Mathematics P1 - Question 8

Question 8

8.1

Given that $\text{XÔP} = \theta$ and the point $P(3; -3\sqrt{3})$.

8.1.1 Determine the value of OP.

$OP = \sqrt{x^2 + y^2} = \sqrt{3^2 + (-3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6$

Therefore, $OP = 6$.

8.1.2 Determine the value of $cos(180° + \theta)$.

First, find $cos(\theta)$ and $sin(\theta)$.

$cos(\theta) = \frac{x}{OP} = \frac{3}{6} = \frac{1}{2}$

$sin(\theta) = \frac{y}{OP} = \frac{-3\sqrt{3}}{6} = \frac{-\sqrt{3}}{2}$

Then, $cos(180° + \theta) = cos(180°)cos(\theta) - sin(180°)sin(\theta) = (-1)(\frac{1}{2}) - (0)(\frac{-\sqrt{3}}{2}) = -\frac{1}{2}$

Therefore, $cos(180° + \theta) = -\frac{1}{2}$.

8.1.3 Determine the value of $-sin(360° - 2\theta)$.

$-sin(360° - 2\theta) = sin(2\theta) = 2sin(\theta)cos(\theta) = 2(\frac{-\sqrt{3}}{2})(\frac{1}{2}) = -\frac{\sqrt{3}}{2}$

Therefore, $-sin(360° - 2\theta) = -\frac{\sqrt{3}}{2}$.

8.2

Given $sin(d) = A$ and $cos(90°-d)cos(23°) + cos(246°)sin(23°) = B$, find the value of $d$ if $A = B$ and $0° ≤ d ≤ 90°$.

Since $cos(90°-d) = sin(d)$, the expression becomes:

$sin(d)cos(23°) + cos(246°)sin(23°) = B$

Also, $cos(246°) = cos(180° + 66°) = -cos(66°) = -sin(24°)$

Therefore, the equation is:

$sin(d)cos(23°) - sin(24°)sin(23°) = B$

Since $A = B = sin(d)$, then
$sin(d) = sin(d)cos(23°) - sin(24°)sin(23°)$

$sin(d) = sin(d + 23)$

$sin(d) = B$

Using $sin(A - B) = sin A cos B - cos A sin B$, let $246 = 180 + 66$. Thus, $cos(246°) = -cos(66°) = -sin(24°)$

So, $sin d = sin d cos 23 - sin 23 sin 24. sin (d+23) = sin d cos 23 + cos d sin 23.$B = sin d cos 23 + cos 246 sin 23 = sin d cos 23 - sin 24 sin 23.

equating A = B$,$sin d = sin (d - 23).
So d = 23°

8.3 Simplify

\frac{cos13° sin 13°}{cos 300° cos 244°}

cos(300°) = cos(360° - 60°) = cos(60°) = \frac{1}{2}

cos(244°) = cos(180° + 64°) = -cos(64°)

So the expression becomes:

\frac{cos(13°)sin(13°)}{(\frac{1}{2})(-cos(64°))} = \frac{2 cos(13°)sin(13°)}{-cos(64°)} = \frac{sin(26°)}{-cos(64°)}

Using cos(90° - x) = sin(x)$, we have$cos(64°) = cos(90° - 26°) = sin(26°)

Therefore, \frac{sin(26°)}{-cos(64°)} = \frac{sin(26°)}{-sin(26°)} = -1

8.4

Given the identity:
\frac{sin 4x - sin 2x}{cos 4x - cos 2x} = tan x

8.4.1 Prove the identity.

Using the sum-to-product formulas:

sin A - sin B = 2 cos(\frac{A+B}{2}) sin(\frac{A-B}{2})

cos A - cos B = -2 sin(\frac{A+B}{2}) sin(\frac{A-B}{2})

\frac{sin 4x - sin 2x}{cos 4x - cos 2x} = \frac{2 cos(\frac{4x+2x}{2}) sin(\frac{4x-2x}{2})}{-2 sin(\frac{4x+2x}{2}) sin(\frac{4x-2x}{2})} = \frac{2 cos(3x) sin(x)}{-2 sin(3x) sin(x)}

= -\frac{cos(3x)}{sin(3x)} = -cot(3x)

= \frac{cos(3x)}{sin(3x)} = \frac{cos(3x)}{sin(3x)}

It should be
\frac{2 cos(3x) sin(x)}{-2 sin(3x) sin(x)} = -\frac{cos(3x) }{sin(3x)}
There seems to have mistake on right hand side

8.4.2 For which values of x in the interval -60° ≤ x ≤ 135° will the identity be undefined?

The identity will be undefined when the denominator is zero.

cos 4x - cos 2x = 0

-2 sin(3x) sin(x) = 0

So, sin(3x) = 0$or$sin(x) = 0

If sin(x) = 0$, then$x = 0°, 180°$etc. In the interval,$x = 0°

If sin(3x) = 0$, then$3x = 0°, 180°, 360°, 540°

x = 0°, 60°, 120°

Considering the range -60° ≤ x ≤ 135°$, the values of$x$that make the identity undefined are$x = 0°, 60°, 120°

8.5 Determine the general solution of the following equation:

sin x + 2 cos(x+30°) = -2 cos^2x

sin x + 2 [cos x cos 30° - sin x sin 30°] = -2 cos^2x

sin x + 2 [cos x (\frac{\sqrt{3}}{2}) - sin x (\frac{1}{2})] = -2 cos^2x

sin x + \sqrt{3} cos x - sin x = -2 cos^2x

\sqrt{3} cos x = -2 cos^2x

2 cos^2x + \sqrt{3} cos x = 0

cos x (2 cos x + \sqrt{3}) = 0

So, cos x = 0$or$2 cos x + \sqrt{3} = 0

If cos x = 0$, then$x = 90° + 180°k, k ∈ Z

If 2 cos x + \sqrt{3} = 0$, then$cos x = -\frac{\sqrt{3}}{2}

x = 150° + 360°k$or$x = -150° + 360°k, k ∈ Z

Therefore, the general solutions are:

x = 90° + 180°k, k ∈ Z

x = 150° + 360°k, k ∈ Z

x = -150° + 360°k, k ∈ Z$$