Question

Hoh to find exact value of sin(3) (degress) ?

Answer 1

Can you use a calculator or not??

Answer 2

not use calcul, but use trignometry concept..

Answer 3

can you find the sine of 12°, 15°

Answer 4

no, i can't..how to get it, please give a example

Answer 5

\[\sin(u\pm v)=\sin (u) \cos(v)\pm \cos(u)\sin(v)\]

Answer 6

\[\sin(3°)=\sin(15°- 12°)=\sin (15°) \cos(12°)- \cos(15°)\sin(12°)\]

Answer 7

yea, i see..so, how to calculate sin15 and sin12 ??

Answer 8

Uncle Rocks I think we have to go with 18 and 15..
It is very difficult to calculate sin12 and cos12 I think..


Or you have a method to find sin12???

Answer 9

i havent worked this through, i was just thinking that 12/360=1/30,
18 might be easier, as 18/360=1/20

Answer 10

I have the value of sin and cos 15 and 18, that is why I am thinking to use that..

Answer 11

go for it waterineyes !

Answer 12

Should I give the values directly or I have to derive them fully here how they come ??

Answer 13

derive them from stuff we should know

Answer 14

Firstly, Let us find sin18..

let x = 18

5x = 90

2x + 3x = 90

2x = 90 - 3x

take sin both sides,

sin2x = sin(90-3x)

as sin(90-y) = cosy

So, sin2x = cos3x

Here we have to use Identities;

\[\sin2x = 2sinx.cosx\]
\[\cos3x = 4\cos^3x - 3cosx\]

So,
\[2sinx.cosx = 4\cos^3x - 3cosx\]

Bringing on one side and taking cosx common,

\[cosx(2sinx - 4\cos^2x + 3) = 0\]
\[2sinx - 4(1-\sin^2x) + 3 = 0\]

\[4\sin^2x + 2sinx - 1 = 0\]

Solving this by quadratic formula we get:

\[sinx = \frac{-2 \pm \sqrt{20}}{8}\]

or,

\[sinx = \frac{-1 \pm \sqrt{5}}{4}\]

One point to be noted here that the x = 18 lies in 1st quadrant so value of sinx will be positive here,.

So,

\[sinx = \frac{-1 +\sqrt{5}}{4} = \frac{\sqrt{5} - 1}{4}\]

Answer 15

Now we know that:

\[cosx = \sqrt{1 - \sin^2x}\]

By this we get the value of:

\[\cos18 = \frac{\sqrt{10 + 2\sqrt{5}}}{4}\]

Answer 16

It is getting more complex now..

Answer 17

values of sin15 and cos15 are easier to find though..

\[\sin(45-30) = \sin45\cos30 - \sin30 \cos45\]

\[\cos(45 - 30) = \cos45. \cos30 + \sin45.\sin30\]

Answer 18

After solving we get,

\[\sin15 = \frac{\sqrt{3} - 1}{2\sqrt{2}}\]
\[\cos15 = \frac{\sqrt{3} + 1}{2\sqrt{2}}\]

Answer 19

Now, lots of work is remaining to find sin3,

\[\sin(18-15) = \sin18.\cos15 - \sin15.\cos18\]

Answer 20

waawwww, very very very long time to finished... amzing,,, i think i will giveus if i do the problem, but i like your solution,, thank you for all...
BTW, from sin3, we can find the value sin 1 and sin 2 isn't it?? hehe

Answer 21

Ha ha ha..
I will definitely use a calculator for this..
Ha ha..

Answer 22

do can you find \(\sin1°, \sin2°\) @tanjung ?

Answer 23

how can you*

Answer 24

hahahaha,,, very simple
i hope i can do it

Answer 25

I think he or she is joking..
Isn't it?? @tanjung

Answer 26

sinx= x- x^3/3!+ x^5/5! - x^7/ 7!+ ...... so on
put 3 and get the value

Answer 27

what is it, Taylor series??

Answer 28

using complex numbers ,,you can get this expression... cosx= x- x^2/2! +x^4/4!.... so on