Identitas penjumlahan sudut

Continuing our discussion on these trigonometric function identities, let's now find a formula for $\sin{(A+B)}$.

The sine area formula for a triangle is:

$A = \frac{1}{2} \cdot ab\sin{C}$

Let $\alpha = \angle{BAD}$ and $\beta = \angle{DAC}$.

Provided this formula, we can use the sum of areas to find the best way to find the formula for $\sin{(\alpha+\beta)}$ in terms of trig functions of $\alpha$ and $\beta$.

It is obvious to see that $[ABC] = [BAD] + [DAC]$. Following from the sine area formula, we get that $$\frac{1}{2} \cdot AB \cdot AC \cdot \sin{(\alpha + \beta)}= \frac{1}{2} \cdot AB \cdot AD \cdot \sin{\alpha} + \frac{1}{2} \cdot AC \cdot AD \cdot \sin{\beta}$$ $$AB \cdot AC \cdot \sin{(\alpha + \beta)}= AB \cdot AD \cdot \sin{\alpha} + AC \cdot AD \cdot \sin{\beta}$$ Now divide by $AB \cdot AC$: $$\sin{(\alpha + \beta)} = \frac{AD}{AC}\sin{\alpha} + \frac{AD}{AB}\sin{\beta}$$

But since $\frac{AD}{AC} = \cos{\beta}$ and $\frac{AD}{AB} = \cos{\alpha}$, we can substitute this in to get our closed form:

$$\boxed{\sin{(\alpha + \beta)} = \sin{\alpha}\cos{\beta} + \sin{\beta}\cos{\alpha}}$$

This is a very great formula to remember. To derive a formula for $\sin{(\alpha - \beta)}$, just use the formula for: $$\sin{(\alpha - \beta)} = \sin{[\alpha + (-\beta)]} = \sin{\alpha}\cos{(-\beta)} + \sin{(-\beta)}\cos{\alpha} \\ = \sin{\alpha}\cos{\beta} - \sin{\beta}\cos{\alpha}$$

Now, let's think about cosine. Well, since $\cos{A} = \sin{(90^{\circ} - A)}$, we have $$\cos{(\alpha + \beta)} = \sin{(90^{\circ} - (\alpha + \beta))}$$

I will leave it to you to come up with the derivations for addition and subtraction for cosine.

$$\cos{(\alpha + \beta)} = \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta} \\ \cos{(\alpha - \beta)} = \cos{\alpha}\cos{\beta} + \sin{\alpha}\sin{\beta}$$

Now let's move on to tangent. $$\tan{(\alpha + \beta)} = \dfrac{\sin{(\alpha + \beta)}}{\cos{(\alpha +\beta)}} = \dfrac{\sin{\alpha}\cos{\beta} + \sin{\beta}\cos{\alpha}}{\cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta}}$$

Now divide numerator and denominator by $\cos{\alpha}\cos{\beta}$: $$\dfrac{\sin{\alpha}\cos{\beta} + \sin{\beta}\cos{\alpha}}{\cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta}} = \dfrac{\frac{\sin{\alpha}\cos{\beta}}{\cos{\alpha}\cos{\beta}} + \frac{\sin{\beta}\cos{\alpha}}{\cos{\alpha}\cos{\beta}}}{\frac{\cos{\alpha}\cos{\beta}}{\cos{\alpha}\cos{\beta}} - \frac{\sin{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta}}} \\ =\dfrac{\tan{\alpha} + \tan{\beta}}{1 - \tan{\alpha}\tan{\beta}}$$

You can do the same for $\alpha - \beta$.

So to recap:

$$\sin{(\alpha + \beta)} = \sin{\alpha}\cos{\beta} + \sin{\beta}\cos{\alpha} \\ \sin{(\alpha - \beta)} = \sin{\alpha}\cos{\beta} - \sin{\beta}\cos{\alpha} \\ \cos{(\alpha + \beta)} = \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta} \\ \cos{(\alpha - \beta)} = \cos{\alpha}\cos{\beta} + \sin{\alpha}\sin{\beta} \\ \tan{(\alpha + \beta)} = \dfrac{\tan{\alpha} + \tan{\beta}}{1 - \tan{\alpha}\tan{\beta}} \\ \tan{(\alpha - \beta)} = \dfrac{\tan{\alpha} - \tan{\beta}}{1 + \tan{\alpha}\tan{\beta}}$$

If an angle $A$ is a "special angle" like $30^{\circ}$ or $45^{\circ}$, or $A$ is an angle in a right triangle whose side lengths we know, then we can evaluate trig functions of the angle.

Let's think about how we can evaluate trig functions of combinations angles we know. For example, if we know the $\cos A, \sin A, \cos B, \sin B$, we can find $\cos (A+B)$ and $\sin (A+B)$. In this lesson we will learn the addition and subtraction identities (via a geometric approach), and later we will use these identities to derive other useful identities such as the double and half angle identities.

Hence, given the hypotenuse $r,s$ of $\triangle ACX$ and $\triangle ADY$ we can find their side lengths:

To find the $\sin$ and $\cos$ of $A+B$, it makes sense to decompose the angle into angles whose $\sin$ and $\cos$ we can compute.

Now we choose $R$ such that $OR=1$. We draw in 2 right triangles $\triangle ORQ$ and $\triangle OQP$ to "complete the picture"; the lengths of these triangles can be computed in terms of $A$ and $B$.

