If sin (π + θ) < 0 and COS (π - θ) < 0, then the quadrant where angle θ is located is

If sin (π + θ) < 0 and COS (π - θ) < 0, then the quadrant where angle θ is located is

third quadrant

Given that sin α > 0 and cos α ≤ 0, it is determined that the angle α is the position of the quadrant angle or final edge

The second quadrant angle or final edge is on the positive half axis of Y axis

Given Tan a = - 4, find the value of (1) sin ^ 2 a (2) 3sin ACOS a (3) cos ^ 2 a-SiN ^ 2 a (4) 4sin a Given Tan a = - 4, find the values of (1) sin ^ 2 a (2) 3sin ACOS a (3) cos ^ 2 a-SiN ^ 2 a (4) (4sin a-2cos a) / (5cos a + 3sin a) a as an arbitrary angle

(1) Tana = - 4 ν COTA = - 1 / 4 CSCA = ± √ (COT? A + 1) = ± √ 17 / 4sina = ± (4 / 17) √ 17 (2) 3sinacosa = (3 / 2) sin2a universal formula: sin2a = 2tana / (1 + tan? A) = - 8 / 173sinacosa = (3 / 2) sin2a = - 12 / 17 (3) cos? A-SiN? A = cos2a universal

Given sin a + cos a = - 1 / 5, find 1) sin ACOS a 2) sin a cos a

Sin a + cos a = - 1 / 5 square (sin a + cos a) ^ 2 = 1 / 25 (sin a) ^ 2 + 2Sin ACOS a + (COS a) ^ 2 = 1 / 25 1 + 2Sin ACOS a = 1 / 25 2Sin ACOS a = - 24 / 25 sin ACOs a = - 12 / 25 (sin a-cos a) ^ 2 = (sin a) ^ 2-2sin ACOS a + (COS a) ^ 2 = 1-2sin ACOS

It is known that sin α = asin β, bcos α = ACOS β, and α, β are acute angles. It is proved that cos α = √ (a ^ 2-1) / (b ^ 2-1) √ is the root sign

It is proved that sin α = asin β, bcos α = ACOS β, (sin α) ^ 2 = a ^ 2 (sin β) ^ 2, B ^ 2 (COS α) ^ 2 = a ^ 2 (COS β) ^ 2, 1 - (COS α) ^ 2 + B ^ 2 (COA α) ^ 2 = a ^ 2 (COS α) ^ 2 = (a ^ 2-1) / (b ^ 2-1 ≠ 0) α is an acute angle, cos α = √ (a ^ 2-1) / (b ^ 2-1)

What is the value of sin ^ 6A + cos ^ 6A + 3sin ^ 2acos ^ 2A?

sin^6a+cos^6a+3sin^2acos^2a=（sin²a＋cos²a）（sin^4a-sin^2acos^2a+cos^4a）+3sin^2acos^2a=sin^4a-sin^2acos^2a+cos^4a+3sin^2acos^2a=sin^4a＋2sin^2acos^2a+cos^4a=（sin²a＋cos²a）²...

It is proved that sin ^ 6A + cos ^ 6A = 1-3sin ^ 2acos ^ a

sin^6A+cos^6A=(sin^2A+cos^2A)(sin^4A+cos^4A-sin^2Acos^2A)=sin^4A+cos^4A+2sin^2Acos^2A-32sin^2Acos^2A
=(sin^2A+cos^2A)^2-3sin^2Acos^A
=1-3sin^2Acos^A

Verification (1-sin ^ 2acos ^ 2a) (cos2a) = cos ^ 6a-sin ^ 6A

Because sin? A + cos? A = 1, the left = [(sin? 2A + cos? A)? - sin? ACOS? A)] cos2a = (sin ⁴ a + 2Sin? ACOS? A + cos ⁴ a-SiN? ACOS? A) (COS? A-SiN? A) = (COS? A-S

Simplification: y = √ 3 cos? X-sin? X - √ 3sin? X and find the period of y = f (x), the X set corresponding to the single increasing interval and the maximum value

In this paper, we obtain the maximum value of π - 2x - 2, where π - 2x - 2 is the maximum

Simplify 1 / sin? X + 1 / cos? X to be equal to

1 / sin? X + 1 / cos? X
(sin²x*cos²x )/sin²x*cos²x
=1/sin²x*cos²x
=1/(sinx*cosx)² (sin2x=2sinxcosx,sinx*cosx=sin2x/2)
=4/sin²(2x)