Without using a calculator, simplify the following expression to a single trigonometry ratio: $$1 - \sin(\theta)(\cos(90^\circ + \theta))$$ $$\cos(\theta - 360^\circ)$$ - NSC Mathematics - Question 5 - 2023 - Paper 2

$\cos(200^\circ) = -\cos(20^\circ)$

$\sin(70^\circ) = \sin(90^\circ - 20^\circ) = \cos(20^\circ) = \sqrt{1 - p}$

Using the double angle identity:

$\sin(10^\circ) = \cos(20^\circ) = \sqrt{1 - p}$

$\cos(A + 55^\circ)\cos(A + 10^\circ) + \sin(A + 55^\circ)\sin(A + 10^\circ) = \cos(A + 45^\circ)$

This uses the angle addition identity, confirming:

$\frac{1}{\sqrt{2}}$

Starting with:

$\cos 2x + \sin 2x - \cos^2 x$

Using the identity for LHS and RHS:

$\text{LHS} = -\sin x$

Thus proving it.

We have:

$\cos 2x + \sin 2x - \cos^2 x = -3\sin^3 x + \sin x$

This can be simplified using substitution and common factors to yield:

$\frac{\cos 2x + \sin 2x - \cos^2 x}{\sin x}$

Using the identity:\n $3\tan 4x = -2\cos 4x$

We simplify to:

$\sin 4x = 2\cos 4x$.

From:

$3\tan 4x = -2$

We deduce solutions based on the standard form leading to general solutions:

$x = 210^{\circ} + k \cdot 360^{\circ}, 330^{\circ} + k \cdot 360^{\circ}, k \in \mathbb{Z}$.