Given that a is the second quadrant angle, then sin square a-SiN square A. can be reduced to?

Given that a is the second quadrant angle, then sin square a-SiN square A. can be reduced to?

A is the second quadrant angle, so Sina > 0
√sin^2a-sin^2a=sina-sin^2a=sina(1-sina)

Sin (a + Pai / 4) = 10 / radical 10 A is the second quadrant angle Tana equals to?

Sin (a + Pai / 4) = sinacos (π / 4) + cosasin (π / 4) = 10 / 10 root, cos (π / 4) = sin (π / 4) = 2 / 2, so Sina + cosa = 5 / 5 root 5 and sum of squares of sina and cosa = 1, so Sina = 2 / 5 root sign 5 or negative 1 / 5 root sign 5 and a is the second quadrant angle, so Sina = minus 5

Given the vector a = (COS α, sin α), B = (radical 3,1), α∈ (0, π), and a ⊥ B, then α is equal to ()

A A kind of B = sin α + radical 3cos α = 2Sin (α + π / 3) = 0
So α + π / 3 = π, so α = 2 π / 3

The vector a = (sin (a + b) / 2, cos (a-b) / 2-3 root sign 2 / 4) vector b = (5 / 4sin (a + b) / 2, cos (a-b) / 2 + 3 root sign 2 / 4), where a B is the inner angle of triangle ABC, and a vector ⊥ B is proved to be Tana × tanb = 1 / 9

Firstly, by simplifying a · B = 0, 5 / 4 * cos (a + b) = cos (a-b) can be obtained;
Then, expand the term sin * sin = 1 / 9cos * cos;
Finally, TGA * TGB = 1 / 9
Learn the formula by yourself, don't ask me!

Given the vector a = (COS θ, sin θ), θ belongs to [0, Wu], vector b = (root 3, - 1), find the maximum and minimum values of 2a-b

2a-b ^ 2 = = 4 | a | 2 - 2 + | B ^ 2 = 4 + 4 - 2 radical sign 3 cos θ + 2Sin θ
=8-4 (0.5 root sign 3 cos θ - 0.5 sin θ) = 8-4 (sin (U / 3) cos θ - cos (Wu / 3) sin θ] = 8-4sin (U / 3 - θ)
Obviously, the maximum is 12 and the minimum is 4

In △ ABC, the opposite sides of angle a B C are a B C C = 60 B = 5 △ ABC area = 10 root sign 3. Find the value of sin (a + 30)

The vertical BH of AC is made from point B, because B = 5, △ ABC area = 10 root sign 3, so the height is 4 root sign 3, and because the C angle is 60 degrees, a = 8, ch = 4, ah = 1, C = 7, Sina = 1 / 7, cosa = 4, root number 3 / 7;
According to the formula of sin (α + β) = sin α cos β + cos α sin β, sin (a + 30) = Sina * cos30 + sin30 * cosa = 5 root sign 3 / 14

How to solve sin (45 + 60)? The answer is root 2 times (2 of 3 plus 1 / 2)

Using the sum difference formula sin (α ± β) = sin α · cos β ± cos α · sin β, we can get the result that sin (45 + 60) = sin45 · cos60 ± cos60 · sin45 = {[(√ 2) / 2] × [1 / 2] + [(√ 2) / 2] × [(√ 3) / 2]} This formula has been introduced in high school

Given sin (75 degrees + a) = 1 / 3, find the value of COS (15 degrees - a)

Cos (15 degrees - a) = sin [90 - (15 degrees - a)] = sin (75 degrees + a) = 1 / 3

It is known that cos (75 ° + α) = 1 3, α is the third quadrant angle. Find the value of COS (15 ° - α) + sin (α - 15 °)

∵ α is the third quadrant angle,

Given cos (75 + α) = 1 / 3, if a is the third quadrant angle, find cos (105-a) + sin (255 + a) - Tan (a-105)

If α is the third quadrant angle, there is:
180°+k*360°<α<270°+k*360°
Then 255 degree + k * 360 degree < 75 degree + α < 345 degree + k * 360 degree
Cos (75 ° + α) = 1 / 3 > 0
It can be seen that 75 ° + α is the fourth quadrant angle
Then sin (75 ° + α) < 0, Tan (75 ° + α) < 0
Therefore, from sin 2 (75 ° + α) + cos 2 (75 ° + α) = 1, it is easy to obtain that: 1
sin(75°+α)=-2√2/3,tan(75°+α)=sin(75°+α)/cos(75°+α)=-2√2
Therefore, cos (105 ° - α) + sin (255 ° + α) - Tan (α - 105 °)
＝cos[180°-(75°+α)]+sin[180°+(75°+α)]-tan[(α+75°)-180°]
＝-cos(75°+α)-sin(75°+α)-tan(α+75°)
＝－(1/3 -2√2/3 -2√2)
＝(8√2 －1)/3