In the triangle ABC, the opposite sides of inner angles a, B and C are a, B, C, B = π / 6, cosa = 4 / 5, B = radical 3, find the value of a, (2) find the value of sin (2a-b) It's today,

In the triangle ABC, the opposite sides of inner angles a, B and C are a, B, C, B = π / 6, cosa = 4 / 5, B = radical 3, find the value of a, (2) find the value of sin (2a-b) It's today,

solution
∵cosA=4/5
And a ∈ (0. π)
∴sinA=3/5
By sine
a=bsinA/sinB=(√3*3/5)/(1/2)=6√3/5
sin2A=2sinAcosA=2*3/5*4/5=24/25
cos2A=2cos²A-1=2*16/25-1=7/25
∴sin(2A-B)
=sin2AcosB-cos2AsinB
=24/25*√3/2-7/25*1/2
=(24√3-7)/50

In the triangle ABC, the sides of angles a, B and C are a, B and C respectively, and cosa = 2 / 5 root sign 5, sin = 10 / 10 root sign 10. (1) find angle c; (2) if A-B =, root sign 2-1, find edge C

In the triangle ABC, the opposite sides of angles a, B and C are a, B, C, and cosa = 2 √ 5 / 5, SINB = √ 10 / 10. (1) find the angle c; (2) if A-B = √ 2-1, find the edge C.1) cosa = 2 √ 5 / 5, = > Sina = 1 / √ 5, a SINB B, a = √ 2, B = 1; C = 135 degrees, = > sinc = 1 / √ 2, = > C / b

In the triangle ABC, B = π / 3, cosa = 4 / 5, B = radical 3, find the value of sinc and the area of triangle Sorry, it's B = radical 3

(1) Because cosa = 4 / 5 and because a, B and C are the inner angles of the triangle ABC, Sina = [under the root sign (5 ^ 2-4 ^ 2)] / 5 = 3 / 5 and because the angle B = 60 degrees, SINB = (root 3) / 2, B = 1 / 2, so we can get sinc = sin [180 degrees - (a + b)] = sin (a + b) = Sina * CoSb + cosa * SINB

In the triangle ABC, B = Pai / 3, cosa = 4 / 5, B = radical 3, find sin value and triangle area

Cosa = 4 / 5, then Sina = 3 / 5
sinC=sin（A+B)
=Sina * CoSb + cosa * SINB = (3 + 4-fold root 3) / 10
S triangle ABC = 1 / 2 * c * bsina
C / sinc = B / SINB C = 4 times root 3 + 3 / 5
So 1 / 2 * c * bsina = (36 + 9 times root 3) / 50

Radical 3tan12 ° - 3 / (4cos40 ° - 2) × sin12 °

=[(√3)tan12°-3]/{[4(cos12°)^2-2]*sin12°}
=[(√3)(sin12°/cos12°)-3]/{2[2(cos12°)^2-1]*sin12°}
=[(√ 3) (sin12 ° / cos12 °) - 3] / (2 * cos24 ° * sin12 °) (denominator with double angle formula)
=[(√ 3) sin12 ° - 3cos12 °] / (2 * cos24 ° * sin12 ° cos12 °) (the numerator and denominator are multiplied by cos12 °)
=-(2 √ 3) sin48 ° / (cos24 ° * sin24 °) (auxiliary angle formula for molecule and double angle formula for denominator)
=-(4 √ 3) sin48 ° / (2 * cos24 ° * sin24 °) (both numerator and denominator multiplied by 2)
=-(4√3)sin48°/sin48°
=-4√3

Radical 3 * tan12 ° - 3 / [(4cos12 ° - 2) * sin12 °] The teacher talked about similar topics for a long time

Radix 3tan12-3 / sin12 (4cos12 square-2)
=(Radix 3sin12-3cos12) / cos12sin12 * 2cos24
=2 root 3 (sin12 / 2-root 3cos12 / 2) / sin24cos24
=4 root numbers 3sin (12-60) / sin48
=4 root numbers 3sin48 / sin48
=3

Sin12 (2-4cos12 ^ 2) / 3-radical 3tan12

Because writing is not convenient, write separately:
Molecule: sin12 [2-2 (1 + cos24)] = - 2sin12cos24
Denominator: 3 - √ 3tan12 = 3 - √ 3sin12 / cos12
In this case, the original formula = - 2sin12cos24 / (3 - √ 3sin12 / cos12)
=-2sin12cos12cos24/(3cos12-√3sin12)
=-sin24cos24/[2√3(√3/2cos12-1/2sin12)]
=-1/2sin48/[2√3cos(12+30)]
=-1/2sin48/2√3cos42
=-1/2sin48/2√3sin48
=(-1/2)/(2√3)
=-√3/12

Find the value of (√ 3tan12-3) / sin12 (4cos12-2)

Is it (√ 3tan12-3) / [sin12 * (4 (cos12) ^ 2-2)]
The original formula = (√ 3sin12-3cos12) / [sin12 * cos12 * (4cos12-2)] = (√ 3sin12-3cos12) / [sin24 * (2 (cos12) ^ 2-1)] = (√ 3sin12-3cos12) / (sin24 * cos24)
=4√3(1/2*sin12-√3/2*cos12)/sin48
=4√3sin48/sin48=4√3

Given that the final edge of angle a falls on the straight line y = 2x, X ≥ 0, find the values of sina, cosa and Tana

According to the meaning of the title, we can take a point on the final edge of angle a, such as point a (1,2), then:
The distance from point a to origin is r = radical 5
Therefore, according to the definition of trigonometric function of arbitrary angle, we can get the following conclusions
Sina = Y / r = 2 / Radix 5 = 2 (Radix 5) / 5;
Cosa = x / r = 1 / Radix 5 = (Radix 5) / 5;
tana=y/x=2/1=2

The final edge of the angle a passes through the point P (x, - 2) (x ≠ 0), and cosa = √ 3 / 6 * x, Sina, Tana

r=√x²+2
cosa=x/√x²+2=√3/6*x
1/√x²+2=√3/6
Square on both sides
1/(x²+2)=1/12
x²=10,r=2√3
(1) When x = √ 10,
cosa =x/r=√10/(2√3)=√30/6
sina= -√2/(2√3)= -√6/6
tana= -√5/5
(2) When x = - √ 10,
cosa =x/r= -√10/(2√3)= - √30/6
sina= -√2/(2√3)= -√6/6
tana=√5/5