Evaluate each __without using a calculator.__

1) \(sin(\frac{\pi}{3})\)

2) \(cos(\frac{\pi}{4})\)

3) \(tan(\frac{\pi}{4})\)

4) \(csc(\frac{\pi}{3})\)

5) \(sec(\frac{\pi}{3})\)

6) \(cot(\frac{\pi}{6})\)

AdamTaurus Jan 16, 2018

#1**+2 **

sin (pi/3) = √3/2

cos(pi/4) = √2/2

tan (pi/4) = 1

csc(pi/3) = 2/√3 = 2√3 / 3

sec (pi/3) = 1 / cos(pi/3) = 1 / (1/2) = 2

cot (pi/6) = 1 / tan (pi/6) = 1 / (1/√3) = √3

CPhill Jan 17, 2018

#2**+2 **

Can you show me the process of how you did at least one of them? Like, how did you jump from sin(pi/3) to √3/2?

AdamTaurus
Jan 17, 2018

#3**+1 **

These are common reference angles, Adam

You really just need to memorize this chart :

http://www.analyzemath.com/trigonometry/special_angles.html

Note that the reciprocal of √2/2 = 2 / √2 = √2*√2 / √ 2 = √2

It's not as bad as it looks...since csc, sec and cot are just reciprocals of sin, cos and tan, respectively.....just learn the values for sin, cos and tan......"flip" these over for csc, sec and tan!!!

This is the way I had to learn them.....

CPhill Jan 17, 2018

#4**+2 **

To find sin( pi/3 ) , we can start with an equilateral triangle with side length 1 .

Now draw a line that bisects an angle and the opposite side. The length of this line is sin(pi/3) .

By the Pythagorean theorem...

(1/2)^{2} + ( sin(pi/3) )^{2} = 1^{2}

Subtract (1/2)^{2} from both sides of the equation.

( sin(pi/3) )^{2} = 1^{2} - (1/2)^{2}

Take the positive sqrt of both sides.

sin(pi/3) = √[1^{2} - (1/2)^{2}]

sin(pi/3) = √[1 - 1/4]

sin(pi/3) = √[3/4]

sin(pi/3) = √3 / 2

hectictar Jan 17, 2018

#5**+2 **

Geez, that was an extremely detailed answer. Thanks, Hectictar! I've learned an easier way to do it though.

First I convert radians to degrees.

\(\frac{\pi}{3}*(\frac{180^o}{\pi})\)

Multiply.

\(\frac{\pi*180^o}{3\pi}\)

The pi cancels.

\(\frac{180^o}{3}=60^o\)

So it is sin(60^{o}), and since I know the values of sin, cos, and tan of 30^{o}, 45^{o}, and 60^{o}, I know that \(sin(60^o)=\frac{\sqrt{3}}{2}\).

AdamTaurus
Jan 18, 2018

#7**+6 **

This method works for anyone, assuming you have five fingers on your hand. This could b**w your mind, literally! This will allow you to know the sin of the following numbers listed on each finger. Keep your hat on, if you have one! Attempt to hold back knocking your socks off!

Lay your hand out like so! Now, let's say you want to know the value of \(\sin60\). Here is our magic formula. \(\sin x=\frac{\sqrt{\text{# of fingers to the left of finger x}}}{2}\). In this case, finger x is 60. The number of fingers to the left is 3, so \(\sin60=\frac{\sqrt{3}}{2}\). Magic, right? I think so!

Wait, don't leave yet! Your fingers can also do the other trigonometric function: cosine! Don't believe me? Well, it is simple, too! Don't fret. Let's find the \(\cos60\). The formula now is \(\cos x=\frac{\sqrt{\text{# of fingers to the right of finger x}}}{2}\). There is 1 finger to the left of finger 60, so \(\cos 60=\frac{\sqrt{1}}{2}=\frac{1}{2}\). Try it out with other values on the hand!

What about tangent? Well, this is where the magic ends, unfortunately. The only thing I can suggest is that \(\tan x=\frac{\sin x}{\cos x}\), which you can get those values using the finger trick already learned! Have fun defeating those trigonometric headaches.

TheXSquaredFactor
Jan 22, 2018

#8**+1 **

Hi TheXSquaredFactor,

That is really cool. I have never seen that trick before!

Melody
Jan 22, 2018

#9**+1 **

I've had to memorize the values of sin, cos, and tan of angles 30^{o}, 45^{o}, and 60^{o}.

AdamTaurus
Jan 23, 2018