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)
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
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)