, according to SiNx cosx = 0, we can get (SiNx cosx) (SiNx cosx) = 0. After simplification, we can get 1-2 SiNx cosx = 0 and replace it with SiNx = cosx