Given the set E = {θ cos θ < sin θ, 0 ≤ θ < 2 π}, f = {θ Tan θ < sin θ}, then E ∩ f =? The answer to the basic training question is not clear. I've been drawing for a long time, but I can't figure it out The answer is (π / 4,3 π / 4). If it's between 45 ° and 135 °, a tan 55 ° is bigger than a sin 55 ° and the sine doesn't exist on the y-axis. Let's let my good brothers and sisters teach me

Given the set E = {θ cos θ < sin θ, 0 ≤ θ < 2 π}, f = {θ Tan θ < sin θ}, then E ∩ f =? The answer to the basic training question is not clear. I've been drawing for a long time, but I can't figure it out The answer is (π / 4,3 π / 4). If it's between 45 ° and 135 °, a tan 55 ° is bigger than a sin 55 ° and the sine doesn't exist on the y-axis. Let's let my good brothers and sisters teach me

Your standard answer is wrong! First solve the set E, from cos θ ＜ sin θ, 0 ≤ θ ＜ 2 π to discuss the situation. When θ = π / 2, cos θ = 0, sin θ = 1, cos θ ＜ sin θ holds; when θ = 3 π / 2, cos θ = 0, sin θ = - 1, cos θ ＜ sin θ does not hold; when 0 ≤ θ ＜ π / 2 or 3 π / 20, both sides of the original inequality

Given that the coordinates of a point P on the terminal edge of acute angle a are (2sin2, - 2cos2), we prove that a = 2 - π / 2

tanA=(-2cos2/2sin2)=-cot2=tan[2-(π/2)],
Then a = 2 - (π / 2)

Application of four trigonometric functions in senior one
"Is the root sign, sin a / 2 + cos A / 2 = 2" 3 "/ 2, find Sina cos2a

Sin (α / 2) + cos (α / 2) = (√ 3) / 2, find sin α Cos2 α
The sum of the squares of the two sides is 1 + sin α = 3 / 4, so sin α = - 1 / 4
cos2α=1-2sin²α=1-2×(-1/4)²=1-1/8=7/8.