Even-odd trig identities

Students should already know the even-odd identities for $\sin$ and $\cos.$ Start this lesson by asking students to determine which of the six basic trig functions are even, which are odd, and which are neither, given a graph of each function. After that, ask students how we could prove each case algebraically, using only our knowledge of the reciprocal identities, and the fact that $\sin$ is odd, and $\cos$ is even? Next, students will use the even-odd identities to evaluate trig functions and simplify trig expressions.

Concude by giving your students this challenge:

Find the area of this triangle:

Our goal is to determine the base and height of the triangle. Draw a segment connecting B to AC, which is perpendicular to AC. Call the point of intersection E, and label BE with h, the height of the triangle. The definition of $\sin$ tells us that $\sin 2x = h / 18$ and $\sin x = h / 30.$ Putting these together, we find

$$h = 18\sin 2x = 30\sin x$$

Using the double angle formula for sine, we get

$$18 \cdot 2\sin x\cos x = 30\sin x$$

Solving for $\cos x,$ we get $\cos x = 5/6.$ We now know $\cos x,$ but our $h$ is in terms of $\sin x.$ From the Pythagorean identity $\sin^2 x + \cos^2 x = 1,$ it's easy to write $\sin$ in terms of $\cos,$ like so:

$$\sin x = \pm\sqrt{1 - \cos^2 x}$$

But since $h = 30\sin x$ and $h$ is a length, we know $\sin x$ can't be negative. After some arithmetic, we find $h = 5\sqrt{11}.$ Now we know the height of the triangle, but we still have to find the length of its base. Remember, we know that $\cos x = 5/6.$ But using the definition of $\cos,$ we also know that $\cos x = EC/30,$ so $5/6 = EC/30,$ and thus $EC = 25.$ If we could just figure out AE, we would have the length of our base. Notice that $x$ is less than $2x.$ Thus, we can construct segment BD, such that $\triangle BDA$ is isosceles. Since $\triangle BDA$ is isosceles, $BD = BA = 18.$ Next, notice that $\angle BDA,$ which has measure $2x,$ is an exterior angle of $\triangle BDC.$ Thus $\angle CBD = x,$ and so $\triangle BDC$ is also isosceles. Because $\triangle BDC$ is isosceles, we know $DC = DB = 18.$ We know $EC = ED + DC,$ but we also know $EC = 25,$ and $DC = 18.$ Thus, $ED$ must be $7.$ From this, it's obvious that $AE$ is also $7,$ and so our triangle has a base of length $7 + 7 + 18 = 32.$ Finally, we plug our base and height into our area formula to get $(1/2)32 \cdot 5\sqrt{11} = 80\sqrt{11}.$

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