It is known that sina = 2 / 3, cosx = - 3 / 4, and a and X are the second quadrant angles. Find sin (A-X), cos (a + x), Tan (a + x)

It is known that sina = 2 / 3, cosx = - 3 / 4, and a and X are the second quadrant angles. Find sin (A-X), cos (a + x), Tan (a + x)

Cosa = ± √ [1 - (Sina) ^ 2] = ± √ 5 / 3sinx = ± √ [1 - (cosx) ^ 2] = ± √ 7 / 4 ∵ a, X is the second quadrant angle ℅ cosa = - √ 5 / 3, SiNx = √ 7 / 4sin (A-X) = sinacosx sinxcosa = (√ 35 - 6) / 12cos (a + x) = cosacosx sinasinx = (3 √ 5 - 2 √ 7) / 12tana = Sina / cos

Given that the final edge of angle a is on the straight line y = - 2x.. find the value of sina

Because the end of angle a is on y = - 2x, Tana is the slope, which means Tana = - 2
So SecA ^ 2 = 1 + Tana ^ 2 = 5
SecA = positive and negative root sign 5
Cosa = 1 / SecA = positive and negative root sign 5 / 5
Sina = cosa * Tana = plus or minus 2 root sign 5 / 5

If sin [π + a] = - 1 / 3 and a is the second quadrant angle, then sin2a

Hello
It is known that sin [π + a] = - Sina = - 1 / 3
sina=1/3
A is the second quadrant angle
cosa=-√（1-sin²a）=-2√3/3
sin2a=2sinacosa=-4√3/9
(the denominator is outside the root)
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If cos (90 ° - a) = 3 / 5 and a is the second quadrant angle, then sin2a=

cos（90°-a）=sina =3/5
Cosa = - radical [1 - (3 / 5) & sup2;] = - 4 / 5 (a is the second quadrant angle)
So sin2a = 2sinacosa = 2 * 3 / 5 * (- 4 / 5) = - 24 / 25