In this section, we will be exploring what happens when we double an angle, specifically how this impacts the Sine, Cosine, and Tangent of the newly doubled angle. To aid in this exploration, let's begin with a typical positive angle in the first quadrant and draw the right triangle:
Let's clean this graph up a bit and remove some of the extra lines:
Now we have a proper triangle with the double angle; we even know the length of all of the sides! However, this is not a right triangle, so we are a bit stuck as to how to get the trig functions and apply them to the double angle. Before we take steps to get a right triangle in the picture, let's first figure out what the other angles are. To do this, let's label them \(\beta\):
Now, we can use the fact that the sum of all the angles in a triangle is \(180^\circ\) or \(\pi\): \[\solve{ 2\theta + 2\beta &=& \pi\\ 2\beta&=& \pi - 2\theta\\ \beta&=&\frac{\pi}{{2}}-\theta }\]
Now that we have the other angles, let's introduce some right triangles by drawing the height of the triangle like this:
From this picture, we can now actually write down an expression for the Sine of the double angle: \[ \sin(2\theta)=\frac{{h}}{{1}}=h \] Well, that is a bit anti-climatic, especially because what the heck is \(h\)? Well, if we look at the other triangle, we can use that angle \(\beta\) to come up with another expression: \[ \solve{ \sin\left(\frac{\pi}{{2}}-\theta\right) &=& \frac{{h}}{2\sin(\theta)} } \]Now, what is particularly interesting about this, is that we actually know that: \[ \solve{ \sin\left(\frac{\pi}{{2}}-\theta\right) &=& \cos\theta } \]Which means that we can actually rewrite the equation as \[ \solve{ \cos\left(\theta\right) &=& \frac{{h}}{2\sin(\theta)}\\ 2\sin\theta\cos\theta &=&h } \]And, here is the best part, since we have two different equations solved for \(h\), we can set them equal to each other!! As a result, we get our first double angle formula: \[ \sin(2\theta) = 2\sin\theta\cos\theta \]
All of this we can put back into our picture: