Compound identities, problem solving strategies

Siyavula textbooks: grade 12 Page 1 / 1

Compound angle identities

Derivation of $sin (α + β)$

We have, for any angles $α$ and $β$ , that

sin (α + β) = sin α cos β + sin β cos α

How do we derive this identity? It is tricky, so follow closely.

Suppose we have the unit circle shown below. The two points $L (a, b)$ and $K (x, y)$ are on the circle.

We can get the coordinates of $L$ and $K$ in terms of the angles $α$ and $β$ . For the triangle $L O K$ , we have that

\begin{matrix} sin β = \frac{b}{1} & \Rightarrow & b = sin β \\ cos β = \frac{a}{1} & \Rightarrow & a = cos β \end{matrix}

Thus the coordinates of $L$ are $(cos β; sin β)$ . In the same way as above, we can see that the coordinates of $K$ are $(cos α; sin α)$ . The identity for $cos (α - β)$ is now determined by calculating $K L^{2}$ in two ways. Using the distance formula (i.e. $d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ or $d^{2} = {(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}$ ), we can find $K L^{2}$ :

\begin{matrix} K L^{2} & = & {(cos α - cos β)}^{2} + {(sin α - sin β)}^{2} \\ = & {cos}^{2} α - 2 cos α cos β + {cos}^{2} β + {sin}^{2} α - 2 sin α sin β + {sin}^{2} β \\ = & ({cos}^{2} α + {sin}^{2} α) + ({cos}^{2} β + {sin}^{2} β) - 2 cos α cos β - 2 sin α sin β \\ = & 1 + 1 - 2 (cos α cos β + sin α sin β) \\ = & 2 - 2 (cos α cos β + sin α sin β) \end{matrix}

The second way we can determine $K L^{2}$ is by using the cosine rule for $▵ K O L$ :

\begin{matrix} K L^{2} & = & K O^{2} + L O^{2} - 2 \cdot K O \cdot L O \cdot cos (α - β) \\ = & 1^{2} + 1^{2} - 2 (1) (1) cos (α - β) \\ = & 2 - 2 \cdot cos (α - β) \end{matrix}

Equating our two values for $K L^{2}$ , we have

\begin{matrix} 2 - 2 \cdot cos (α - β) & = & 2 - 2 (cos α cos β + sin α \cdot sin β) \\ \Rightarrow cos (α - β) & = & cos α \cdot cos β + sin α \cdot sin β \end{matrix}

Now let $α \to 90^{\circ} - α$ . Then

\begin{matrix} cos (90^{\circ} - α - β) & = & cos (90^{\circ} - α) cos β + sin (90^{\circ} - α) sin β \\ = & sin α \cdot cos β + cos α \cdot sin β \end{matrix}

But $cos (90^{\circ} - (α + β)) = sin (α + β)$ . Thus

sin (α + β) = sin α \cdot cos β + cos α \cdot sin β

Derivation of $sin (α - β)$

We can use

sin (α + β) = sin α cos β + cos α sin β

to show that

sin (α - β) = sin α cos β - cos α sin β

We know that

sin (- θ) = - sin (θ)

and

cos (- θ) = cos θ

Therefore,

\begin{matrix} sin (α - β) & = & sin (α + (- β)) \\ = & sin α cos (- β) + cos α sin (- β) \\ = & sin α cos β - cos α sin β \end{matrix}

Derivation of $cos (α + β)$

We can use

sin (α - β) = sin α cos β - sin β cos α

to show that

cos (α + β) = cos α cos β - sin α sin β

We know that

sin (θ) = cos (90 - θ) .

Therefore,

\begin{matrix} cos (α + β) & = & sin (90 - (α + β)) \\ = & sin ((90 - α) - β)) \\ = & sin (90 - α) cos β - sin β cos (90 - α) \\ = & cos α cos β - sin β sin α \end{matrix}

Derivation of $cos (α - β)$

We found this identity in our derivation of the $sin (α + β)$ identity. We can also use the fact that

sin (α + β) = sin α cos β + cos α sin β

to derive that

cos (α - β) = cos α cos β + sin α sin β

cos (θ) = sin (90 - θ),

we have that

\begin{matrix} cos (α - β) & = & sin (90 - (α - β)) \\ = & sin ((90 - α) + β)) \\ = & sin (90 - α) cos β + cos (90 - α) sin β \\ = & cos α cos β + sin α sin β \end{matrix}

Derivation of $sin 2 α$

We know that

sin (α + β) = sin α cos β + cos α sin β

When $α = β$ , we have that

\begin{matrix} sin (2 α) = sin (α + α) & = & sin α cos α + cos α sin α \\ = & 2 sin α cos α \\ = & sin (2 α) \end{matrix}

Derivation of $cos 2 α$

We know that

cos (α + β) = cos α cos β - sin α sin β

When $α = β$ , we have that

\begin{matrix} cos (2 α) = cos (α + α) & = & cos α cos α - sin α sin α \\ = & {cos}^{2} α - {sin}^{2} α \\ = & cos (2 α) \end{matrix}

However, we can also write

cos 2 α = 2 {cos}^{2} α - 1

and

cos 2 α = 1 - 2 {sin}^{2} α

by using

{sin}^{2} α + {cos}^{2} α = 1 .

The $cos 2 α$ Identity

Use

{sin}^{2} α + {cos}^{2} α = 1

to show that:

\begin{matrix} cos 2 α & = & \{\begin{matrix} 2 {cos}^{2} α - 1 \\ 1 - 2 {sin}^{2} α \end{matrix}) \end{matrix}

Problem-solving strategy for identities

The most important thing to remember when asked to prove identities is:

Trigonometric Identities

When proving trigonometric identities, never assume that the left hand side is equal to the right hand side. You need to show that both sides are equal.

A suggestion for proving identities: It is usually much easier simplifying the more complex side of an identity to get the simpler side than the other way round.

Prove that $sin 75^{\circ} = \frac{\sqrt{2} (\sqrt{3} + 1)}{4}$ without using a calculator.

Identify a strategy
We only know the exact values of the trig functions for a few special angles ( $30^{\circ}$ , $45^{\circ}$ , $60^{\circ}$ , etc.). We can see that $75^{\circ} = 30^{\circ} + 45^{\circ}$ . Thus we can use our double-angle identity for $sin (α + β)$ to express $sin 75^{\circ}$ in terms of known trig function values.
Execute strategy
$\begin{matrix} sin 75^{\circ} & = & sin (45^{\circ} + 30^{\circ}) \\ = & sin (45^{\circ}) cos (30^{\circ}) + cos (45^{\circ}) sin (30^{\circ}) \\ = & \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \\ = & \frac{\sqrt{3} + 1}{2 \sqrt{2}} \\ = & \frac{\sqrt{3} + 1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\ = & \frac{\sqrt{2} (\sqrt{3} + 1)}{4} \end{matrix}$

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

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Source: OpenStax, Siyavula textbooks: grade 12 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11242/1.2

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Compound identities, problem solving strategies

Compound angle identities

Derivation of sin ( α + β )

Derivation of sin ( α - β )

Derivation of cos ( α + β )

Derivation of cos ( α - β )

Derivation of sin 2 α

Derivation of cos 2 α

The cos 2 α Identity

Problem-solving strategy for identities

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Derivation of $sin (α + β)$

Derivation of $sin (α - β)$

Derivation of $cos (α + β)$

Derivation of $cos (α - β)$

Derivation of $sin 2 α$

Derivation of $cos 2 α$

The $cos 2 α$ Identity