If sin θ= Five 3，θ∈(π 2, π), then Tan θ =___ .

If sin θ= Five 3，θ∈(π 2, π), then Tan θ =___ .

∵θ∈(π
2，π)，∴cosθ＜0
And ∵ sin θ=
Five
3，
∴cosθ=-
1-sin2θ=-2
3, Tan θ = sin θ
cosθ=-
Five
Two
So the answer is-
Five
Two

sin(-11π/6)+cos12π/5·tan3π+cos(-π/4)

The solution sin (- 11 π / 6) = sin (2 π - 11 π / 6) = sin π / 6 = 1 / 2cos12 π / 5 · tan3 π = cos12 π / 5 × Tan π = cos12 π / 5 × 0 = 0cos (- π / 4) = cos (π / 4) = √ 2 / 2, so sin (- 11 π / 6) + cos12 π / 5 · tan3 π + cos (- π / 4) = 1 / 2 + 0 + √ 2 / 2 = (√ 2 + 1) / 2

Comparison size: sin (- 15 π / 8) ---- sin (- 14 π / 9) tan89°-----tan91°

Sin (- 15 π / 8) is greater than sin (- 14 π / 9)
Tan 89 ° is greater than Tan 91 °

How to find the sin, cos, Tan of 15,75 degrees? How much How to ask? Then if the result is a decimal (fraction), write it in fractional form, not decimal form Oh, my God. I don't understand. It's like sin30 = 1 / 2 If you are a teacher, can you make sure that the whole class can read it? If you write clearly, it is impossible for the whole class to understand it, but at least half of the students can understand it

Sin (75) = sin (30 + 45) = sin30cos45 + sin45cos30sin15 = sin (45-30) = sin45cos30-sin30cos45cos75 = cos (45 + 30) = cos45cos30-sin30sin45cos15 = cos (45-30) = cos45cos30 + sin30sin45tan. There is no simple formula but to find it slowly

In mathematics, Tan sin, etc. What are these?

These are called trigonometric functions

The value of SiN4 π / 3 × cos25 π / 6 × tan5 π / 4 is

1)
sin4π/3×cos25π/6×tan5π/4
=0/3*(-1)/6*0/4
=0
2)
sin4π/3×cos25π/6×tan5π/4
=(sin4π/3)×cos(25π/6)×tan(5π/4)
=(-√3/2)(√3/2)(1)
= - 3/4

sin4π 3•cos25π 6•tan5π The value of 4 is______ .

The original formula = sin (π + π)
3）•cos（4π+π
6）•tan（π+π
4）=-
Three
2 x
Three
2×1=-3
4．
So the answer is: - 3
Four

Evaluate cos25 / 3 π + Tan (- 15 / 4 π)

cos25/3*π+tan（－15/4*π）
=cos(8+1/3)π+tan（4－1/4)π
=cos(1/3 *π)+tan(－1/4 *π)
=1/2 -1=-1/2

Given that π / 6 is less than α and less than 2 π / 3, cos (α + π / 3) = m (m ≠ 0), the value of Tan (2 π / 3 - α) is calculated

cos²(α+π/3)+sin²(α+π/3)=1
From the value range, sin (α + π / 3) > 0
sin(α+π/3)=√(1-m²)
So the original formula = - Tan (α + π / 3) = - √ (1-m 2) / m

Given that cos (α - π / 3) = 2 / 3, find the value of COS (2 π / 3 + α) + Tan ^ 2 (4 π / 3 - α)

Given that cos (α - π / 3) = 2 / 3, cos (2 π / 3 + α) + Tan ^ 2 (4 π / 3 - α) = cos (π - π / 3 + α) + Tan ^ 2 (π + π / 3 - α)
=-cos(α-π/3)+sin^2(α-π/3)/cos^2(α-π/3)=-2/3+5/4=7/12