Question
If sin(x + y) = 1 and sin(x - y) = (1/2), then what is
the value of sinx cosx + 2sin2y + cos2y?Solution
We have given, sin(x + y) = 1 and sin(x - y) = 1/2 So, sin(x + y) = 1 Or, sin(x + y) = sin90° Or, (x + y) = 90° ......(i) Again, sin(x - y) = 1/2 Or, sin(x - y) = sin30° Or, (x - y) = 30° ......(ii) By adding (i) and (ii) we get, 2x = 120° Or, x = 60° Putting the value of 'x' in the eq (i) we get, y = 30° Now, sinx cosx + 2sin2y + cos2y = sin60° X cos60° + 2 X sin230° + cos230° = = (1/2) X (√3/2) + 2 X (1/2)2 + (√3/2)2 = (1/2) X (√3/2) + 2 X (1/4) + (3/4) = (√3/4) + (1/2) + (3/4) = {(√3 + 2 + 3)/4} = {(√3 + 5)/4}
12.5% of (100 + ?) = 40
2/9 of 5/8 of 3/25 of ? = 40
24 × √? + 4008 ÷ 24 = 40% of 200 + 327
7(1/2) – 3(5/6) = ? − 2(7/12)
280 ÷ 14 + 11 × 12 – 15 × 6 = ?
1550 ÷ 62 + 54.6 x 36 = (? x 10) + (28.5 x 40)
25% of 1000 + 10% of 150 – 22 × ? = 45
√ 729 × 5 – 220 % of 15 + ? = 120% of 160
What will come in the place of question mark (?) in the given expression?
(40% of ? × 43 ) – 232 = 751
180 % of 45 + √144 × 8 = ?2 + 80 % of 70