It is known that cos (75 ° + α) = 5 / 13 α is the value of sin (195 ° - α) + (α - 15 °) in the third quadrant

It is known that cos (75 ° + α) = 5 / 13 α is the value of sin (195 ° - α) + (α - 15 °) in the third quadrant

sin(75°＋α)=-12/13
Sin (195 ° - α) = sin [(75 ° + a) + 120 °] = sin (75 ° + a) cos120 ° + cos (75 ° + a) sin120 ° = (5 root number 3 + 12) / 26
sin(a-15°)=sin[(75°+a)-90°]=-cos(75°+a)=-5/13

If cos α = - 3 / 5, α∈ (π / 2, π), sin β = - 12 / 13, β is the third quadrant angle, then the value of COS (β - α) is () if a > 0, b > 0, a + B = 1, then the minimum value of y = 1 / A + 4 / B is () if all the items of the proportional sequence {an} are positive, and a3a3a8 + a5a6 = 18, log3a1 + log3a2 +. + log3a10 = ()

63/65；7；3+log3（2）

It is known that sin α = 2 / 3, cos β = - 3 / 4 α belongs to (π / 2, π) β is the third quadrant angle, and the values of COS (α + β) and sin (α - β) are calculated

Don't you know what's under here_ Because sin α = 2 / 3, cos β = - 3 / 4 α belongs to (π / 2, π) β is the third quadrant angle, so cosa = - radical 5 / 3, sin β = - radical 7 / 4, so cos (a + β) = cosacos β - sinasin β = radical 5 / 4 - radical 7 / 6sin (a - β) = sinacos β - cosa

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, ᙽ K · 360 ° + 255 ° < α + 75 ° < K · 360 ° + 345 ° (K ∈ z),
∵cos(75°+α)＝1
3, να + 75 ° is the fourth quadrant angle,
∴sin(75°+α)＝−
1−(1
3)2＝−2
Two
3，
The original formula = cos (15 ° − α) − sin (15 ° − α) = sin (α + 75 °) − cos (α + 75 °) = − 2
2+1
3．

Given that sin [A-B] cos a-cos [B-A] sin a = 3 / 5, B is the third quadrant angle, find the value of sin [B + 5 π / 4] 1 / 2cos X - √ 3 / 2Sin x Question 2 √ 3sin x + cos x Question 3 √ 2 [sin x-cos x] Question 4 √ 2cos X - √ 6sin x

sin(a-b)cos a -cos(b-a)sin a= -[sin(b-a)cos a + cos(b-a) sina]= - sin(b-a + a)= -sinb=> sin b = -3/5=> cos b = -4/5sin(b+5π/4) = sin b cos(5π/4) + cos b sin(5π/4) = -3/5 * (-√2/2) - 4/5 * (-√2/2)...

Sin (- α) = 5 / 4, α is the third quadrant angle, you can find cos α

If sin α is greater than - 1 and less than 1, it can't be equal to 5 / 4. According to 4 / 5, sin (- α) = - sin α = 4 / 5, sin α = - 4 / 5. From α is the third quadrant angle, cos α is a negative number, cos 2 α = 1 - (4 / 5) 2 = 9 / 25cos α = - 9 / 25 = - 3 / 3 under the root sign

Given that a is an acute angle and COS ^ 4a-sin ^ 4A = 3 / 5, find sin2a （2） (sin ^ 2A + 3sinacosa cos ^ 2a) / (2Sin ^ 2A + cos ^ 2a)

cos^4a-sin^4a=3/5
(cos^2a-sin^2a)(cos^2a-sin^2a)=3/5
1*(cos^2a-sin^2a)=3/5
cos^2a-sin^2a=3/5
cos2a=3/5
A is an acute angle, ν 2A ∈ (0, π)
∴sin2a = √{1- (sin2a)^2} = √{1- (3/5)^2} = 4/5

If sin2a = 3 / 5, then sin ^ 4A + cos ^ 4a=

Because sin2a = 3 / 5 = 2sinacosa, sin? 2A + cos? A = 1
So sin ^ 4A + cos ^ 4A = (sin? 2A + cos? A)? - 2Sin? ACOS? A
=1²-sin2a/2
=1-9/50
=41/50

Given that a is the third quadrant angle and sin ^ 2 a-cos ^ 4 a = 5 / 9, the value of sin2a is

From the deformation of the above formula, cos ^ 4 A + cos ^ 2 A-4 / 9 = 0
If cos ^ 2 a = 1 / 3 or - 4 / 3 (R), then sin ^ 2 a = 1-cos ^ 2 a = 2 / 3
Since a is the third quadrant angle, cosa = - √ 3 / 3, Sina = √ 6 / 3,
sin2a=2sinacosa=2√2/3.

Given Tan α = - radical 3, find sin α, cos α,

∵ Tan α = - √ 3 ᙽ α = k π - π / 3 (k is an integer), so sin α = sin (K π - π / 3) = sin (K π) * cos (π / 3) - sin (π / 3) * cos (K π) = - (√ 3 / 2) * (- 1) ^ k = (√ 3 / 2) * - 1) ^ (K + 1) cos α = cos (K π - π / 3)