# Elementary trigonometry

I have come up with a simple trig problem. It says which of the following is positive? \begin{align} A&)\cos 3 \\ B&)\sin 4 \\ C&)\sin 2 \\ D&)\tan 2 \\ E&)\cos 9 \end{align}. The answer is written as follows: "Since $\frac{\pi}{2}< 2< \pi$, number 2 lies in the second quadrant and, therefore, $\sin2$ is positive". I can't understand why/how 2 lies in the second quadrant? I have tried to fathom it by converting radians to degrees, but the situation remained the same. Was the idea of converting right? Could anybody explain the answer given above for me in details?

• What is tg2? is it $\tan 2$ – Jaideep Khare Apr 19 '17 at 18:15
• Jaiideep Most probably it means $\tan(2)$ where $2$ means $2$ radians. – Namaste Apr 19 '17 at 18:16
• Yes, tg2 is tan2. – user36339 Apr 19 '17 at 18:21
• Yoqubjon, please fromat your question using LaTeX, see this math.meta.stackexchange.com/questions/5020/… – Jaideep Khare Apr 19 '17 at 18:22
• Yoqubjon Estimate the value of $\frac\pi{2}$. Then estimate $\pi$. (Those values are the bounds of the second quadrant). You see that $\pi/2 \lt 2\lt \pi.$ That is, the angle 2 radians has its terminal angle fall in the second quadrant. – Namaste Apr 19 '17 at 18:24

# 2,3,4 etc. in the arguement of sine,cosine and tangent are all radians, not degrees.

Coming to solution ;

Since, $\dfrac{\pi}{2} \approx 1.5707$ and $\pi \approx 3.1416$ and

$$1.5707 < 2 < 3.1416 \implies \frac{\pi}{2} < 2 < \pi$$

Also, all the angles between $\dfrac{\pi}{2} (=90^{\circ})$ and $\pi(=180^{\circ})$ lie in second quadrant.Therefore $2$ lies in second quadrant.

$2 ~\text{radians}= \Big(2 \times \dfrac{180}{\pi}\Big) \approx~ 114.6^{\circ}~~\text{And}~~\sin \theta >0 ~\forall ~\theta \in (90^{\circ},180^{\circ}) \implies \sin 2 >0$

• What about sin4? I think that 4 is greater than π, then it is in the third quadrant, where sin is negative. Is this expression right: -sin4=sin4? – user36339 Apr 19 '17 at 18:40
• What , $-\sin 4 = \sin 4$ ? What do you mean by that? And yes, $\sin 4 < 0$ – Jaideep Khare Apr 19 '17 at 18:42
• I meant to say that since 4, coming from my understanding, is in the 3rd quadrant (in terms of sin) it must be negative. So, sin4 is equal to -sin4. But I am NOT also sure about it. – user36339 Apr 19 '17 at 18:48
• You cannot say since a is negative, it is equal to -a. You can say since a is negative, it is equal to -|a|, where |.| denotes absolute value – Jaideep Khare Apr 19 '17 at 18:50
• How did you know that the arguments are in radians, not in degrees? Please, explain. – user36339 Apr 19 '17 at 20:00

Just for the heck of it, here is a proof that $\pi/2 < 2 < \pi$ without resorting to numerical approximation.

The first inequality is equivalent to $\pi < 4$. Note that $\pi$ is the area of the circle with unit radius, and hence diameter $2$. This circle can be therefore inscribed in a square with side 2, and the area of the circle ($\pi$) is smaller than the area of the square ($4$), yielding the result.

For the second inequality $\pi > 2$, remember that $\pi$ is also the length of a semicircular arc in a circle of diameter $2$. Since diameter is a line segment between 2 points, it is the shortest such connection, and thus must be less than the semicircular arc connecting the same 2 points, which implies $\pi > 2$ as desired...

Today, I had been taught to solve this equation very differently by converting radians to degrees. For example, let's take sin2. Since 1 rad is approximately 59.23, I will round it to 60 deg. just to make multiplication process easier. sin(2*60)=sin120, which is in the second quadrant, henceforth it is positive.