is it required to memorize trig ratios such as sin pi/3 = sqrt3 / 2?
will such ratios be given in examinations?
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.Using Pythagorus, the diagonal is $\displaystyle \sqrt{2}.$Code:* - - - - - - * | * | | _ * | | √2 * | 1 | * | | * | | * 45° | * - - - - - - * 1
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.Using Pythagorus, the altitude is $\displaystyle \sqrt{3}.$Code:* /|\ / | \ / | \ 2 /30°| _ \ 2 / |√3 \ / | \ / 60° | \ * - - - * - - - * 1 1
Then: . $\displaystyle \begin{Bmatrix}\sin30^o &=& \dfrac{opp}{hyp} &=& \dfrac{1}{2} \\ \\[-4mm] \cos30^o &=& \dfrac{adj}{hyp} &=& \dfrac{\sqrt{3}}{2} \\ \\[-4mm] \tan30^o &=& \dfrac{opp}{adj} &=& \dfrac{1}{\sqrt{3}} \end{Bmatrix}$ . . . $\displaystyle \begin{Bmatrix}\sin60^o &=& \dfrac{opp}{hyp} &=& \dfrac{\sqrt{3}}{2} \\ \\[-4mm] \cos60^o &=& \dfrac{adj}{hyp} &=& \dfrac{1}{2} \\ \\[-4mm] \tan60^o &=& \dfrac{opp}{adj} &=& \sqrt{3} \end{Bmatrix}$
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
Divide each by 4Code:0 30 45 60 90 0 1 2 3 4
Take the square roots for each..Code:0 30 45 60 90 0 1 2 3 4 -- -- -- -- -- 4 4 4 4 4
We have the sine ratio.. For co-sine ratio take sin 90 as cos 0 and you are done..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