## anonymous one year ago Sine, Cosine, and Tangent of 45 degrees?

1. UnkleRhaukus

|dw:1439360650195:dw|

2. UnkleRhaukus

|dw:1439360684499:dw|

3. UnkleRhaukus

we need to set the lengths of the sides of the triangle for simplicity we can set the adjacent side to be 1

4. UnkleRhaukus

what will the other sides of the triangle be if the adjacent side is 1?

5. anonymous

Wait what do you mean?

6. anonymous

This is actually a timed test and I'm super tired

7. UnkleRhaukus

|dw:1439361178381:dw|

8. UnkleRhaukus

(notice it is an isosceles triangle )

9. anonymous

I forgot how to do this and I don't have a calculator with me. Is there another way?

10. UnkleRhaukus

this way doesn't need a calculator

11. UnkleRhaukus

[isosceles triangle have a pair of sides of equal length]

12. UnkleRhaukus

[the sides of equal length, will be opposite to the angles of equal measure]

13. anonymous

Ohhhhhhh so 1

14. UnkleRhaukus

yes

15. UnkleRhaukus

|dw:1439361833342:dw|

16. UnkleRhaukus

now $\tan\theta = \frac{\text{opp}}{\text{adj}}$

17. UnkleRhaukus

so tan 45° = / =

18. UnkleRhaukus

What do you get?

19. anonymous

so confused but 1/ over something, I'm not sure what the hypotenuse is

20. UnkleRhaukus

we don't need the hypotenuse for tan (we will need it for sine and cosine)

21. anonymous

22. anonymous

sorry

23. anonymous

1/1

24. UnkleRhaukus

Great! which simplifies to 1 So we have tan 45° = 1

25. UnkleRhaukus

now we need to work out the length of that hypotenuse. we have a right angled triangle, so we can use Pythagorus a^2 + b^2 = c^2 where a and b are the adj and opp, and c is the hypo

26. UnkleRhaukus

so hypo = √( adj^2 + opp^2 )

27. UnkleRhaukus

plug in adj = 1, opp = 1

28. UsukiDoll

that's one of the standard triangles. |dw:1439362733572:dw|

29. UnkleRhaukus

why remember, when you can work it out

30. UsukiDoll

$\tan 45 = 1 , \sin 45 = \frac{1}{\sqrt{2}}, \cos 45 = \frac{1}{\sqrt{2}}$

31. UnkleRhaukus

why bother to learn pythag. and triangle geometry ?

32. anonymous

Thank you both :)