the sin of 150 degrees

Z. used the Rad mode. Come on Z, quit goofing off! Do the reverse quadratic for x= 9 and x=15. You said you don’t have time, but you had time to give the wrong answer for this very easy one here. I know Rosala wouldn’t have made that mistake.

CPhill’s method is very cool!

Jedithious may also be interested in this. It’s the Taylor series form. The signs alternate and the exponent and factorial are odd and the same. Convert the degrees to rads and plug into x.

This is how a calculator computes the value.

$$\sin x = x- \frac{x^3}{3!}+ \frac{x^5}{5!}- \frac{x^7}{7!}...$$

Ironically, you might need a calculator to do it. This is doable by pencil and paper at the cost of writer's cramp. You will need to do about 7 steps for accuracy to about 0.3 of a degree over 360 degrees.

 7UP

Isnt it 0.71 with a calculator?

Note Jedithious, that 150 degrees is just a 30 degree reference angle in the second quadrant. And the sine of 30 degrees is just 1/2. And since the sine is positive in the second quadrant, then the sin of 150 degrees is just 1/2.

Does that make sense??

Hi Jedithious,

It's nice to see you back again,  

Sine 150°

First this is the second quadrant, sine is positive in the second quad.

Acute angle equivalent is 180-150=30degrees   so

$$sin 150^0=+sin30^0$$

Now some people just memorise these (30, 60, 45 degrees) ratios but I do not.

For 30 and 60 degrees

I think of an equilateral triangle of side length 2 units.  (All angles are 60 degrees)

I then drop a perpendicular down the middle to cut it in half.  The top angle is also cut in half so it becomes 30 degrees.

Now you can read straight off the triangle that  sin30°= 1/2

This might seem slow but I draw this in a about 1 second.  And I couldnot just memorise all the 60 and 30 degrees ratios!

Zegroes got the wrong answer.  I am not sure how he managed that but he is correct in that you could just put sin150 degrees into any calculator and get 0.5 immediately.

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{150}}^\circ\right)} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}} = {\mathtt{0.5}}$$

Greetings and felicitations,

CPhil: Yes, indeed it does. Thanks .

Melody: Hello, and same to you! Your explanation was precisely what I was looking for. Thank you! 

SevenUP: Hi, 

I don't know anything about Taylor series but it does in fact look fascinating.  I may have to look into it.