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

Page 2 / 3

Deduce a formula for $tan (α + β)$ in terms of $tan α$ and $tan β$ .

Hint: Use the formulae for $sin (α + β)$ and $cos (α + β)$

Identify a strategy
We can express $tan (α + β)$ 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 $tan α$ and $tan β$ .
Execute strategy
$\begin{matrix} tan (α + β) & = & \frac{sin (α + β)}{cos (α + β)} \\ = & \frac{sin α \cdot cos β + cos α \cdot sin β}{cos α \cdot cos β - sin α \cdot sin β} \\ = & \frac{\frac{sin α \cdot cos β}{cos α \cdot cos β} + \frac{cos α \cdot sin β}{cos α \cdot cos β}}{\frac{cos α \cdot cos β}{cos α \cdot cos β} - \frac{sin α \cdot sin β}{cos α \cdot cos β}} \\ = & \frac{tan α + tan β}{1 - tan α \cdot tan β} \end{matrix}$

Prove that

\frac{sin θ + sin 2 θ}{1 + cos θ + cos 2 θ} = tan θ

In fact, this identity is not valid for all values of $θ$ . Which values are those?

Identify a strategy
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 $2 θ$ dependance. Thus we will need to use the double-angle formulas to simplify the $sin 2 θ$ and $cos 2 θ$ on the LHS. We know that $tan θ$ is undefined for some angles $θ$ . Thus the identity is also undefined for these $θ$ , and hence is not valid for these angles. Also, for some $θ$ , we might have division by zero in the LHS, which is not allowed. Thus the identity won't hold for these angles also.
Execute the strategy
$\begin{matrix} L H S & = & \frac{sin θ + 2 sin θ cos θ}{1 + cos θ + (2 {cos}^{2} θ - 1)} \\ = & \frac{sin θ (1 + 2 cos θ)}{cos θ (1 + 2 cos θ)} \\ = & \frac{sin θ}{cos θ} \\ = & tan θ \\ = & R H S \end{matrix}$

We know that $tan θ$ is undefined when $θ = 90^{\circ} + 180^{\circ} n$ , where $n$ is an integer.The LHS is undefined when $1 + cos θ + cos 2 θ = 0$ . Thus we need to solve this equation.

$\begin{matrix} 1 + cos θ + cos 2 θ & = & 0 \\ \Rightarrow cos θ (1 + 2 cos θ) & = & 0 \end{matrix}$

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

Solve the following equation for $y$ without using a calculator:

\frac{1 - sin y - cos 2 y}{sin 2 y - cos y} = - 1

Identify a strategy
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.
Execute the strategy
$\begin{matrix} \frac{1 - sin y - (1 - 2 {sin}^{2} y)}{2 sin y cos y - cos y} & = & - 1 \\ \Rightarrow \frac{2 {sin}^{2} y - sin y}{cos y (2 sin y - 1)} & = & - 1 \\ \Rightarrow \frac{sin y (2 sin y - 1)}{cos y (2 sin y - 1)} & = & - 1 \\ \Rightarrow tan y & = & - 1 \\ \Rightarrow y = 135^{\circ} + 180^{\circ} n; n \in Z \end{matrix}$

Applications of trigonometric functions

Problems in two dimensions

For the figure below, we are given that $B C = B D = x$ .

Show that $B C^{2} = 2 x^{2} (1 + sin θ)$ .

Identify a strategy
We want $C B$ , and we have $C D$ and $B D$ . If we could get the angle $B \hat{D} C$ , then we could use the cosine rule to determine $B C$ . This is possible, as $▵ A B D$ 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 $▵ A B D$ , we know $A \hat{D} B$ and hence $B \hat{D} C$ . Using the cosine rule, we can get $B C^{2}$ .
Execute the strategy
$A \hat{D} B = 180^{\circ} - θ - 90^{\circ} = 90^{\circ} - θ$

Thus

$\begin{matrix} B \hat{D} C & = & 180^{\circ} - A \hat{D} B \\ = & 180^{\circ} - (90^{\circ} - θ) \\ = & 90^{\circ} + θ \end{matrix}$

Now the cosine rule gives

$\begin{matrix} B C^{2} & = & C D^{2} + B D^{2} - 2 \cdot C D \cdot B D \cdot cos (B \hat{D} C) \\ = & x^{2} + x^{2} - 2 \cdot x^{2} \cdot cos (90^{\circ} + θ) \\ = & 2 x^{2} + 2 x^{2} [sin (90^{\circ}) cos (θ) + sin (θ) cos (90^{\circ})] \\ = & 2 x^{2} + 2 x^{2} [1 \cdot cos (θ) + sin (θ) \cdot 0] \\ = & 2 x^{2} (1 - sin θ) \end{matrix}$

For the diagram on the right,
1. Find $A \hat{O} C$ in terms of $θ$ .
2. Find an expression for:
  1. $cos θ$
  2. $sin θ$
  3. $sin 2 θ$
3. Using the above, show that $sin 2 θ = 2 sin θ cos θ$ .
4. Now do the same for $cos 2 θ$ and $tan θ$ .
$D C$ is a diameter of circle $O$ with radius $r$ . $C A = r$ , $A B = D E$ and $D \hat{O} E = θ$ . Show that $cos θ = \frac{1}{4}$ .
The figure below shows a cyclic quadrilateral with B C C D = A D A B .
1. Show that the area of the cyclic quadrilateral is $D C \cdot D A \cdot sin \hat{D}$ .
2. Find expressions for $cos \hat{D}$ and $cos \hat{B}$ in terms of the quadrilateral sides.
3. Show that $2 C A^{2} = C D^{2} + D A^{2} + A B^{2} + B C^{2}$ .
4. Suppose that $B C = 10$ , $C D = 15$ , $A D = 4$ and $A B = 6$ . Find $C A^{2}$ .
5. Find the angle $\hat{D}$ using your expression for $cos \hat{D}$ . Hence find the area of $A B C D$ .

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Source: OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1

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