# Tutor profile: Samantha M.

## Questions

### Subject: Trigonometry

One of the trigonometric identities states: cos^2 (x) + sin^2 (x) = 1. Using this identity, solve this equation: 2 cos^2 (x) + 3 sin (x) = 3, for all x: [0 , 2pi].

Firstly, we want the expression in terms of cos or sin, but not both. Using our identity, it is easier to eliminate cos^2 (x), giving: 2(1 - sin^2 (x) ) + 3 sin (x) = 3 This can be further simplified to: 2 sin^2 (x) - 3 sin (x) + 1 = 0 Does this form look familiar? It should, as its now a quadratic equation. Let's change that sin (x) to an X to make it easier. 2 X^2 - 3 X + 1 = 0 Which, when factorised as a quadratic, gives: (2X - 1) (X - 1) = 0. With the two solutions: X = 1/2 and X = 1. Going back, to where we said that sin (x) = X, we can now work out the values for x. 1). If X = 1/2, then sin (x) = 1/2. Since sin is positive in the 1st and 2nd quadrants, and that sin (pi/6) = 1/2, we can see that x = (pi/6) (Q1) and x = (5pi/6) (Q2). 2). If X = 1, then sin (x) = 1. Since sin (pi/2) = 1, this gives us a single solution: x = (pi/2) Therefore, we have 3 solutions for x, between 0 and 2pi: x = (pi/6) , x = (5pi/6) , x = (pi/2)

### Subject: Calculus

The Product Rule states that if two functions are differentiable, then the product of those two functions is also differentiable. This gives us the general formula: y' = ( uv )' = u' v + u v'. Using this rule, where y = uv is the product of the two functions u and v, find y'(3) if u(3) = 3, u'(3) = -6, v(3) = 2, and v'(3) = 4.

From the Product Rule, we have y' = ( uv )' = u' v + u v'. Hence, y'(3) = u'(3)v(3) = u(3)v'(3) = (-6)(2) + (3)(4) = -12 + 12 = 0

### Subject: Algebra

In many algebraic problems, such as factorising an expression, we use our knowledge of each of the numerical coefficients and their factors, to solve it by hand. For example, factorise: ac + 3bc - a - 3b

Immediately, we can see that there is no common factor to all four of our terms. So we take them in pairs: ac + 3bc - a - 3b = ac - a + 3bc - 3b = a(c-1) + 3b(c-1) = (a + 3b)(c - 1)

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