Ok, I'm having a bad of a time understanding identities for my pre-calc class...all of trig is hard for me to grasp since this is the first time learning any of it (didn't take trig in high school), so basically my question is with subtraction and addition identities. Say I have a problem that says find the exact value of sin 15 degrees and I break it up with sin (x + y) = sinxsiny + cosxcosy (I think that's the right formula?) so it's sin (45 D + 30 D) = sin45sin30 + cos45cos30...ok that's easy enough but I need to get the exact values for each part to find the final solution and I know for instance that sin45 is square root 2/2 (I think anyways) but how would I find that out without already knowing? If it's relevant, I'm using a 84 SE, thanks.
You should know that 45 degrees is the angle in an equilateral right triangle,
Originally Posted by premedtim
and that the sine of an angle in a right triangle is the side opposite the angle
divided by the hypotenuse (that’s the longest side).
In the case of an equilateral right triangle we apply Pythagoras's theorem
to get the length of the longest side as sqrt(2)s, where s is the length
of one of the other sides.
So sin(45) = cos(45) = s/(sqrt(2)s)=1/sqrt(2)=sqrt(2)/2.
You should also know that if you take an equilateral triangle and cut it into
two congruent right triangles you have two 30, 60, 90 degree angle triangles
with sides 1/2, 1, sqrt(3)/2 times the side of the original equilateral triangle.
Which should be enough to allow you to deduce that sin(30)=1/2, and
Also you should be using:
sin(15) = sin(45-30) = sin(45)cos(30) - cos(45)sin(30)
You're expected to know the trig values for 30°, 45°, and 60°.
I'll add diagrams to what Captain Black said.
For the 45° angle, we have "half a square".
+ - - - - - *
: * *
: x * *
: * * 1
: * *
: * 45° *
* * * * * * *
If the sides of the square are of length 1,
. . we use Pythagorus to find the length of the diagonal.
For a 45° angle, the sides of the right triangle are: .
If you memorize that, you can write all the trig values for 45°.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
For the 30° and 60° angles, we have an equilateral triangle.
Draw an altitude, bisecting the vertex angle and the base.
* : *
* : *
1 *30*: *
* : *
* : _ *
* :½√3 *
* : *
* 60* : *
* * * * * * * * * * *
We have a 30-60 right triangle with
Pythagorus says that:
The sides are: .
Multiply by 2: .
. . . . . . . . . opp 30° . opp 60° .hyp
If you memorize that, you can recreate these triangles any time.
2 * *
* * 1
* 30° *
* * * * * * * * *
2 * * _
* 60° *
* * * * *
Ahh ok, thanks guys...so outside 30, 45 and 60 degree angles, it's just repeats of those for say something like 180 degrees?