Given that - 90 degree is less than X and less than 0 degree, SiNx + cosx = 0.2, find SiNx cosx =?

Given that - 90 degree is less than X and less than 0 degree, SiNx + cosx = 0.2, find SiNx cosx =?

-7/5

Given the vector a = (SiNx, cosx-2sinx), B = (1,2) (1) if a / / B, find the value of TaNx. (2) if | a | = | B |, 0

Because the A is parallel B, so (cosx - 2 SiNx) / SiNx = 2 / 1, that is (cosx - 2 (TaNx * cosx) / (TaNx * cosx) = 2 (cosx * (1 - 2 TaNx) / TaNx = 2 (1 - 2 TaNx) / TaNx = 2 (cosx - 2 SiNx) / SiNx = 2, so (cosx - 2 (TaNx - 2 SiNx) / SiNx = 2 / 2 / 1 / 2 / 1 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 4 2 / 4 2 / 4 2. Because because the [a [a] is because [a [a [a] is the result of the [b [124 [124 [b] where the [b] is the (the [b] where the (the [b] is cosx+ 4 (SiNx) ^ 2 = 5 - 4sinx * cosx + 4 (SiNx) ^ 2 = 4 - SiNx * cosx + (SiNx) ^ 2 = 1 - SiNx * cosx = 1 - (SiNx) ^ 2 - SiNx * cosx = (cosx) ^ 2 TaNx = (cosx) ^ 2 / (cosx) ^ 2 when (cosx) ^ 2 > 0, TaNx = - 1; X = - 45 ° when (cosx) ^ 2 = 0, (cosx) ^ 2 / (cosx) ^ 2 has no meaning, that is, TaNx does not exist, that is, x = k * PI + 90 ° (k is an integer), and x = 90 ° because 0 < x < PI

Given the vector a = (SiNx, 2). B = (1, - COS).. and a perpendicular to b.. Find the values of TaNx and Tan (x-pie / 4)
By the way, add 20

A * b = sinx-2cosx = 0, so SiNx / cosx = TaNx = 2, Tan (x-pie / 4) = {TaNx + Tan (- Pai / 4)] / [1-tanx * Tan (- Pai / 4)] = 1 / 3
Adopt

Given that the vector a = (SiNx, 1) B = (cosx, 1), X belongs to R. when x = quarter, find the coordinates of the vector a + B

a+b=(sinx+cosx,2)=(√2sin(x+π/4),2)
When x = π / 4, √ 2Sin (x + π / 4) = √ 2Sin (π / 2) = √ 2
That is, a + B = (√ 2,2), that is, the coordinates of vector a + B are (√ 2,2)

The set of X with cos3x / 2 * cosx / 2 + sin3x / 2 * (- SiNx / 2) = 0
A detailed explanation of the question is helpful for the respondent to give an accurate answer

cos3x/2*cosx/2+sin3x/2*(-sinx/2)=0
cos3x/2*cosx/2-sin3x/2*sinx/2=0
cos(3x/2+x/2)=0
cos2x=0
2x=kπ+π/2
So x ∈ {x | x = k π / 2 + π / 4, K ∈ Z}

Calculate TaNx + Tan (PAI / 4-x) + TaNx * Tan (PAI / 4-x)=

tanπ/4=1
tan[(π/4-x)+x]=1
[tan(π/4-x)+tanx]/[1-tan(π/4-x)tanx]=1
tan(π/4-x)+tanx=1-tan(π/4-x)tanx
So tan (π / 4-x) + TaNx + Tan (π / 4-x) TaNx = 1

tan（x+pai/4）=1+tanx/1-tanx

tan(x+pai/4)
=(tanx+tan(pai/4))/(1-tanxtan(pai/4))
=(tanx+1)/(1-tanx)

Given that Tan (α + β) = 1 / 2, Tan (α - Pai / 4) = - 1 / 3, then the value of Tan (β + Pai / 4) is

The original formula = Tan [(α + β) - (α - Pai / 4)]
=[tan(α+β)-tan(α-pai/4)]/[1+tan(α+β)tan(α-pai/4)]
=1

It is known that Tan (PAI / 4 + a) = 1 / 2
(1) Find the value of Tan; (2) find the value of (sin2a cos ^ 2a) / (1 + cos2a)

tan(pai/4+a)=(tanpai/4+tana)/1-tanpai/4tana=(1+tana)/(1-tana)
From Tan (PAI / 4 + a) = 1 / 2
There are (1 + Tana) / (1-tana) = 1 / 2
The solution is Tana = - 1 / 3
(2) (sin2a-cos^2a)/(1+cos2a)=(2sinacosa-cosa^2)/(1+2cosa^2-1)
=(2sina-cosa)/2cosa
=tana-1/2
=-1/3-1/2
=-5/6

It is known that the ellipse x2a2 + y2b2 = 1 (a > b > 0) and the parabola y2 = 2px (P > 0) have the same focus F. P and Q are the intersection points of the ellipse and the parabola. If PQ passes through the focus F, the eccentricity of the ellipse x2a2 + y2b2 = 1 (a > b > 0) is 0______ ．

Since the focus F of the parabola y2 = 2px (P > 0) is (P2, 0), let the other focus of the ellipse be e. when x = P2, we enter the parabola equation and get y = ± P. because PQ passes through the focus F, so p (P2, P) and PF ⊥ of. So | PE | = (P2 + P2) 2 + P2 = 2p, | PF | = P. | EF | = P. so 2A = 2p + P, 2C = P. E = 2c2