Solve for A: (0° ≤ A ≤ 90°) cos(A − 20) = sin(A + 30)

here are the solutions to the questions. QUESTION 1 1.1 Solve for A (0^ A 90^) (A - 20^) = (A + 30^) Step 1: Convert (A + 30^) to a cosine function using the co-function identity = (90^ - ). (A + 30^) = (90^ - (A + 30^)) (A + 30^) = (90^ - A - 30^) (A + 30^) = (60^ - A) Step 2: Equate the arguments of the cosine functions. (A - 20^) = (60^ - A) For X = Y, the general solution is X = ± Y + k · 360^, where k is an integer. Considering the positive case: A - 20^ = 60^ - A + k · 360^ 2A = 80^ + k · 360^ A = 40^ + k · 180^ Step 3: Find the value of A within the given range (0^ A 90^). For k=0: A = 40^ + 0 · 180^ = 40^ This value is within the specified range. (If we consider the negative case A - 20^ = -(60^ - A) + k · 360^, it leads to -20^ = -60^ + k · 360^, which simplifies to 40^ = k · 360^, meaning k = (1)/(9), which is not an integer, so no solutions from this case.) The solution is A = 40^. 1.2 Prove the following identity ( 2x)/(1 + 2x) = x Step 1: Start with the Left Hand Side (LHS) and apply double angle identities. Recall the identities: 2x = 2 x x and 2x = 2 ^2 x - 1. LHS = ( 2x)/(1 + 2x) Substitute the identities into the expression: LHS = (2 x x)/(1 + (2 ^2 x - 1)) Step 2: Simplify the expression. LHS = (2 x x)/(2 ^2 x) Cancel out 2 x from the numerator and denominator: LHS = ( x)/( x) Step 3: Recognize the tangent identity. LHS = x Since LHS = RHS, the identity is proven. 1.3 Simplify without using a calculator, (1)/( 75^) Step 1: Use the reciprocal identity for secant. Recall that = (1)/( ). (1)/( 75^) = 75^ Step 2: Use the angle addition formula for cosine. Recall (A+B) = A B - A B. We can write 75^ as 45^ + 30^. 75^ = (45^ + 30^) 75^ = 45^ 30^ - 45^ 30^ Step 3: Substitute the known exact values of trigonometric functions. 45^ = sqrt(2)2 30^ = sqrt(3)2 45^ = sqrt(2)2 30^ = (1)/(2) 75^ = (sqrt(2)2)(sqrt(3)2) - (sqrt(2)2)((1)/(2)) 75^ = sqrt(6)4 - sqrt(2)4 Step 4: Combine the terms. 75^ = sqrt(6) - sqrt(2)4 1.4 Simplify: ((-) (90^ - ) ^2(90^ - ))/((180^ - ) (360^ - )) Step 1: Apply reduction formulas and co-function identities to each term. (-) = - (90^ - ) = (90^ - ) = ^2(90^ - ) = ^2 (180^ - ) = (360^ - ) = Step 2: Substitute these simplified terms into the expression. Expression = ((- )( )(^2 ))/(( )( )) Step 3: Simplify the expression by canceling common terms and using = ( )/( ). Expression = (-^2 ^2 )/( ) Cancel one from the numerator and denominator: Expression = (- ^2 )/( ) Substitute ^2 = (^2 )/(^2 ): Expression = (- (^2 )/(^2 )) Expression = (-^2 )/( ) Expression = (-^2 )/( ) Cancel one from the numerator and denominator: Expression = (- )/( ) Step 4: Recognize the cotangent identity. Expression = - QUESTION 2 2.1 Differentiate using the first principle: y = 3x^2 - 2x The first principle of differentiation is given by f'(x) = _h 0 (f(x+h) - f(x))/(h). Given f(x) = 3x^2 - 2x. Step 1: Find f(x+h). f(x+h) = 3(x+h)^2 - 2(x+h) f(x+h) = 3(x^2 + 2xh + h^2) - 2x - 2h f(x+h) = 3x^2 + 6xh + 3h^2 - 2x - 2h Step 2: Find f(x+h) - f(x). f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 2x - 2h) - (3x^2 - 2x) f(x+h) - f(x) = 3x^2 + 6xh + 3h^2 - 2x - 2h - 3x^2 + 2x f(x+h) - f(x) = 6xh + 3h^2 - 2h Step 3: Divide by h. (f(x+h) - f(x))/(h) = (6xh + 3h^2 - 2h)/(h) (f(x+h) - f(x))/(h) = (h(6x + 3h - 2))/(h) (f(x+h) - f(x))/(h) = 6x + 3h - 2 Step 4: Take the limit as h 0. f'(x) = _h 0 (6x + 3h - 2) f'(x) = 6x + 3(0) - 2 f'(x) = 6x - 2 2.2 Expand to four terms by using the binomial theorem. (2x - 4)^(1)/(2) Step 1: Rewrite the expression to fit the form (a+b)^n or (1+u)^n. Factor out 2x from the expression: (2x - 4)^(1)/(2) = [2x(1 - (4)/(2x))]^(1)/(2) = (2x)^(1)/(2) (1 - (2)/(x))^(1)/(2) Let n = (1)/(2) and u = -(2)/(x). The generalized binomial theorem is (1+u)^n = 1 + nu + (n(n-1))/(2!) u^2 + (n(n-1)(n-2))/(3!) u^3 + Step 2: Calculate the first four terms of the expansion for (1 - (2)/(x))^(1)/(2). Term 1: 1 Term 2: nu = (1)/(2)(-(2)/(x)) = -(1)/(x) Term 3: (n(n-1))/(2!) u^2 = (1)/(2)((1)/(2)-1)2 (-(2)/(x))^2 = (1)/(2)(-(1)/(2))2 ((4)/(x^2)) = (-1)/(4)2 ((4)/(x^2)) = -(1)/(8) · (4)/(x^2) = -(1)/(2x^2) Term 4: (n(n-1)(n-2))/(3!) u^3 = (1)/(2)((1)/(2)-1)((1)/(2)-2)6 (-(2)/(x))^3 = (1)/(2)(-(1)/(2))(-(3)/(2))6 (-(8)/(x^3)) = (3)/(8)6 (-(8)/(x^3)) = (3)/(48) (-(8)/(x^3)) = (1)/(16) (-(8)/(x^3)) = -(1)/(2x^3) So, (1 - (2)/(x))^(1)/(2) = 1 - (1)/(x) - (1)/(2x^2) - (1)/(2x^3) + Step 3: Multiply the expansion by (2x)^(1)/(2) = sqrt(2x). (2x - 4)^(1)/(2) = sqrt(2x) (1 - (1)/(x) - (1)/(2x^2) - (1)/(2x^3) + ) (2x - 4)^(1)/(2) = sqrt(2x) - sqrt(2x)x - sqrt(2x)2x^2 - sqrt(2x)2x^3 + To simplify the terms, we can write sqrt(2x)x = sqrt(2x)sqrt(x)xsqrt(x) = sqrt(2)xxsqrt(x) = sqrt(2)sqrt(x) = sqrt((2)/(x)). Alternatively, sqrt(2x)x = sqrt(2)sqrt(x)x = sqrt(2)sqrt(x). Let's keep the form sqrt(2x)x or rationalize the denominator for each term. sqrt(2x)x = sqrt(2x) · sqrt(2x)x · sqrt(2x) = (2x)/(xsqrt(2x)) = (2)/(sqrt(2x)). sqrt(2x)2x^2 = sqrt(2x) · sqrt(2x)2x^2 · sqrt(2x) = (2x)/(2x^2sqrt(2x)) = (1)/(xsqrt(2x)). sqrt(2x)2x^3 = sqrt(2x) · sqrt(2x)2x^3 · sqrt(2x) = (2x)/(2x^3sqrt(2x)) = (1)/(x^2sqrt(2x)). So the expansion is: sqrt(2x) - (2)/(sqrt(2x)) - (1)/(xsqrt(2x)) - (1)/(x^2sqrt(2x)) + 2.3 Differentiate by using quotient rule y = (3)/(2x+1) The quotient rule states that if y = (u)/(v), then (dy)/(dx) = (u'v - uv')/(v^2). Here, u = 3 and v = 2x+1. Step 1: Find the derivatives of u and v. u = 3 u' = 0 v = 2x+1 v' = 2 Step 2: Apply the quotient rule formula. (dy)/(dx) = ((0)(2x+1) - (3)(2))/((2x+1)^2) (dy)/(dx) = (0 - 6)/((2x+1)^2) (dy)/(dx) = (-6)/((2x+1)^2) 2.4 Given the function y = x^3 - 2x^2 - 5x + 6. Determine, with the aid of differentiation, the co-ordinates of maximum and minimum turning points an distinguish between the maximum and minimum turning points by using the second derivative. Step 1: Find the first derivative ((dy)/(dx)) and set it to zero to find the x-coordinates of the turning points. y = x^3 - 2x^2 - 5x + 6 (dy)/(dx) = 3x^2 - 4x - 5 Set (dy)/(dx) = 0: 3x^2 - 4x - 5 = 0 Use the quadratic formula x = -b ± sqrt(b^2 - 4ac)2a: x = -(-4) ± sqrt((-4)^2 - 4(3)(-5))2(3) x = 4 ± sqrt(16 + 60)6 x = 4 ± sqrt(76)6 x = 4 ± 2sqrt(19)6 x = 2 ± sqrt(19)3 The x-coordinates of the turning points are x_1 = 2 + sqrt(19)3 and x_2 = 2 - sqrt(19)3. Step 2: Find the second derivative ((d^2y)/(dx^2)). (d^2y)/(dx^2) = (d)/(dx)(3x^2 - 4x - 5) (d^2y)/(dx^2) = 6x - 4 Step 3: Use the second derivative to distinguish between maximum and minimum turning points. For x_1 = 2 + sqrt(19)3: (d^2y)/(dx^2) |_x_1 = 6(2 + sqrt(19)3) - 4 = 2(2 + sqrt(19)) - 4 = 4 + 2sqrt(19) - 4 = 2sqrt(19) Since 2sqrt(19) > 0, this is a minimum turning point. For x_2 = 2 - sqrt(19)3: (d^2y)/(dx^2) |_x_2 = 6(2 - sqrt(19)3) - 4 = 2(2 - sqrt(19)) - 4 = 4 - 2sqrt(19) - 4 = -2sqrt(19) Since -2sqrt(19) < 0, this is a maximum turning point. Step 4: Calculate the y-coordinates for each turning point. Substitute the x-values back into the original function y = x^3 - 2x^2 - 5x + 6. A simplified form for y can be derived from 3x^2 - 4x - 5 = 0 x^2 = (4x+5)/(3): y = x(x^2) - 2x^2 - 5x + 6 y = x((4x+5)/(3)) - 2((4x+5)/(3)) - 5x + 6 y = (4x^2+5x - 8x-10 - 15x + 18)/(3) y = (4x^2 - 18x + 8)/(3) Substitute x^2 = (4x+5)/(3) again: y = (4(4x+5)/(3)) - 18x + 83 y = (16x+20)/(3) - 18x + 83 y = (16x+20 - 54x + 24)/(9) y = (-38x + 44)/(9) For the minimum turning point (x_1 = 2 + sqrt(19)3): y_1 = -38(2 + sqrt(19)3) + 449 y_1 = -76 - 38sqrt(19) + 13227 y_1 = 56 - 381 ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._