(sin10 π / 11 * sin8 π / 11 * sin6 π / 11) / (sin π / 11 * SIN3 π / 11 * sin5 π / 11) * 1 / 2 ^ 5 = 1 / 2 ^ 5, how to get this?

(sin10 π / 11 * sin8 π / 11 * sin6 π / 11) / (sin π / 11 * SIN3 π / 11 * sin5 π / 11) * 1 / 2 ^ 5 = 1 / 2 ^ 5, how to get this?

Well, the key is to use the [complement formula]:
Sin (π - a) = Sina. So sin (10 π / 11) = sin (π / 11). And so on
The upper limit is divided by each other, soldiers to soldiers, generals to generals
As a result, there are only "hard bones" left. I am equal to myself. 1 / 2 ^ 5 = 1 / 2 ^ 5

sinπ/12-sin5π/12+2sinπ/8sin3π/8

The steps are as follows: sin (15 degrees) = sin (45 degrees - 30 degrees) = (√ 6 - √ 2) / 4sin (75 degrees) = sin (45 degrees + 20 degrees) = (√ 6 + √ 2) / 4 because sin (22.5 degrees) = sin (90 degrees - 67.5 degrees) = cos (67.5 degrees), so 2Sin (22.5 degrees) sin

What's the sum of sin divided by two times cos divided by two
[sin^2(x/2)]*[cos^2(x/2)]

Is it [sin (x / 2)] ^ 2 * [cos (x / 2)] ^ 2?
If so, then
[sin(x/2)]^2*[cos(x/2)]^2
=[sin(x/2)cos(x/2)]^2
=[(1/2)*sin(x/2)cos(x/2)]^2
=(1/4)[sin(2*x/2)}^2
=(1/4)(sinx)^2

It is known that (sin α plus cos α) divided by (sin α minus cos α) is equal to 2

= =.
（sinα+cosα）/（sinα-cosα）=2
（sinα+cosα）^2/（sinα-cosα）^2=4
(sin α + 2Sin α cos α + cos α) / (sin α - 2Sin α cos α + cos α) = 4
2sinαcosα=3/5
sinαcosα=3/10

Given that sin a divided by cos a equals 3, how much is sin a multiplied by cos a?

Sin a = 3 cos a, sin ` 2 + cos ` 2 = 1
Then 10 cos a ` 2 = 1
cos `2=1/10
sin a * cos a=3cos a ^2=3/10

Known 0

Cosa = under the positive and negative root sign (the square of 1-sin α), because 0

It is known that cos Alfa divided by 1 minus sin alfa is equal to the root sign 3 plus 1 divided by 2
It is known that cos a'erfa divided by 1 minus sin a'erfa equals the root sign 3 plus 1 divided by 2. Find 1 plus sin a'erfa divided by cos a'erfa

The numerator denominator on the left side of the formula is multiplied by (1 + sin α)
cosα / (1 - sinα)
=cosα（1+sinα）/{ (1-sinα)（1+sinα）}
=cosα(1+sinα）/ (1- sin^2α)
=cosα(1+sinα）/{cos^2α
=(1 + sin α) / cos α = (radical 3 + 1) / 2

Simplifying sin [(K + 1) π + θ] * cos [(K + 1) π - θ] / sin (K π - θ) * cos (K π + θ) k ∈ Z

When k is even, i.e. k = 2n, n ∈ Z,
sin[(k+1)π+θ]*cos[(k+1)π-θ]/sin（kπ-θ）*cos（kπ+θ）=sinθ*cosθ/sin（-θ）*cosθ=-1
When k is odd, i.e. k = 2n + 1, n ∈ Z,
sin[(k+1)π+θ]*cos[(k+1)π-θ]/sin（kπ-θ）*cos（kπ+θ）=sinθ*cosθ/sinθ*（-cosθ）=-1
To sum up, we can get the conclusion
sin[(k+1)π+θ]*cos[(k+1)π-θ]/sin（kπ-θ）*cos（kπ+θ）=-1 ,（k∈z）

Simplification {sin [α + (K + 1) π] + sin [α - (K + 1) π]} / [sin (α + K π) * cos (α - K π)], K ∈ Z

The formula sin [a + b] + sin [A-B] = 2sina * CoSb
A simplification of 2Sin α cos [(K + 1) π] / [sin (α + K π) * cos (α - K π)]
When k is odd
2sinα/[(-sinα)*(-cosα)]= 2/cosα
When k is even
-2sinα/[(-sinα)*(-cosα)]= -2/cosα

Simplification: cos [(K + 1) π - α] * sin (K π - α) / cos (K π + α) * sin [(K + 1) π + α] please

K is odd sin (K π - θ) × cos (K π θ) = - sin θ cos θ sin [(K 1) π θ] × cos [(K 1) π - θ] = sin θ cos θ K is even sin (K π - θ) × cos (K π θ) = - sin θ cos θ sin [(K 1) π θ] × cos [(K 1) π - θ] = sin θ cos θ, so sin [(K 1) π - θ] = sin θ cos θ