## farah2318 3 years ago why is 9sin(pi/3)= 9sqrt3/2 and 3tan(pi/4)=3 and 2 sec(pi/3)=4 and 5 cos(pi/4)= 5sqrt2 how would i figure these out?

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1. timo86m

try using the unit circle

2. farah2318

how would i memorize all of that

3. NotTim

Continuously draw the unit circle?

4. farah2318

lol

5. NotTim

That's how it works for martial arts, biology and combat situations :|

6. farah2318

do they have unit circle for tan and sec?

7. NotTim

eh...leemme check.

8. NotTim

here's what it says in my book

9. NotTim

|dw:1347081765068:dw| That suffice?

10. farah2318

lol wth

11. hartnn

i would advice you to atleast remember these 3 values : $$sin (\pi/6)=0.5$$ or sin 30 =0.5 $$sin (\pi/3)=\sqrt{3}/2$$ or sin 60 =$$\sqrt{3}/2$$ $$sin (\pi/4)=1/\sqrt{2}$$ or sin 45 =$$1/\sqrt{2}$$ then you can easily derive any ratio with angles 30,45,60 degrees, using standard formulas like sin x= cos (90-x) or cot x =cos x / sin x say cos 60 = sin (90-60)= sin 30 = 0.5 tan 45 = sin 45 / cos 45 = sin 45 / sin (90-45) = sin 45/sin 45 =1

12. farah2318

thank you!

13. hartnn

from your question 2 sec $$\pi/3$$= 2/(cos$$\pi/3$$ )= 2/(sin ($$\pi/2-\pi/3))$$=2/sin $$\pi/6$$=2/0.5=2*2=4 5 cos (pi/4)=5 sin (pi/2-pi/4)= 5 sin (pi/4) = 5/sqrt 2

14. hartnn

welcome :)