Question

question 4
in the diagram below, the graph of $f(x)=a\sin bx + 1$ and $g(x)=2\cos(x + p)$ are shown
for the interval $-90^{circ}\leq x\leq180^{circ}$.
4.1 write down the amplitude of $g$.
4.2 determine the period of $f$.
4.3 determine the values of $a,b$ and $p$.
4.4 determine the values of $x$ such that:
4.4.1 $g(x)>0$
4.4.2 $f(x)\geq g(x)$
4.4.3 $f(x).g(x)<0$
4.5 given $h(x)=g(x + t)$, determine all values of $t$ where $h(x)=h(-x)$.

Explanation:

Step1: Recall amplitude formula for cosine

For $y = A\cos(Bx + C)+D$, amplitude is $|A|$. For $g(x)=2\cos(x + p)$, amplitude $= 2$.

Step2: Recall period formula for sine

For $y = a\sin(bx)+1$, period $T=\frac{360^{\circ}}{|b|}$. From the graph of $f(x)$, the period is $120^{\circ}$, so $\frac{360^{\circ}}{|b|}=120^{\circ}$, then $|b| = 3$. Since the graph is moving as a normal sine - like curve, $b = 3$.

Step3: Find $a$ using a point on $f(x)$

The point $(48.05^{\circ},0.41)$ lies on $f(x)=a\sin(3x)+1$. Substitute $x = 48.05^{\circ}$ and $y = 0.41$:
$0.41=a\sin(3\times48.05^{\circ})+1$.
$\sin(144.15^{\circ})\approx0.588$.
$0.41=a\times0.588 + 1$.
$a\times0.588=0.41 - 1=-0.59$.
$a=\frac{-0.59}{0.588}\approx - 1$.

Step4: Find $p$ using a point on $g(x)$

The graph of $g(x)=2\cos(x + p)$ passes through $(0,1)$. Substitute $x = 0$ and $y = 1$:
$1 = 2\cos(0 + p)$.
$\cos(p)=\frac{1}{2}$. In the given domain, $p = 60^{\circ}$.

Step5: Solve $g(x)>0$

$g(x)=2\cos(x + 60^{\circ})>0$.
$\cos(x + 60^{\circ})>0$.
$-90^{\circ}$-150^{\circ}

Step6: Solve $f(x)\geq g(x)$

From the graph, $f(x)\geq g(x)$ when $-90^{\circ}\leq x\leq - 30^{\circ}$ or $60^{\circ}\leq x\leq120^{\circ}$.

Step7: Solve $f(x)\cdot g(x)<0$

$f(x)\cdot g(x)<0$ when $f(x)$ and $g(x)$ have opposite signs. From the graph, this is when $-30^{\circ}

Step8: Analyze $h(x)=h(-x)$

$h(x)=g(x + t)=2\cos((x + t)+60^{\circ})$, $h(-x)=2\cos((-x + t)+60^{\circ})$.
Since $h(x)=h(-x)$, the function $h(x)$ is even. For $y = \cos(Ax + B)$, it is even when $B = k\cdot180^{\circ},k\in\mathbb{Z}$.
$(x + t)+60^{\circ}=(-x + t)+60^{\circ}+k\cdot360^{\circ}$ or $(x + t)+60^{\circ}=-( - x + t)-60^{\circ}+k\cdot360^{\circ}$.
The first equation gives $0 = k\cdot360^{\circ}$ (not useful for finding $t$).
The second equation:
$x + t+60^{\circ}=x - t - 60^{\circ}+k\cdot360^{\circ}$.
$2t=-120^{\circ}+k\cdot360^{\circ}$.
$t=-60^{\circ}+k\cdot180^{\circ},k\in\mathbb{Z}$. In the domain (assuming we are looking for values in a reasonable range related to the given domain of $x$), when $k = 0,t=-60^{\circ}$; when $k = 1,t = 120^{\circ}$.

Answer:

4.1. $2$
4.2. $120^{\circ}$
4.3. $a=-1,b = 3,p = 60^{\circ}$
4.4.1. $-90^{\circ}\leq x<30^{\circ}$
4.4.2. $-90^{\circ}\leq x\leq - 30^{\circ}$ or $60^{\circ}\leq x\leq120^{\circ}$
4.4.3. $-30^{\circ}4.5. $t=-60^{\circ},120^{\circ}$