Mathematics Grade 12 Hech zars QUESTION 2 2.1 Simplify the following: (6) $$ \frac{\cos \left(x-360^{\circ}\right) \cdot \sin \left(360^{\circ}-x\right) \cdot \cos \left(90^{\circ}+x\right)}{\sin ^{2}\left(90^{\circ}-x\right) \cdot \sin (-x)} $$ 2.2 Given: $$ \cos (x-y)=\cos x \cdot \cos y+\sin x \cdot \sin y $$ Derive $$ \sin (x+y) $$ (3) 2.3 Consider: $$ \sin \left(2 x+40^{\circ}\right) \cos \left(x+30^{\circ}\right)-\cos \left(2 x+40^{\circ}\right) \sin \left(x+30^{\circ}\right) $$ 2.3.1 Write as a single trigonometric form in its simplest form. (2) 2.3.2 Hence determine the general solution of the equation: $$ \sin \left(2 x+40^{\circ}\right) \cos \left(x+30^{\circ}\right)-\cos \left(2 x+40^{\circ}\right) \sin \left(x+30^{\circ}\right)=-\frac{1}{2} $$ [15]

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**2.1 Simplify**
  
We need to simplify

$$
\frac{\cos \left(x-360^{\circ}\right) \cdot \sin \left(360^{\circ}-x\right) \cdot \cos \left(90^{\circ}+x\right)}{\sin ^{2}\left(90^{\circ}-x\right) \cdot \sin (-x)}
$$

1. Use the periodicity of cosine:
   $$
   \cos \left(x-360^{\circ}\right)=\cos x
   $$
2. Use the sine subtraction formula by writing
   $$
   \sin(360^{\circ}-x)=\sin360^{\circ}\cos x-\cos360^{\circ}\sin x=0\cdot\cos x-1\cdot\sin x=-\sin x.
   $$
3. Use the co-function identity:
   $$
   \cos(90^{\circ}+x)=-\sin x.
   $$
4. In the denominator note that
   $$
   \sin(90^{\circ}-x)=\cos x\quad \text{so}\quad \sin^2(90^{\circ}-x)=\cos^2 x.
   $$
5. Also, using the odd property of sine:
   $$
   \sin(-x)=-\sin x.
   $$
   
Substitute all these into the original expression:

$$
\frac{\cos x\cdot \left(-\sin x\right)\cdot \left(-\sin x\right)}{\cos^2 x \cdot \left(-\sin x\right)}
$$

Simplify the numerator:

$$
(-\sin x)(-\sin x)=\sin^2 x,\quad \text{so the numerator becomes}\quad \cos x \cdot \sin^2 x.
$$

Thus the expression is

$$
\frac{\cos x \cdot \sin^2 x}{-\cos^2 x \cdot \sin x}.
$$

Now, cancel a common factor of $$\sin x$$ (assuming $$\sin x\neq 0$$) and $$\cos x$$:

$$
=\frac{\sin x}{-\cos x}=-\tan x.
$$

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**2.2 Derive $$\sin (x+y)$$ from $$\cos (x-y)=\cos x \cdot \cos y+\sin x \cdot \sin y$$**

To derive $$\sin (x+y)$$, we use the co-function identity together with the cosine addition formula. Recognize that

$$
\sin (x+y)=\cos\Bigl(90^{\circ}-(x+y)\Bigr),
$$

but a more direct method is as follows:

Recall the identity for the sine of a sum:

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

One way to see this using the given cosine formula is to replace one of the angles with its complement. Note that

$$
\cos \left[ x - \left(90^{\circ}-y \right)\right] =\cos x\cos (90^{\circ}-y)+\sin x\sin (90^{\circ}-y).
$$

Since

$$
\cos (90^{\circ}-y)=\sin y\quad \text{and}\quad \sin (90^{\circ}-y)=\cos y,
$$

the above becomes

$$
\cos \left(x-90^{\circ}+y\right)=\cos x\sin y+\sin x\cos y.
$$

Noticing that

$$
\cos \left(x-90^{\circ}+y\right)=\cos \Bigl[(x+y)-90^{\circ}\Bigr]=\sin(x+y),
$$

we conclude that

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

---

**2.3 Simplify and Solve**

*2.3.1 Write as a Single Trigonometric Function*

We are given

$$
\sin \left(2x+40^{\circ}\right)\cos \left(x+30^{\circ}\right)-\cos \left(2x+40^{\circ}\right) \sin \left(x+30^{\circ}\right).
$$

Recognize the sine subtraction formula:

$$
\sin A\cos B-\cos A\sin B=\sin(A-B).
$$

Let

$$
A=2x+40^{\circ}\quad \text{and}\quad B=x+30^{\circ}.
$$

Thus, the expression becomes

$$
\sin\Bigl[(2x+40^{\circ})-(x+30^{\circ})\Bigr]=\sin(x+10^{\circ}).
$$

*2.3.2 Determine the General Solution of*

$$
\sin (x+10^{\circ})=-\frac{1}{2}.
$$

We know that for any angle $$\theta$$, if

$$
\sin\theta=-\frac{1}{2},
$$

then the general solutions are

$$
\theta=210^{\circ}+360^{\circ}k\quad \text{or}\quad \theta=330^{\circ}+360^{\circ}k,\quad k\in \mathbb{Z}.
$$

Setting $$\theta=x+10^{\circ}$$, we have

$$
x+10^{\circ}=210^{\circ}+360^{\circ}k\quad \text{or}\quad x+10^{\circ}=330^{\circ}+360^{\circ}k.
$$

Solving for $$x$$:

1. $$x=200^{\circ}+360^{\circ}k,$$
2. $$x=320^{\circ}+360^{\circ}k,$$

where $$k$$ is any integer.

---

**Final Answers:**

**2.1** $$-\tan x$$.

**2.2** $$\sin(x+y)=\sin x\cos y+\cos x\sin y$$.

**2.3.1** $$\sin(x+10^{\circ})$$.

**2.3.2** $$x=200^{\circ}+360^{\circ}k$$ or $$x=320^{\circ}+360^{\circ}k,\quad k\in \mathbb{Z}$$.
**2.1 Simplify:** $$ -\tan x $$ **2.2 Derive $$\sin(x+y)$$:** $$ \sin(x+y) = \sin x \cos y + \cos x \sin y $$ **2.3.1 Simplify:** $$ \sin(x + 10^\circ) $$ **2.3.2 General Solution:** $$ x = 200^\circ + 360^\circ k \quad \text{or} \quad x = 320^\circ + 360^\circ k, \quad k \in \mathbb{Z} $$

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