11.2 Compound angle identities and applications  (Page 2/3)

We can express $tan(\alpha +\beta )$ in terms of cosines and sines, and then use the double-angle formulas for these. We then manipulate the resulting expression in order to get it in terms of $tan\alpha$ and $tan\beta$ .

The right-hand side (RHS) of the identity cannot be simplified. Thus we should try simplify the left-hand side (LHS). We can also notice that the trig function on the RHS does not have a $2\theta$ dependance. Thus we will need to use the double-angle formulas to simplify the $sin2\theta$ and $cos2\theta$ on the LHS. We know that $tan\theta$ is undefined for some angles $\theta$ . Thus the identity is also undefined for these $\theta$ , and hence is not valid for these angles. Also, for some $\theta$ , we might have division by zero in the LHS, which is not allowed. Thus the identity won't hold for these angles also.

We know that $tan\theta$ is undefined when $\theta ={90}^{\circ }+{180}^{\circ }n$ , where $n$ is an integer.The LHS is undefined when $1+cos\theta +cos2\theta =0$ . Thus we need to solve this equation.

The above has solutions when $cos\theta =0$ , which occurs when $\theta ={90}^{\circ }+{180}^{\circ }n$ , where $n$ is an integer. These are the same values when $tan\theta$ is undefined. It also has solutions when $1+2cos\theta =0$ . This is true when $cos\theta =-\frac{1}{2}$ , and thus $\theta =...-{240}^{\circ },-{120}^{\circ },{120}^{\circ },{240}^{\circ },...$ . To summarise, the identity is not valid when $\theta =...-{270}^{\circ },-{240}^{\circ },-{120}^{\circ },-{90}^{\circ },{90}^{\circ },{120}^{\circ },{240}^{\circ },{270}^{\circ },...$

Before we are able to solve the equation, we first need to simplify the left-hand side. We do this by using the double-angle formulas.

We want $CB$ , and we have $CD$ and $BD$ . If we could get the angle $B\widehat{D}C$ , then we could use the cosine rule to determine $BC$ . This is possible, as $▵ABD$ is a right-angled triangle. We know this from circle geometry, that any triangle circumscribed by a circle with one side going through the origin, is right-angled. As we have two angles of $▵ABD$ , we know $A\widehat{D}B$ and hence $B\widehat{D}C$ . Using the cosine rule, we can get $B{C}^2$ .

Thus

Now the cosine rule gives