Given: sin A = 2p and cos A = p 5.1.1 Determine the value of tan A - NSC Mathematics - Question 5 - 2017 - Paper 2

Since A is in the range [180°, 270°], we know that sin A is negative and cos A is also negative in this quadrant. Using the Pythagorean identity:

$sin^2 A + cos^2 A = 1$

Substituting the values:

$(2p)^2 + p^2 = 1$

which simplifies to:

$4p^2 + p^2 = 1 \Rightarrow 5p^2 = 1 \Rightarrow p^2 = \frac{1}{5} \Rightarrow p = -\frac{1}{\sqrt{5}}$

The negative value is chosen since both sine and cosine are negative in the third quadrant.

To solve the equation, we will apply the quadratic formula. Letting $s = sin x$, we rewrite the equation as:

$2s^2 - 5s + 2 = 0.$
Using the quadratic formula:

$s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\Rightarrow s = \frac{5 \pm \sqrt{(-5)^2 - 4 * 2 * 2}}{2 * 2} \Rightarrow s = \frac{5 \pm \sqrt{25 - 16}}{4}$

This gives:

$s = \frac{5 \pm 3}{4} \Rightarrow s_1 = 2 \quad s_2 = \frac{1}{2}.$
Thus, we have $s = sin x = 2$ with no solution in real numbers and $s = sin x = \frac{1}{2}$.
The general solutions would be:

$x = 30° + k \cdot 360° \text{ or } 150° + k \cdot 360°, \text{ where } k \in \mathbb{Z}.$

Using the sine addition formula, we have:

$sin(x + 300°) = sin x \cdot cos(300°) + cos x \cdot sin(300°).$
Knowing that:

$cos(300°) = \frac{1}{2} \quad \text{and} \quad sin(300°) = -\frac{\sqrt{3}}{2},$
we then write:

$sin(x + 300°) = sin x \cdot \frac{1}{2} - cos x \cdot \frac{\sqrt{3}}{2}.$

From the previous expansion, we have:

$sin(x + 300°) = \frac{1}{2} sin x - \frac{\sqrt{3}}{2} cos x.$
Now to find $cos(x - 150°)$, using the cosine subtraction formula gives:

$cos(x - 150°) = cos x \cdot cos(150°) + sin x \cdot sin(150°)$

where:

$cos(150°) = -\frac{\sqrt{3}}{2}\quad ext{and} \quad sin(150°) = \frac{1}{2}.$
So,

$cos(x - 150°) = -\frac{\sqrt{3}}{2} cos x + \frac{1}{2} sin x.$
Thus,

$sin(x + 300°) - cos(x - 150°) = \frac{1}{2} sin x - \frac{\sqrt{3}}{2} cos x - (-\frac{\sqrt{3}}{2} cos x + \frac{1}{2} sin x)$

This simplifies to:

$= 0.$

Starting with the left-hand side (LHS):

$LHS = tan x + 1 = \frac{sin x}{cos x} + 1 = \frac{sin x + cos x}{cos x}.$
For the right-hand side (RHS):

$RHS = \frac{sin x + cos x}{(sin x tan x) + cos x} = \frac{sin x + cos x}{\frac{sin^2 x}{cos x} + cos x} = \frac{sin x + cos x}{\frac{sin^2 x + cos^2 x}{cos x}}.$
As we know that:

$sin^2 x + cos^2 x = 1,$
this becomes:

$RH = \frac{sin x + cos x}{\frac{1}{cos x}} = sin x + cos x.$
Thus, both sides are equal, proving the identity.

Starting from the equation:

$sin x + cos x = \sqrt{1 + k}.$
Square both sides gives:

$(sin x + cos x)^2 = 1 + k.$
Expanding the left side:

$sin^2 x + 2sin x cos x + cos^2 x = 1 + k.$
Using the Pythagorean identity, we have:

$1 + 2sin x cos x = 1 + k \Rightarrow k = 2sin x cos x.$

Given:

$k = 2sin x cos x,$
Using the formula:

$sin x + cos x = \sqrt{1 + k}.$
The maximum value of $k$ occurs when both $sin x$ and $cos x$ are at their maximum at angles of 45°:

$k_{max} = 1 + 1 = 2.$
Thus, the maximum value of $sin x + cos x = \sqrt{2}.$