Trigonometric functions HSC question (2 Viewers)

BlueGas · May 25, 2015

I need help with ii) and iii). I just want to know for ii) why in the solution was y=1.35. For iii), what is the formula T = 2Pi/a? I've never seen it before.

InteGrand · May 25, 2015

BlueGas said:
I need help with ii) and iii). I just want to know for ii) why in the solution was y=1.35. For iii), what is the formula T = 2Pi/a? I've never seen it before.

$The formula $T=\frac{2\pi}{\omega}$ is the formula for the period of the general simple sine wave $y=A\sin(\omega t+ \varphi)+C$ (the physical significance of $\omega$ is that it is the \textit{angular frequency} of the motion).$

$i) Clearly $A=1.73$ (in metres) is the amplitude (we can just read this off the given graph).$

$ii) When $t=0$, $y=1.35$, so $1.35 = \underbrace{A}_{1.73}\sin(0+b)\Rightarrow \sin b = \frac{1.35}{1.73}\Rightarrow b=\sin^{-1} \frac{1.35}{1.73}\approx 0.895220231 \rad$.$

$iii) We can see that the period is $T = 12.42$ h, so $\frac{2\pi}{a}=12.42\Rightarrow a=\frac{2\pi}{12.42}\approx 0.505892537$ (units being rad/h).$

BlueGas · May 25, 2015

InteGrand said:
$The formula $T=\frac{2\pi}{\omega}$ is the formula for the period of the general simple sine wave $y=A\sin(\omega t+ \varphi)+C$ (the physical significance of $\omega$ is that it is the \textit{angular frequency} of the motion).$

$i) Clearly $A=1.73$ (in metres) is the amplitude (we can just read this off the given graph).$

$ii) When $t=0$, $y=1.35$, so $1.35 = \underbrace{A}_{1.73}\sin(0+b)\Rightarrow \sin b = \frac{1.35}{1.73}\Rightarrow b=\sin^{-1} \frac{1.35}{1.73}\approx 0.895220231 \rad$.$

$iii) We can see that the period is $T = 12.42$ h, so $\frac{2\pi}{a}=12.42\Rightarrow a=\frac{2\pi}{12.42}\approx 0.505892537$ (units being rad/h).$

How do you know for ii) that y = 1.35 not 1.73?

InteGrand · May 25, 2015

BlueGas said:
How do you know for ii) that y = 1.35 not 1.73?

$Because we can see from the given graph that the $y$-value is 1.35 when $t=0$. (The value of the function $y(t)$ when $t=0$ is just the $y$-intercept.)$

BlueGas · May 25, 2015

Also another question: I was finding the area of a sector using the formula 1/2 r^2 theta (man I want to use latex) and my calculator was set to degrees because I was using 120', but then I got a completely different answer when I did 2Pi/3 but in radians, what caused the answers to be different when 120' and 2Pi/3 are the same but set in appropriate modes for each?

InteGrand · May 25, 2015

BlueGas said:
Also another question: I was finding the area of a sector using the formula 1/2 r^2 theta (man I want to use latex) and my calculator was set to degrees because I was using 120', but then I got a completely different answer when I did 2Pi/3 but in radians, what caused the answers to be different when 120' and 2Pi/3 are the same but set in appropriate modes for each?

$We need to use radians if we use the formula $\frac{1}{2}r^2 \theta$. The angle $\theta$ must be typed into your calculator in radians (we don't need radian mode because we are not taking the sine of anything or something, we're just typing in a number. So if we type in 120 into the calculator, that's like saying the area is $\frac{1}{2}r^2 \times 120$). If we want to use the angle in degrees (say $\alpha ^\circ$), the formula for the area of the sector would be $A=\frac{\alpha}{360}\times \pi r^2$.$

BlueGas · May 25, 2015

InteGrand said:
$We need to use radians if we use the formula $\frac{1}{2}r^2 \theta$. The angle $\theta$ must be typed into your calculator in radians (we don't need radian mode because we are not taking the sine of anything or something, we're just typing in a number. So if we type in 120 into the calculator, that's like saying the area is $\frac{1}{2}r^2 \times 120$). If we want to use the angle in degrees (say $\alpha ^\circ$), the formula for the area of the sector would be $A=\frac{\alpha}{360}\times \pi r^2$.$

So what formulas should always be set in radians mode and what formulas should be set in degrees?

InteGrand · May 25, 2015

BlueGas said:
So what formulas should always be set in radians mode and what formulas should be set in degrees?

It's generally best to keep your calculator in radians mode, and only switch to degrees mode if you want to calculate something involving evaluating a trig. or inverse trig. function with degrees (e.g. if you want the answer to $\sin ^{-1} \frac{1}{\sqrt{2}}$ in degrees (i.e. you want to find the (acute) angle that has a sine of $\frac{1}{\sqrt{2}}$ , you should have your calculator in degrees mode before typing in $\sin ^{-1} \frac{1}{\sqrt{2}}$ ). In general, if you evaluate an inverse trig. function, the answer is an angle, so you'll get the answer in degrees if you're calculator is in degree mode, and radians if in radians mode.

If you want to evaluate a trig. function at a particular value in degrees, use degree mode (e.g. if you use degree mode, typing $\texttt{sin 30}$ into the calculator will return the value of sin 30º, which is $\frac{1}{2}$ ; if you'd used radian mode, typing $\texttt{sin 30}$ would return the value of $\sin (30 \text{ rad})$ ).

If using non-trigonometric/inverse trigonometric functions, it shouldn't matter whether you're in degrees mode or radians mode. E.g. typing $\texttt{1+1}$ or $\texttt{e}^\texttt{2}$ will give the same value regardless of the mode.

BlueGas · May 25, 2015

So basically any sort of question involving arc length, area of sector, etc. It should always be in radians mode and degrees are in radians?

applebeeby · Jun 11, 2015

Yep. Pretty much

Trigonometric functions HSC question (2 Viewers)

BlueGas

Well-Known Member

InteGrand

Well-Known Member

BlueGas

Well-Known Member

InteGrand

Well-Known Member

BlueGas

Well-Known Member

InteGrand

Well-Known Member

BlueGas

Well-Known Member

InteGrand

Well-Known Member

BlueGas

Well-Known Member

applebeeby

New Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)