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question:Let x and y be angles such that sin (x + y) = frac{5}{8} and sin (x y) = frac{1}{4}. Find frac{tan x}{tan y}.
answer:Using the product-to-sum identities, we have: [ 2 sin x cos y = sin (x + y) + sin (x - y) = frac{5}{8} + frac{1}{4} = frac{3}{4} ] [ 2 cos x sin y = sin (x + y) - sin (x - y) = frac{5}{8} - frac{1}{4} = frac{3}{8} ] Now, dividing these equations to solve for frac{tan x}{tan y}: [ frac{sin x cos y}{cos x sin y} = frac{frac{3}{4}}{frac{3}{8}} = 2 ] Thus, frac{tan x}{tan y} = boxed{2}.
question:Given that a,b in mathbb{R}, and ab<0, then ( ). A: |a+b|>|ab| B: |a+b|<|ab| C: |ab|<|a||b| D: |ab|<|a|+|b|
answer:Since ab < 0, a and b must have opposite signs. Without loss of generality, suppose a > 0 and b < 0 (if a < 0 and b > 0, the argument will follow similarly with the roles of a and b swapped). There are two cases to consider for the absolute value expressions: **Case 1**: If a+b geq 0, then both |a+b| and a+b represent the same number. Since a > 0 and b < 0, we have that a > -b, so |a+b| = a+b < a-b. As a-b is positive, |a-b| = a-b. Therefore, in this case: |a+b| < |a-b|. **Case 2**: If a+b < 0, then |a+b| = -(a+b). Since a > -b, -(a+b) < -(-b+b) = 0. However, a-b is still positive, as a is greater in magnitude than b, which means |a-b| = a-b > 0. Thus: |a+b| = -(a+b) < 0 < a-b = |a-b|. In both cases, we have |a+b| < |a-b|, therefore the correct choice is B: |a+b|<|a-b|. [boxed{B: |a+b|<|a-b|}]
question:Suppose ( z ) and ( w ) are complex numbers such that [ |z| = sqrt{2}, |w| = 2, text{ and } z overline{w} + overline{z} w = 1. ] Find the largest possible value of the real part of ( z + w ).
answer:Let ( z = a + bi ) and ( w = c + di ), where ( a, b, c, ) and ( d ) are real numbers. Then from ( |z| = sqrt{2} ), [ a^2 + b^2 = 2, ] and from ( |w| = 2 ), [ c^2 + d^2 = 4. ] Also, from ( z overline{w} + overline{z} w = 1 ), [ (a + bi)(c - di) + (a - bi)(c + di) = 1, ] so ( 2ac + 2bd = 1 ). Then consider the square of the sum of ( z ) and ( w ): [ (a + c)^2 + (b + d)^2 = a^2 + 2ac + c^2 + b^2 + 2bd + d^2 = (a^2 + b^2) + (c^2 + d^2) + 2ac + 2bd = 2 + 4 + 1 = 7. ] The real part of ( z + w ) is ( a + c ), which can be at most ( sqrt{7} ). Equality occurs when ( a + c ) and ( b + d ) are maximized under the given conditions, potentially needing specific values for ( a, b, c, ) and ( d ) to meet these conditions. The largest possible value of ( a + c ) is ( boxed{sqrt{7}} ).
question:Jerome has 20 classmates on his cell phone contact list. He has half as many out of school friends as classmates on his contact list. He also has some family members on his contact list. Jerome has 33 people on his contact list. How many family members does Jerome have on his contact list?
answer:Jerome has 20 classmates on his contact list. He has half as many out-of-school friends as classmates, so he has 20 / 2 = 10 out-of-school friends on his contact list. The total number of classmates and out-of-school friends on his contact list is 20 + 10 = 30. Jerome has 33 people on his contact list in total. To find out how many family members he has on his contact list, we subtract the number of classmates and out-of-school friends from the total number of contacts. 33 - 30 = boxed{3} family members on Jerome's contact list.