# trig ratios

• Jun 25th 2009, 04:07 AM
furor celtica
trig ratios
is it required to memorize trig ratios such as sin pi/3 = sqrt3 / 2?
will such ratios be given in examinations?
• Jun 25th 2009, 04:26 AM
Khonics89
I'd doubt any exams will require you to memorise these ratios, what I can say is just make sure you are able to re-produce them in exam conditions rather than memorising them individually :)
• Jun 25th 2009, 04:38 AM
furor celtica
are you sure of this, and what exactly is 'reproduce them' supposed to mean? i can get up to a certain point but one has to know the exact value where calculator results are not appropriate.
• Jun 25th 2009, 04:44 AM
Soroban
Hello, furor celtica!

The bad news: Yes, we're expected to know these values.

The good news: They're not that hard to learn.

For 45°, draw a square (side 1), draw a diagonal.
Code:

      * - - - - - - *       |          * |       |    _  *  |       |    √2 *    | 1       |    *      |       |  *        |       | * 45°      |       * - - - - - - *             1
Using Pythagorus, the diagonal is $\displaystyle \sqrt{2}.$

And we have: .$\displaystyle \begin{Bmatrix}\sin45^o &=& \dfrac{opp}{hyp} &=& \dfrac{1}{\sqrt{2}} \\ \\[-4mm] \cos45^o &=& \dfrac{adj}{hyp} &=& \dfrac{1}{\sqrt{2}} \\ \\[-4mm] \tan45^o &=& \dfrac{opp}{adj} &=& 1 \end{Bmatrix}$

For 30° and 60°, draw an equilateral triangle (side 2) and an alitutude.
Code:

              *             /|\             / | \           /  |  \         2 /30°| _ \ 2         /    |√3  \         /    |    \       / 60°  |      \       * - - - * - - - *           1      1
Using Pythagorus, the altitude is $\displaystyle \sqrt{3}.$

Then: . $\displaystyle \begin{Bmatrix}\sin30^o &=& \dfrac{opp}{hyp} &=& \dfrac{1}{2} \\ \\[-4mm] \cos30^o &=& \dfrac{adj}{hyp} &=& \dfrac{\sqrt{3}}{2} \\ \\[-4mm] \tan30^o &=& \dfrac{opp}{adj} &=& \dfrac{1}{\sqrt{3}} \end{Bmatrix}$ . . . $\displaystyle \begin{Bmatrix}\sin60^o &=& \dfrac{opp}{hyp} &=& \dfrac{\sqrt{3}}{2} \\ \\[-4mm] \cos60^o &=& \dfrac{adj}{hyp} &=& \dfrac{1}{2} \\ \\[-4mm] \tan60^o &=& \dfrac{opp}{adj} &=& \sqrt{3} \end{Bmatrix}$

• Jun 25th 2009, 08:59 AM
great_math
A trick which i used to remember ratios in my earlier classes(when i was new to trigonometry..)

For sine ratio

First, write 0 30 45 60 90 which are the basic ratios.

Write from 0 to 4 under each degree

Code:

0  30  45  60  90 0  1    2    3  4
Divide each by 4

Code:

0  30  45  60  90 0  1    2    3  4    -- --  --  --  -- 4  4    4    4  4
Take the square roots for each..

Code:

0  30  45  60  90 0  1    2    3  4    -- --  --  --  -- 4  4    4    4  4 0  1    1  $\displaystyle \sqrt3$  1     --  --  --  --     2  $\displaystyle \sqrt2$    2  1
We have the sine ratio.. For co-sine ratio take sin 90 as cos 0 and you are done.. :)