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In space quadrilateral ABCD, if vector AB = (- 3,5,2), vector CD = (- 7, - 1, - 4), points E and F are the midpoint of edge BC and ad respectively, and vector EF is equal to?
EF=EB+BA+AF
EF=EC+CD+DF
∵EB=-EC AF=-DF
∴2EF=BA+CD
∴EF=1/2(BA+CD)
=1/2[(3,-5,-2)+(-7,-1,-4)]
=(-2,-3,-3)
How does f (x) = - √ 3 sin2x + cos2x + 1 become the standard form of y = asin (ω x + φ) + B?
The answer I got from the auxiliary angle formula is: y = 2Sin (2x - sextile) + 1
But the correct answer is: y = 2Sin (2x + 5 / 6) + 1
Q1: y = 2Sin (2x - 5 / 6) + 1 and y = 2Sin (2x + 5 / 6) + 1
Q2: what's wrong with me? Y = 2Sin (2x + 5 / 6) + 1. What's the right answer?
Q1: different. Observe the image of trigonometric function Q2: mobile party is not easy to call! F (x) = - √ 3 sin2x + cos2x + 1 = 2 (- √ 3 / 2 sin2x + 1 / 2cos2x) + 1 = 2 (sin2x5 / 6 pie + cos2x5 / 6 pie) + 1 = y = 2Sin (2x + 5 / 6 pie) + 1 ps: sin150 ° = 1 / 2 cos150 ° = - cos30 ° = - root sign 3 / 2 use induction
How can f (x) = - √ 3 sin2x + cos2x + 1 be transformed into the standard form of y = asin (ω x + φ) + B by auxiliary diagonalization?
How does f (x) = - √ 3 sin2x + cos2x + 1 become the standard form of y = asin (ω x + φ) + B?
The answer I got from the auxiliary angle formula is: y = 2Sin (2x - sextile) + 1
But the correct answer is: y = 2Sin (2x + 5 / 6) + 1
What's wrong with me!
Q1: different. Observe the image of trigonometric function Q2: F (x) = - √ 3 sin2x + cos2x + 1 = 2 (- √ 3 / 2 sin2x + 1 / 2cos2x) + 1 = 2 (sin2x5 / 6 pie + cos2x5 / 6 pie) + 1 = y = 2Sin (2x + 5 / 6 pie) + 1 ps: sin150 ° = 1 / 2 cos150 ° = - cos30 ° = - root sign 3 / 2
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