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Powerful Quadratics

This comes in two parts, with the first being less fiendish than the second. It’s great for practising both quadratics and laws of indices, and you can get a lot from making sure that you find all the solutions. For a real challenge (requiring a bit more knowledge), you could consider finding the complex solutions.

Discriminating

You're invited to decide whether statements about the number of solutions of a quadratic equation are always, sometimes or never true.

Factorisable Quadratics

This will encourage you to think about whether all quadratics can be factorised and to develop a better understanding of the effect that changing the coefficients has on the factorised form.

Inspecting Identities

Age 16 to 18
Challenge Level

This resource is from Underground Mathematics.
 

 

Gururvignesh from Hymers College in the UK sent in this solution:

1. We expand $\sin^2\frac x2$ and $\cos^2\frac x2$ so that it equals $\frac12(1-\cos x)$ and $\frac12(1+\cos x)$ respectively

2. We find out the amplitude, period, vertical shift and face shift by substituting the equation $y=a\cos(bx+c)+k$ into the expanded identity.

3. After finding the important points we can sketch the cos or sine graph and label the points.

 

$y=a\cos(bx+c)+k$

$a$ = amplitude (the height of the graph)

$b$ = period (how long it takes for the graph to repeat)

$c$ = period shift/phase shift

$k$ = vertical shift

$\cos^2\frac x2 = \frac12(1+\cos x)$

$\sin^2\frac x2= \frac12(1-\cos x)$

$\begin{split}y&=\cos^2\tfrac x2\\ &= \frac12+\frac{\cos x}2\end{split}$

$a=\frac12$

period = $\frac{2\pi}b = \frac{2\pi}1=2\pi$

vertical shift = $\frac12$

period shift = $0$

 

$y = \sin^2\frac x2$

$y =\frac12(1-\cos x)$

$a = -\frac12$ (which means the graph of cosine has to be flipped.)

period = $\frac{2\pi}b = \frac{2\pi}1=2\pi$

vertical shift = $\frac12$

period shift = $0$

Click here to see a larger image.

The orange graph represents $y= \cos^2 \frac x2$

the red graph represents $y= \sin^2 \frac x2$