We let $S$ be the foot of the perpendicular from $R$ to $OP$. Then we have $$\begin{align} OS&=\cos(A+B)\\ RS&=\sin(A+B). \end{align}$$

We want to calculate these lengths. We first fill in as much of the picture as we can.

Here's what we have so far.

We know $PQ$ and want to find $RS$. We know $OP$ and want to find $OS$. To relate them, we draw in point $T$ below. The colored lengths are equal because $PQTS$ is a rectangle, and opposite sides of a rectangle are equal.

It now suffices to find the side lengths of $\triangle RQT$. For this, we need the angles in the triangle.

We find this using some angle chasing: we have $\angle ORQ=90^{\circ}-B$, $\angle ORS=90^{\circ}-(A+B)$, so $\angle TRQ=\angle ORQ-\angle ORS=A$. Alternatively, if you have learned about circles, you can note $OSQR$ is a cyclic quadrilateral because the right angles imply that $S,Q$ are on the circle with diameter $OR$, and it follows that $\angle SRQ=\angle SOQ=A$.

Now we can finish the computation.

This is the final picture.

Note that this computation still works even if the angles are not in $[0^{\circ},90^{\circ}]$. However, then we will have to use directed lengths, i.e., keep track of which are positive and negative. (If you have not learned about trig functions of angles outside $[0^{\circ},90^{\circ}]$, ignore this statement.)

To find these, one way is to substitute $-B\leftarrow B$ in the angle addition formula, and note $\sin (-B)=-\sin B$ ($\sin$ is odd) and $\cos(-B)=\cos B$ ($\cos$ is even). Another is to draw a diagram emulating the above proof, but with $\triangle OQR$ reversed. (Add the picture!) The angle subtraction formulas are the same as the angle addition formulas, just with their signs switched.

To summarize, we have found $$\begin{align} \sin(A+B)&=\sin A\cos B+\cos A\sin B\\ \sin(A-B)&=\sin A\cos B-\cos A\sin B\\ \cos(A+B)&=\cos A\cos B-\sin A\sin B\\ \cos(A-B)&=\cos A\cos B+\sin A\sin B. \end{align}$$

A set of helpful identities when working with sines and cosines:

$\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)$

$\sin(\alpha - \beta) = \sin(\alpha)\cos(\beta) - \sin(\beta)\cos(\alpha)$

$\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$

$\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$

The first two are the formulas for sine, and the second two are the formulas for cosine.

One can prove another important identity: $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$.

$$\begin {align*} \sin(2\theta) &= \sin(\theta + \theta) \\&= \sin(\theta)\cos(\theta) + \sin(\theta)\cos(\theta) \\&= 2\sin(\theta)\cos(\theta) \end {align*}$$

This uses the angle addition formula for sine.

In addition, one can also prove the identity $\cos(2\theta)=2\cos^2\theta-1=1-2\sin^2\theta$.

$$\begin{align*}\cos(2\theta)&= \cos(\theta+\theta)\\ &= \cos\theta\cos\theta-\sin\theta\sin\theta \\ &=\cos^2\theta-\sin^2\theta\\ &= 2\cos^2\theta-1\tag{By $\sin^2\theta+\cos^2\theta=1$} \\ &= 1-2\sin^2\theta\end{align*}$$

Here are some useful identities pertaining to sums of angles:

$\sin(x + y) = \sin(x)\cos(y) + \cos(x)\sin(y)$

$\cos(x + y) = \cos(x)\cos(y) - \sin(x)\sin(y)$

Using these two, we can prove other identities.

Proof 1:

$\tan(x + y) = \frac{\sin(x + y)}{\cos(x + y)} = \frac{\sin(x)\cos(y) + \cos(x)\sin(y)}{\cos(x)\cos(y) - \sin(x)\sin(y)}$

Divide the numerator and denominator of the fraction by $\cos(x)\cos(y)$:

$\tan(x + y) = \frac{\tan(x) + \tan(y)}{1 - \tan(x)\tan(y)}$

Proof 2:

$\sin(x - y) = \sin(x)\cos(-y) + \cos(x)\sin(-y) = \sin(x)\cos(y) - \cos(x)\sin(y)$

Here it is important to remember that sine is an odd function and cosine is an even function.

Proof 3:

$\cos(x - y) = \cos(x)\cos(-y) - \sin(x)\sin(-y) = \cos(x)\cos(y) + \sin(x)\sin(y)$

Proof 4:

$\tan(x - y) = \frac{\tan(x) + \tan(-y)}{1 - \tan(x)\tan(-y)}$

Remembering that tangent is an odd function is important here--$\tan(-y) = -\tan(y)$:

$\tan(x - y) = \frac{\tan(x) - \tan(y)}{1 + \tan(x)\tan(y)}$

These six identities are very useful for the wide variety of problems involving sums and differences of angles.

Angle Sum & Difference Identities

There are six negative angle identities you need to memorize!

• Negative angle identities
• Co-function identities
• Angle and sum differences identities (this lesson)
• Double angle identities
• Half angle identities

The identities are as follows:

$$\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)$$ $$\sin(\alpha - \beta) = \sin(\alpha)\cos(\beta) - \sin(\beta)\cos(\alpha)$$ $$\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$$ $$\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$$

(Video) Sum and Difference Identities

The Organic Chemistry Tutor

In this video by The Organic Chemistry Tutor, you will learn about sum and difference identities in trigonometry.