Identities

• Nov 8th 2006, 10:41 PM
premedtim
Identities
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.
• Nov 8th 2006, 10:54 PM
CaptainBlack
Quote:

Originally Posted by premedtim
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,
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
that cos(30)=sqrt(3)/2.

Also you should be using:

sin(15) = sin(45-30) = sin(45)cos(30) - cos(45)sin(30)

RonL
• Nov 9th 2006, 04:02 AM
Soroban
Hello, premedtim!

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".
Code:

      + - - - - - *       :        * *       :    x  *  *       :    *    * 1       :  *      *       : * 45°    *       * * * * * * *             1

If the sides of the square are of length 1,
. . we use Pythagorus to find the length of the diagonal.
Hence:. $x^2\,=\,1^2+1^2\quad\Rightarrow\quad x = \sqrt{2}$

For a 45° angle, the sides of the right triangle are: . $\{1,\,1,\,\sqrt{2}\}$

If you memorize that, you can write all the trig values for 45°.

. . $\begin{array}{cccc}\sin45^o \:= \\ \cos45^o\:= \\ \tan45^o\:= \\ \vdots \end{array}
\begin{array}{cccc}\frac{1}{\sqrt{2}}\:= \\ \frac{1}{\sqrt{2}}\:= \\ \frac{1}{1} \:\:=\\ \vdots\end{array}
\begin{array}{cccc}\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} \\ 1 \\ \vdots\end{array}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

For the 30° and 60° angles, we have an equilateral triangle.
Draw an altitude, bisecting the vertex angle and the base.
Code:

                *               *:*               * : *             *  :  *           1 *30*:  *           *    :    *           *    :  _  *         *      :½√3  *         *      :      *       * 60*    :        *       * * * * * * * * * * *           ½

We have a 30-60 right triangle with $base \,= \,\frac{1}{2},\:hyp\,=\,1$
Pythagorus says that: $height\,=\,\frac{\sqrt{3}}{2}$

The sides are: . $\frac{1}{2},\:\frac{\sqrt{3}}{2},\:1$

Multiply by 2: . $\underbrace{1}_\uparrow\:\underbrace{\sqrt{3}}_\up arrow \:\underbrace{2}_\uparrow$
. . . . . . . . .
opp 30° . opp 60° .hyp

If you memorize that, you can recreate these triangles any time.
Code:

                      *             2    *  *               *      * 1           * 30°      *       * * * * * * * * *               √3               *             **             * *         2 *  * _           *  *√3         *    *         *    *       * 60°  *       * * * * *           1
• Nov 9th 2006, 04:07 PM
premedtim
Ahh ok, thanks guys...so outside 30, 45 and 60 degree angles, it's just repeats of those for say something like 180 degrees?