Inspecting Identities
Problem
Take a look at these identities.
$$ \cos^2 \frac{\theta}{2} \equiv \frac{1}{2}(1+\cos \theta) \quad \quad \quad \sin^2 \frac{\theta}{2} \equiv \frac{1}{2}(1-\cos \theta)$$
How could you use these identities to help you sketch graphs of $y=\cos^2 \frac{\theta}{2}$ and $y=\sin^2 \frac{\theta}{2}$?
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Student Solutions
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 $y= \sin^2 \frac x2$