Write 15 = 45 - 30. Using sin(A - B) = sin A cos B - cos A sin B: sin 15 = sin 45 cos 30 - cos 45 sin 30 = (1/sqrt2)(sqrt3/2) - (1/sqrt2)(1/2) = (sqrt3 - 1)/(2 sqrt2) = (sqrt6 - sqrt2)/4.

sin(A - B) = sin A cos B - cos A sin B

Write sin 3x = sin(2x + x) = sin 2x cos x + cos 2x sin x. Substitute sin 2x = 2 sin x cos x and cos 2x = 1 - 2 sin^2 x: = (2 sin x cos x) cos x + (1 - 2 sin^2 x) sin x = 2 sin x cos^2 x + sin x - 2 sin^3 x. Replace cos^2 x = 1 - sin^2 x: = 2 sin x (1 - sin^2 x) + sin x - 2 sin^3 x = 2 sin x - 2 sin^3 x + sin x - 2 sin^3 x = 3 sin x - 4 sin^3 x. Hence proved.

sin 3x = 3 sin x - 4 sin^3 x

Write tan A = sin A/cos A and cot A = cos A/sin A. First term: cos A/(1 - sin A/cos A) = cos A/((cos A - sin A)/cos A) = cos^2 A/(cos A - sin A). Second term: sin A/(1 - cos A/sin A) = sin A/((sin A - cos A)/sin A) = sin^2 A/(sin A - cos A) = -sin^2 A/(cos A - sin A). Adding: [cos^2 A - sin^2 A]/(cos A - sin A) = [(cos A - sin A)(cos A + sin A)]/(cos A - sin A) = cos A + sin A. Hence proved.

cos^2 A - sin^2 A = (cosA - sinA)(cosA + sinA)

(1 + i)(2 - i) = 1 x 2 + 1 x (-i) + i x 2 + i x (-i) = 2 - i + 2i - i^2. Since i^2 = -1, this is 2 + i + 1 = 3 + i. So a = 3 and b = 1.

i^2 = -1

The multiplicative inverse of z is 1/z = conjugate(z)/|z|^2. Here conjugate of (4 - 3i) is (4 + 3i) and |z|^2 = 4^2 + (-3)^2 = 16 + 9 = 25. So the inverse = (4 + 3i)/25 = 4/25 + (3/25) i.

z^-1 = conjugate(z)/|z|^2

Modulus |z| = sqrt(1^2 + (sqrt3)^2) = sqrt(1 + 3) = sqrt4 = 2. For the argument, tan(theta) = sqrt3/1 = sqrt3, and since both real and imaginary parts are positive (first quadrant), theta = pi/3. Polar form: z = 2(cos(pi/3) + i sin(pi/3)).

z = r(cos theta + i sin theta)

Multiply numerator and denominator by the conjugate of the denominator (1 + 2i): [(3 + 4i)(1 + 2i)]/[(1 - 2i)(1 + 2i)]. Numerator = 3 + 6i + 4i + 8i^2 = 3 + 10i - 8 = -5 + 10i. Denominator = 1^2 + 2^2 = 5. So the result = (-5 + 10i)/5 = -1 + 2i.

multiply by conjugate of denominator

Using the quadratic formula x = [-b +/- sqrt(b^2 - 4ac)]/(2a) with a = 1, b = 1, c = 1: discriminant = 1 - 4 = -3. So x = [-1 +/- sqrt(-3)]/2 = [-1 +/- i sqrt3]/2. The roots are (-1 + i sqrt3)/2 and (-1 - i sqrt3)/2.

x = [-b +/- sqrt(b^2 - 4ac)]/(2a)

Here a = 1, b = -2, c = 5. Discriminant = b^2 - 4ac = 4 - 20 = -16. So x = [2 +/- sqrt(-16)]/2 = [2 +/- 4i]/2 = 1 +/- 2i. The roots are 1 + 2i and 1 - 2i.

x = [-b +/- sqrt(b^2 - 4ac)]/(2a)

Powers of i repeat with period 4 (i, -1, -i, 1). i^9 = i^(8+1) = (i^4)^2 x i = 1 x i = i. i^19 = i^(16+3) = (i^4)^4 x i^3 = 1 x (-i) = -i. So i^9 + i^19 = i + (-i) = 0.

i^4 = 1

Let sqrt(-5 + 12i) = a + i b. Squaring: a^2 - b^2 + 2ab i = -5 + 12i. So a^2 - b^2 = -5 and 2ab = 12 (ab = 6). Also |z| = sqrt((-5)^2 + 12^2) = sqrt(25 + 144) = 13, so a^2 + b^2 = 13. Adding a^2 - b^2 = -5 and a^2 + b^2 = 13 gives 2a^2 = 8, a^2 = 4, a = +/-2; then b^2 = 9, b = +/-3. Since ab = 6 is positive, a and b have the same sign. Hence sqrt(-5 + 12i) = +/- (2 + 3i).

(a + ib)^2 = a^2 - b^2 + 2ab i

3x - 5 < 7 gives 3x < 12, so x < 4. The solution set is {x : x < 4, x is real} = (-infinity, 4). On the number line, this is shown by an open circle at 4 with shading to the left (all values less than 4).

5x - 3 >= 3x - 5. Subtract 3x from both sides: 2x - 3 >= -5. Add 3: 2x >= -2. Divide by 2: x >= -1. The solution set is {x : x >= -1} = [-1, infinity). On the number line, a closed (filled) circle at -1 with shading to the right.

First inequality: 2x - 1 < 5 gives 2x < 6, so x < 3. Second inequality: 3x + 2 > -4 gives 3x > -6, so x > -2. The common solution is the intersection: -2 < x < 3, i.e. the interval (-2, 3).

Let the shortest side = x cm. Then the longest side = 3x and the third side = 3x - 2. Perimeter = x + 3x + (3x - 2) = 7x - 2. Given perimeter is at least 61: 7x - 2 >= 61, so 7x >= 63, giving x >= 9. Hence the minimum length of the shortest side is 9 cm.

perimeter = sum of sides

Write 8! = 8 x 7 x 6!. So 8!/(6! x 2!) = (8 x 7 x 6!)/(6! x 2!) = (8 x 7)/2! = 56/2 = 28.

n! = n x (n-1)!

The hundreds place can be filled in 9 ways (any of 1 to 9), the tens place in 8 ways (remaining digits) and the units place in 7 ways. By the multiplication principle, total numbers = 9 x 8 x 7 = 504.

total = product of choices at each place

MATHEMATICS has 11 letters in which M appears 2 times, A appears 2 times and T appears 2 times; the rest (H, E, I, C, S) are distinct. Number of distinct arrangements = 11! / (2! x 2! x 2!) = 39916800 / 8 = 4989600.

arrangements = n!/(p! q! r!)

nC2 = n(n-1)/2 = 15, so n(n-1) = 30. We need two consecutive integers whose product is 30: 6 x 5 = 30, so n = 6. (n = -5 is rejected since n must be a positive integer.)

nC2 = n(n-1)/2

Since order does not matter in a committee, we use combinations. Number of ways = 10C3 = 10!/(3! x 7!) = (10 x 9 x 8)/(3 x 2 x 1) = 720/6 = 120.

nCr = n!/(r!(n-r)!)