If 25 sin θ = -7 and 0° < θ < 270°, determine with the aid of a diagram, the values of the following: 1.1 tan θ 1.2 sec (180° – θ)

The value of $\tan \theta$ is \boxed{\text{}\frac{7}{24}}}.

3.1.2 sec $(180^\circ - \theta)$

The value of $\sec(180^\circ - \theta)$ is \boxed{\text{}\frac{25}{24}}}.

3.2 Determine the values of $\beta$, if $\cos(\beta - 60^\circ) = 0.25$ and $0^\circ \le \beta \le 360^\circ$.

So, $\cos A = 0.25$.

General solutions for $A$:
$A = \alpha + k \cdot 360^\circ$ or $A = 360^\circ - \alpha + k \cdot 360^\circ$, where $k$ is an integer.

Case 1: $A = 75.52^\circ + k \cdot 360^\circ$

For $k=0$, $\beta = 135.52^\circ$. (This is within $0^\circ \le \beta \le 360^\circ$)

Case 2: $A = 360^\circ - 75.52^\circ + k \cdot 360^\circ = 284.48^\circ + k \cdot 360^\circ$

For $k=0$, $\beta = 344.48^\circ$. (This is within $0^\circ \le \beta \le 360^\circ$)

The values of $\beta$ are \boxed{\text{135.52^\circ and 344.48^\circ}}.

QUESTION 4

4.1 Simplify:

• $\cot(180^\circ+\theta) = \cot \theta$ (cot is positive in Q3)
• $\sec(360^\circ-\theta) = \sec \theta$ (sec is positive in Q4) Numerator: $\cot \theta \cdot \sec \theta$

• $\sin(360^\circ-\theta) = -\sin \theta$ (sin is negative in Q4)
• $\sin(180^\circ+\theta) = -\sin \theta$ (sin is negative in Q3)
• $\cos(540^\circ+\theta) = \cos(180^\circ + 360^\circ + \theta) = \cos(180^\circ+\theta) = -\cos \theta$ (cos is negative in Q3)
• $\cos(180^\circ-\theta) = -\cos \theta$ (cos is negative in Q2)

The simplified expression is \boxed{\text{}\csc \theta}}.

4.2 Prove that:

The identity is proven.

QUESTION 5

Given the functions defined by $f(x) = \tan 3x$ and $g(x) = \sin 6x$ for $x \in [0^\circ; 90^\circ]$.

5.1 On the same axes, given in your SPECIAL ANSWER BOOK, draw the graphs of $f$ and $g$. Clearly show the turning points, asymptotes, endpoints, and the intercepts with the axes.

For $f(x) = \tan 3x$:

• Period: $\frac{180^\circ}{3} = 60^\circ$.
• Asymptotes: $x = 30^\circ$ and $x = 90^\circ$.
• x-intercepts: $(0^\circ, 0)$ and $(60^\circ, 0)$.
• y-intercept: $(0^\circ, 0)$.
• Endpoints: $(0^\circ, 0)$. The graph approaches infinity as $x \to 90^\circ$ from the left.
• Key points for sketching: $(15^\circ, 1)$, $(45^\circ, -1)$, $(75^\circ, 1)$.

For $g(x) = \sin 6x$:

• Period: $\frac{360^\circ}{6} = 60^\circ$.
• Amplitude: 1.
• x-intercepts: $(0^\circ, 0)$, $(30^\circ, 0)$, $(60^\circ, 0)$, $(90^\circ, 0)$.
• y-intercept: $(0^\circ, 0)$.
• Turning points: Maximum at $(15^\circ, 1)$ and $(75^\circ, 1)$. Minimum at $(45^\circ, -1)$.
• Endpoints: $(0^\circ, 0)$ and $(90^\circ, 0)$.

5.2 Write down the period of $f$ and $g$.

The period of $f(x)$ is \boxed{\text{60^\circ}}.
The period of $g(x)$ is \boxed{\text{60^\circ}}.

5.3 Use your graphs to determine for which values of $x$ is $\tan 3x = \sin 6x$. Give any TWO values.

By observing the key points and intercepts from the graph analysis in 5.1, the graphs intersect at several points within the given domain.

• At $x=0^\circ$: $f(0^\circ) = \tan(0^\circ) = 0$, $g(0^\circ) = \sin(0^\circ) = 0$. So $x=0^\circ$ is a solution.
• At $x=15^\circ$: $f(15^\circ) = \tan(45^\circ) = 1$, $g(15^\circ) = \sin(90^\circ) = 1$. So $x=15^\circ$ is a solution.
• At $x=45^\circ$: $f(45^\circ) = \tan(135^\circ) = -1$, $g(45^\circ) = \sin(270^\circ) = -1$. So $x=45^\circ$ is a solution.
• At $x=60^\circ$: $f(60^\circ) = \tan(180^\circ) = 0$, $g(60^\circ) = \sin(360^\circ) = 0$. So $x=60^\circ$ is a solution.
• At $x=75^\circ$: $f(75^\circ) = \tan(225^\circ) = 1$, $g(75^\circ) = \sin(450^\circ) = \sin(90^\circ) = 1$. So $x=75^\circ$ is a solution.

Any two of these values can be given.
Two values for which $\tan 3x = \sin 6x$ are \boxed{\text{0^\circ and 15^\circ}}. (Other valid answers include $45^\circ, 60^\circ, 75^\circ$)