5.1 Simplify the expression to a single trigonometric term: tan(-x)cosxsin(-x-180°)-1 5.2 Given: cos 35° = m Without using a calculator, determine the value of EACH of the following in terms of m: 5.2.1 cos 215° 5.2.2 sin 20° 5.3 Determine the general solution of: cos 4x.cos x + sin 4x.sin x = -0.7 5.4 Prove the identity: sin 4x.cos 2x - 2cos 4x.sin x.cos x tan 2x = cos^3 x - sin^3 x - NSC Mathematics - Question 5 - 2021 - Paper 2

To find $cos(215°)$, we can use the reduction formula:

1. Rewrite $215°$ as $180° + 35°$.
2. Therefore, $cos(215°) = -cos(35°) = -m$

To compute $sin(20°)$ in terms of $m$:

1. Use the co-function identity, which gives $sin(20°) = cos(70°)$.

2. Rewrite $70°$ as $90° - 20°$:

   $sin(20°) = cos(35° + 35°)$

3. Using the double angle formulas, we expand:

   $= 2cos(35°)sin(35°) = 2m ext{sqrt{1-m^2}}$

To solve the equation $cos 4x.cos x + sin 4x.sin x = -0.7$

1. Express the left side using the cosine identity:

   $cos(4x - x) = cos(3x) = -0.7$

2. Finding the angles for $3x$:

   $3x = 180° + 45.57° + k imes 360° ext{ or } 3x = 180° - 45.57° + k imes 360°$

3. This simplifies to two cases:

   Case 1: $3x = 134.43° + k imes 360° ext{ or } 3x = 225.57° + k imes 360°$

   Thus,

   $x = \frac{134.43°}{3} + k imes 120° ext{ or } x = \frac{225.57°}{3} + k imes 120°$

4. The solutions are:

   $x = 44.81° + k imes 120°$ and $x = 75.19° + k imes 120°$

To prove the identity:

$sin 4x.cos 2x - 2cos 4x.sin x = cos^3 x - sin^3 x$

1. Start with the left-hand side (LHS):

   $sin 4x.cos 2x - 2cos 4x.sin x = sin(4x-2x)$

2. Using the sine identity:

   $= sin(2x)$

3. Now, focus on $RHS$:

   $cos^3 x - sin^3 x = (cos x - sin x)(cos^2 x + cos x sin x + sin^2 x)$

   Since $cos^2 x + sin^2 x = 1$, we simplify:

   $= (cos x - sin x)(1 + cos x sin x)$

4. Both sides equal, thus:

   $LHS = RHS$, confirming the identity.