5.1 Vereenvoudig (SONDER DIE GEBRUIK VAN 'n SAKREKENAAR) sin(180° - x) - NSC Mathematics - Question 5 - 2016 - Paper 2

To prove the identity, we examine the left-hand side (LHS):

$$LHS = \frac{sin x}{1 + cos x} + \frac{1 + cos x}{sin x}$$

Combining the fractions:

$$= \frac{sin^2 x + (1 + cos x)(1 + cos x)}{sin x(1 + cos x)}$$

This simplifies to:

$$= \frac{sin^2 x + 2cos x + cos^2 x}{sin x(1 + cos x)}$$

Using the Pythagorean identity ( sin^2 x + cos^2 x = 1 ), we find:

$$= \frac{1 + 2cos x}{sin x(1 + cos x)}\ = \frac{2}{sin x}$$

This verifies the identity.

To prove:

$$2 cos 2x = 2 cos^2 x - 1$$

We can utilize the double angle identity for cosine:

cos 2x = 2cos^2 x - 1\ $$(Thus,\ 2cos 2x = 2(2cos^2 x - 1) = 2cos^2 x - 1)$$

Given:

cos 2x + 3 sin 2x = 2\

Rearranging gives:

cos 2x = 2 - 3 sin 2x\

Using the double angle identity and solve:

This leads to using inverse functions to find possible solutions.

Since ( A + B = 90° ), we can utilize the co-function identity:

sin A = cos B \rightarrow sin A \cdot cos B = sin(90°) = 1$$