Prove the identity: sin 2x / 2 cos2 x + sin x / tan x [1 + cos(90 - x)] = 1 / cos x

QUESTION 5 5.1.1 Use the formula given above to derive a formula for (A+B). Step 1: Start with the given formula. We are given: (A-B) = A B - A B Step 2: Substitute -B for B in the formula. To derive (A+B), we can replace B with -B: (A-(-B)) = A (-B) - A (-B) Step 3: Apply trigonometric identities for negative angles. We know that (-B) = B and (-B) = - B. Substitute these into the equation: (A+B) = A B - A (- B) (A+B) = A B + A B (A+B) = A B + A B 5.1.2 Hence, without using a calculator, solve for x in the interval x [0^; 360^] if sqrt(3) x = (50^+x)(10^-x) + (50^+x)(10^-x). Step 1: Simplify the right-hand side (RHS) using the compound angle formula. The RHS matches the formula for (A+B) derived in 5.1.1, where A = 50^+x and B = 10^-x. RHS = ((50^+x) + (10^-x)) RHS = (50^+10^+x-x) RHS = (60^) Step 2: Substitute the value of (60^). We know that (60^) = sqrt(3)2. So the equation becomes: sqrt(3) x = sqrt(3)2 Step 3: Solve for x. Divide both sides by sqrt(3): x = (1)/(2) Step 4: Find the values of x in the given interval. The reference angle for x = (1)/(2) is 60^. Cosine is positive in the first and fourth quadrants. In the first quadrant: x = 60^ In the fourth quadrant: x = 360^ - 60^ = 300^ Both solutions are within the interval [0^; 360^]. x = 60^ or x = 300^ 5.2 Prove the identity: ( 2x)/(2 ^2 x) + ( x)/( x [1 + (90^ - x)]) = (1)/( x). Step 1: Simplify the first term of the Left Hand Side (LHS). ( 2x)/(2 ^2 x) = (2 x x)/(2 ^2 x) (using 2x = 2 x x) = ( x)/( x) Step 2: Simplify the second term of the LHS. Use the identities (90^ - x) = x and x = ( x)/( x). ( x)/( x [1 + (90^ - x)]) = ( x)/( x) x [1 + x] = ( x)/( x (1 + x)) x = x × ( x)/( x (1 + x)) = ( x)/(1 + x) Step 3: Add the simplified terms. LHS = ( x)/( x) + ( x)/(1 + x) Find a common denominator: = ( x (1 + x) + x ( x))/( x (1 + x)) = ( x + ^2 x + ^2 x)/( x (1 + x)) Step 4: Use the Pythagorean identity. We know ^2 x + ^2 x = 1. = ( x + 1)/( x (1 + x)) = (1 + x)/( x (1 + x)) Step 5: Cancel common factors. = (1)/( x) Since LHS = RHS, the identity is proven. ( 2x)/(2 ^2 x) + ( x)/( x [1 + (90^ - x)]) = (1)/( x) 5.3 Show, without using a calculator, that 260^ = -sqrt(( 20^ + 1)/(2)). Step 1: Simplify the left-hand side (LHS). Use reduction formula for 260^: 260^ = (180^ + 80^) Since sine is negative in the third quadrant: 260^ = - 80^ Step 2: Use the co-function identity. 80^ = -(90^ - 80^) = - 10^ So, LHS = - 10^. Step 3: Simplify the right-hand side (RHS). The expression sqrt(( 20^ + 1)/(2)) resembles the half-angle formula for cosine. Recall the identity ^2 = (1 + 2)/(2). Let = 10^. Then 2 = 20^. ^2 10^ = (1 + 20^)/(2) Taking the square root of both sides: 10^ = sqrt((1 + 20^)/(2)) (Since 10^ is in the first quadrant, 10^ is positive, so we take the positive square root). Step 4: Compare LHS and RHS. The RHS of the original problem is -sqrt(( 20^ + 1)/(2)). Substituting the identity from Step 3: -sqrt(( 20^ + 1)/(2)) = - 10^ Since LHS = - 10^ and RHS = - 10^, the identity is shown. 260^ = -sqrt(( 20^ + 1)/(2)) QUESTION 6 6.1.1 Without using a calculator, simplify the above expression to a single trigonometric ratio. The expression is 330^ 365^ + 175^ 210^. Step 1: Reduce each trigonometric term to an acute angle. 330^ = (360^ - 30^) = 30^ 365^ = (360^ + 5^) = 5^ 175^ = (180^ - 5^) = 5^ 210^ = (180^ + 30^) = - 30^ Step 2: Substitute the reduced terms into the expression. 30^ 5^ + 5^ (- 30^) = 30^ 5^ - 30^ 5^ Step 3: Apply the compound angle formula. This expression matches the compound angle formula for (A+B) = A B - A B. Here, A = 30^ and B = 5^. = (30^ + 5^) = 35^ 35^ 6.1.2 Hence, determine 70^ in terms of p if 330^ 365^ + 175^ 210^ = p. Step 1: Use the result from 6.1.1. From 6.1.1, we know that 330^ 365^ + 175^ 210^ = 35^. Given that this expression equals p, we have: 35^ = p Step 2: Relate 70^ to 35^. We need to find 70^. We can use the double angle identity 2 = 2 . Let = 35^. Then 2 = 70^. 70^ = 2 35^ 35^ Step 3: Express 35^ in terms of p. We know 35^ = p. Using the Pythagorean identity ^2 + ^2 = 1: ^2 35^ = 1 - ^2 35^ ^2 35^ = 1 - p^2 Since 35^ is in the first quadrant, 35^ is positive: 35^ = sqrt(1 - p^2) Step 4: Substitute into the double angle formula. 70^ = 2 (sqrt(1 - p^2)) (p) 70^ = 2psqrt(1 - p^2) 70^ = 2psqrt(1 - p^2) 6.2 It is given that 2^3^2 x4^2 x = 1. Determine the general solution for x if x < 0. Step 1: Simplify the exponential equation. Rewrite the base 4 as 2^2: 2^3^2 x(2^2)^2 x = 1 2^3^2 x2^4 x = 1 Step 2: Combine the terms using exponent rules. 2^3^2 x - 4 x = 1 Since 1 = 2^0, we can equate the exponents: 3^2 x - 4 x = 0 Step 3: Solve the quadratic equation for x. Factor out x: x (3 x - 4) = 0 This gives two possible solutions for x: 1. x = 0 2. 3 x - 4 = 0 3 x = 4 x = (4)/(3) Step 4: Evaluate the validity of the solutions for x. The range of the sine function is [-1, 1]. Therefore, x = (4)/(3) is not possible. So, the only valid solution from the equation is x = 0. Step 5: Apply the given condition. The problem states that we need to determine the general solution for x if x < 0. Our only solution for x is x = 0. However, 0 is not less than 0. The condition x < 0 is not satisfied by x = 0. Therefore, there is no value of x that satisfies both the given equation and the condition x < 0. No solution exists 3 done, 2 left today. You're making progress.