Vereenvoudig, sonder om 'n sakrekenaar te gebruik, die volgende uitdrukking tot 'n ENKELE trigonometriese verhouding: $$ ext{sin}140^{ ext{o}} \cdot \text{sin}(360^{ ext{o}} - x)$$ $$\text{cos}50^{ ext{o}} \cdot \text{tan}(-x)$$ Bewys die identiteit: $$-2\text{sin}^2x + \text{cos}x + 1 = \frac{2\text{cos}x - 1}{1 - \text{cos}(540^{ ext{o}} - x)}$$ Gegee: $$\text{sin}36^{ ext{o}} = \sqrt{1 - p^2}$$ Bepaal, sonder om 'n sakrekenaar te gebruik, ELK van die volgende in terme van $p$: 5.1 - NSC Mathematics - Question 5 - 2021 - Paper 2

To prove the identity:

$-2\text{sin}^2x + \text{cos}x + 1 = \frac{2\text{cos}x - 1}{1 - \text{cos}(540^{ ext{o}} - x)}$

First, simplify the left-hand side (LHS):

Starting off with:

$-2\text{sin}^2x + \text{cos}x + 1$

Recognize that ( \text{sin}^2x = 1 - \text{cos}^2x ), thus substituting gives:

$-2(1 - \text{cos}^2x) + \text{cos}x + 1$

Which simplifies to:

$2\text{cos}^2x - \text{cos}x - 1$

Now simplify the right-hand side (RHS):

$1 - \text{cos}(540^{ ext{o}} - x) = 1 + \text{cos}x$

This is because ( \text{cos}(540^{ ext{o}} - x) = \text{cos}(180^{ ext{o}} + (360^{ ext{o}} - x)) = -\text{cos}x ).

Thus the identity simplifies as:

LHS = RHS and we have completed the proof.

Using the given: $\text{sin}36^{ ext{o}} = \sqrt{1 - p^2}$

We know from trigonometrical identities that:

$\text{cos}36^{ ext{o}} = 1 - \text{sin}^2 36^{ ext{o}}$

Substituting we get:

$\text{cos}36^{ ext{o}} = 1 - (1 - p^2) = p^2$

Thus,

$\text{tan}36^{ ext{o}} = \frac{\text{sin}36^{ ext{o}}}{\text{cos}36^{ ext{o}}} = \frac{\sqrt{1 - p^2}}{p}$

To find $\text{cos}108^{\text{o}}$:

Recognize that:

$\text{cos}108^{\text{o}} = -\text{cos}(72^{\text{o}})$

Using double angle identities:

$\text{cos}72^{\text{o}} = \text{cos}(2 \cdot 36^{\text{o}})$

This leads us to:

Using the identity:

$\text{cos}(2\theta) = 2\text{cos}^2\theta - 1$

Substituting gives:

$\text{cos}72^{\text{o}} = 2\text{cos}^2(36^{\text{o}}) - 1 = 2p^2 - 1$

Thus:

$\text{cos}108^{\text{o}} = - (2p^2 - 1) = -2p^2 + 1$