5.1 Vereenvoudig (SONDER DIE GEBRUIK VAN 'n SAKREKENAAR) $$ ext{sin}(180^ ext{o} - x) imes ext{cos}(x - 360^ ext{o}) imes ext{tan}(180^ ext{o} + x) imes ext{cos}(-x) imes ext{tan}(-x) imes ext{cos}(90^ ext{o} - x) imes ext{sin}(90^ ext{o} - x)$$ 5.2 Bewys die identiteit: $$rac{ ext{sin} x}{1 + ext{cos} x} + rac{1 + ext{cos} x}{ ext{sin} x} = rac{2}{ ext{sin} x}$$ 5.3 Gebruik saamgestelde hoeke om aan te toon dat: $$ ext{cos} 2x = 2 ext{cos}^2 x - 1$$ 5.4 Bepaal die algemene oplossing vir $x$, as: $$ ext{cos} 2x + 3 = 2$$ 5.5 In $ riangle ABC$: $ ext{A} + ext{B} = 90^ ext{o}$ - NSC Mathematics - Question 5 - 2016 - Paper 2

To prove the identity, we start with the left-hand side:

1. Write down the identity:

   rac{ ext{sin} x}{1 + ext{cos} x} + rac{1 + ext{cos} x}{ ext{sin} x}

2. Find a common denominator, which is $(1 + ext{cos} x) imes ext{sin} x$:

   = rac{ ext{sin}^2 x + (1 + ext{cos} x)(1 + ext{cos} x)}{(1 + ext{cos} x) ext{sin} x}

3. Expand:

   = rac{ ext{sin}^2 x + 1 + 2 ext{cos} x + ext{cos}^2 x}{(1 + ext{cos} x) ext{sin} x}

4. Recognize that $ext{sin}^2 x + ext{cos}^2 x = 1$:

   = rac{2(1 + ext{cos} x)}{(1 + ext{cos} x) ext{sin} x} = rac{2}{ ext{sin} x}.

To prove that:

$ext{cos} 2x = 2 ext{cos}^2 x - 1$

1. Recall the double angle formula for cosine:

   $ext{cos} 2x = ext{cos}^2 x - ext{sin}^2 x$

2. Substitute $ext{sin}^2 x$ with $1 - ext{cos}^2 x$:

   $= ext{cos}^2 x - (1 - ext{cos}^2 x)$

3. Simplify:

   $= 2 ext{cos}^2 x - 1.$

To determine the general solution for:

$ext{cos} 2x + 3 = 2$

1. Rearrange the equation:

   $ext{cos} 2x = -1$

2. The cosine function equals -1 at:

   $2x = 180^ ext{o} + k imes 360^ ext{o}, ext{ where } k ext{ is an integer.}$

3. Therefore:

   $x = 90^ ext{o} + k imes 180^ ext{o}.$

In triangle ABC where A + B = 90°:

1. We can use the identity:

   $ext{sin} A imes ext{cos} B = ext{sin}(A + B).$

2. Since $A + B = 90^ ext{o}$, it follows that:

   $ext{sin}(A + B) = ext{sin} 90^ ext{o} = 1.$