Trigonometry/Graph of Sine Squared

Sin Squared

The graph of $\displaystyle (\sin \theta)^2$, or $\displaystyle\sin^2\theta$ as it is more usually written, is shown below:

This function

• Must be non-negative, since the square of a negative number is always positive.
• Cannot exceed 1 since $\displaystyle \sin \theta$ always lies between -1 and 1.

It looks like a sine or cosine wave shifted and compressed. It is. We will show this is true later when we look at double angle formulae and prove that $\displaystyle\sin^2\theta = \frac{1}{2}(1-\cos(2\theta))$.

 Exercise: Spot check this Is the graph correct? Try some actual avalues to see if they have been plotted in the correct place. Any value where $\displaystyle\sin\theta=0$ will be zero in this graph too. Any value where $\displaystyle\sin\theta=\pm 1$ will be 1 in this graph. What about values that have $\displaystyle y=0.5$ on this graph? What values of $\displaystyle\sin\theta$ will work?
 Amplitude, Frequency and Phase Going from $\displaystyle \sin \theta$ to $\displaystyle \sin^2 \theta$ The amplitude is halved. (the y values lie between 0 and +1, previously they were between -1 and +1). The frequency is doubled. (we have more complete cycles in the same x distance. The phase is... well we could argue that the phase is changed, but it really only makes sense to compare the phase of two waves of the same frequency, so it is hard to say what has happened to the phase.

Cos Squared

The graph below does the same thing for $\displaystyle\cos^2\theta$

Once again, this function:

• Must be non-negative, since the square of a negative number is always positive.
• Cannot exceed 1 since $\displaystyle \cos \theta$ always lies between -1 and 1.

Comparing the two graphs it looks like they would sum to one. They do. This is a graphical way to see what we have already seen earlier, that:

$\displaystyle \cos^2\theta + \sin^2\theta = 1$

 Formula for cos squared Using the fact that we have established that: $\displaystyle \cos^2\theta + \sin^2\theta = 1$ and assuming the result we will prove later that: $\displaystyle\sin^2\theta = \frac{1}{2}(1-\cos(2\theta))$ work out an expression for: $\displaystyle\cos^2\theta =$ Be careful with the signs and brackets as you are taking the minus of a minus. Be sure to simplify the formula - your answer should be at least as simple as the formula for $\displaystyle \sin^2 \theta$. Does the formula you come up with look like it is consistent with the graph we have drawn for $\displaystyle\cos^2\theta$?