If Sina cosa = radical 2, a belongs to (0, Wu), then Tana =?

If Sina cosa = radical 2, a belongs to (0, Wu), then Tana =?

Sina cosa = radical 2
sin^2 a+cos^2 a=1
So (Sina COSA) ^ 2 = 2
1-2sinacosa=2
-2sinacosa=-1
sin2a=1
2A = Wu / 2
A = Wu / 4
So Tana = 1

In the triangle ABC, if cosa = (radical 3) / 2, then sin (pai-a) =?

Sin (pai-a)
=sinA
=√ 1-cosa
=1/2

It is known that sin (45 degree - a) = root 2 / 10 and 0 degree Math homework for users on October 31, 2017 report Use this app to check the operation efficiently and accurately!

If sin (45 ° - a) = √ 2 / 10, then 0
Job help users 2017-10-31
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In the triangle ABC, a = 45 degrees, cosa, CoSb are the two roots of equation 4x ^ 2-2 (1 + radical 2) x + M = 0, AC = radical 2, find ∠ B, find the length of BC

According to Veda's theorem, the sum of two is equal to - B / A
This question has cosa + CoSb = 1 / 2 + √ 2 / 2 and a = 45 ° cosa = √ 2 / 2
The angle of the triangle is between 0 and 180, so B = 60 degrees
AC=b=√2 BC=a b/cosB=a/sinA
∴a=2

In △ ABC, cosa = Five 5，cosB＝ Ten 10． (I) find the angle C; (II) let AB = 2. Find the area of △ ABC

(1) from cosa = 55, CoSb = 1010, we get a, B ∈ (0, π 2), so Sina = 25, SINB = 310. (3 points) because COSC = cos [π - (a + b)] = - cos (a + b) = - cosacosb + sinasinb = 22, (6 points) and 0 < C < π, C = π 4. (7 points) (II) according to the sine theorem, absinc = a

It is known that in the triangle ABC, cosa = root six / 3, a and C are the sides opposite the angles a, B, C respectively (1) find sin2a; (2) if sin (3 μ / 2 + b) = - two and root two / 3, C = two and two, find the area of triangle ABC

The number of words is limited, so I only write ideas
(1) Sin (a) = under radical (1-cos (a) ^ 2)
Sin2a = 2sinacosa
(2) Sin (3 μ g / 2 + b) = cos (b)
So SINB is calculated
Sin (c) = sin (a + b) can be calculated
According to the sine theorem, C / sinc = A / Sina, a is obtained
So the area is s = 1 / 2 * a * c * SINB

In the triangle ABC, the opposite sides of a, B and C are a, B, C respectively, and cosa = 1 / 3. If a = radical 3, find the maximum value of BC I know the sine and cosine theorem, but I don't know how to apply it to this problem!

According to the cosine theorem, cosa = (B? + C? - a?) / 2BC = (B? + C? - 3) / 2BC = 1 / 3, then B? + C? - 3 = 2 / 3bC, so B? + C? = 3 + 2 / 3bC

In the triangle ABC, ABC is the opposite side of angle ABC, and cosa = 1.3, find the square of sin B + C + 2 + cos2a, if a = radical 3, angle c = 45, find the edge B

Sina (B + C) = Sina = 2 √ 2 / 3cos (B + C) = - cosa = - 1 / 3cos [(B + C) / 2] = [(1-1 / 3 / 3) / 2] = √ 3 / 3sin [(B + C) / 2] = √ 6 / 3cos 2A = 2 (COSA) (COSA) ^ 2-1 = - 7 / 9, so [sin (B + C) / 2] ^ 2 + cos2a = 2 / 3-7 / 9 = - 1 / 9, use the sine theorem to find C = 3 √ 3 / 4, B = √ 6 / 2 / 2 + (6 / 2 + (2 + (2) / 1 / 9, use the sine theorem to find C = 3 √ 3 / 4, B = 6 / 2 + (6 / 2 + (2 + (2 + (√ 3 / 4

In △ ABC, cosa = 1 / 3, find: (1) the value of sin ^ 2 [(B + C) / 2] + cos2a; and (2) if a = √ 3, find the maximum value of BC RT

Also in the △ ABC, cosa = 1 / 3, so (COSA / 2) ^ 2 = 1 / 3, so (COSA / 2) ^ 2 is in the △ ABC, cosa = 1 / 3, so (COSA / 2) ^ 2 = 2 / 3 (1) sin ^ 2 [(B + C) / 2] + cos2a = (COSA / 2) ^ 2 + 2 (COSA / 2) ^ 2 + 2 (COSA) ^ 2 = 2 / 3 + 2 * 1 / 9-1 = - 1 / 9 (9) if a = √ 3, from the theorem of cosine: a ^ 2 = B ^ 2 + C ^ 2 ^ 2-2-2-2-2-2-2-2 = B ^ 2 = B ^ 2 + C ^ 2-2-2-2-2-2 = B ^ 2 = B ^ 2 = B bccosa = B ^ 2 + C ^ 2-2bc

In the triangle ABC, the opposite sides of angles a, B and C are a, B, C, B = 60 °, cosa = 4 / 5, B = radical 3 (1) Find the value of sinc (2) Find the area of triangle ABC

1 sinB=√3/2,cosB=1/2,cosA=4/5,sinA=3/5
sinC=sin(A+B)=sinAcosB+cosAsinB=(3+4√3)/10
2 S=b*sinC/sinB*sinA*b/2=……………………