Toolbox question

[MBBS]

Exercise 1B Question 4A, I can do this question but will solve one B in the process, so I'm confused why it says HENCE find exact values for sin(15) and cos (15). Is there a way to solve this questions without finding sin (15) and cos (15) in the process.


Thanks guys!

[Chavi]

First you solve individual angles and sides.
Hence, use these answer to deduce the exact angles.

eg. Sin(15) = sin(z) = y/x

[MBBS]

How do you find y without using sin(15)?

[Chavi]

Pythagoras

[BubbleWrapMan]

I'll just show you how I did it, it might help.

Since the side of the 45-degree triangle is 1, then the bottom length is 1, and a = root2 (Pythagoras)

Since the top angle is 30 degrees, the bottom angle (45 + z) is 60 degrees. Therefore, z is 15 degrees.

If we let the vertical length = v, then we can solve for v using tangent.

tan(60) = O/A = v/1
root3 = v/1
v = root3

The hypotenuse of the triangle at the top is therefore root3 - 1

Using cosine to obtain w:

cos(30) = A/H = w/root3 - 1
root3/2 = w/root3 - 1
w = (3 - root3)/2

And sine to obtain y:

sin(30) = O/H = y/root3 - 1
1/2 = y/root3 - 1
y = (root3 - 1)/2

Using Pythagoras, we can find x + w
(root3)^2 + 1^2 = (x + w)^2
x + w = 2
x = 2 - (3 - root3)/2 = (1 + root3)/2

For part b:

sin(15) = O/H = y/a = [(root3 - 1)/2]/root2
sin(15) = (root3 - 1)/2 x root2/2
sin(15) = (root6 - root2)/4

cos(15) = A/H = x/a = [(1 + root3)/2]/root2
cos(15) = (1 + root3)/2 x root2/2
cos(15) = (root2 + root6)/4

tan(15) = sin(15)/cos(15) = [(root6 - root2)/4]/[(root2 + root6)/4]
tan(15) = (root6 - root2)/(root2 + root6) x (root2 - root6)/(root2 - root6) <---- rationalise the denominator
tan(15) = (2root3 - 6 - 2 + 2root3)/2 - 2root3 + 2root3 - 6
tan(15) = (4root3 - /-4 = 2 - root3

Part c:

Since 15 and 75 are complementary angles

sin(75) = cos(15) = (root2 + root6)/4
cos(75) = sin(15) = (root6 - root2)/4
tan(75) = cot(15) = 1/(2 - root3)
Rationalise the denominator to tan(75) = 2 + root3

[MBBS]

WOW, thanks mate. I feel like an idiot for not looking at the larger triangle!
Thank you so much for your help!

[BubbleWrapMan]

It is a bitch, yeah. No worries.

[MBBS]

Could anybody show me how to work out Questions 3,5,7,9 from Exercise 1C from the toolbox?

I keep getting the wrong answer :S


NB: Scratch that guys, calc was messed up and upon resetting it, all answers are coming up correctly.

[Greatness]

huh... im cas is playing up, the answers that i get are wrong with taht but correct with a scientific :/

[MBBS]

Yeh, I spent legit around about an 1 hour looking at myself thinking I'm going to get like 29 for my study score in Specialist and less in Methods, then I realised my CAS was stuffed for the time being.

[Greatness]

what was wrong with yours?? because i just got the new OS and ive been gettting weird answers for a lot of things :/ ive tried changing radians -> degrees and vice versa :S

[MBBS]

Same thing, I just bought a new calc, updated to new software, I did it on my old one with  cracked screen and the correct answers are coming up!

Just cbf explaining it before so said I reset it.

[Greatness]

LOL wtf... i reset my calc (ti nspire) and  now it's giving me another completely different answer.... so lame, sooooo lame


EDIT: Dont worry i reset calc and changed to degrees, giving me right answers now thanks
pretty dodgy that we had to reset the calcs :/

[cohen]

Couldn't we use Sin(45-30)
and then do:
sin(45)cos(30)-cos(45)sin(30)?

[Andiio]

Couldn't we use Sin(45-30)
and then do:
sin(45)cos(30)-cos(45)sin(30)?

Simply using sin(z) = sin(15) is much easier and less complicated; it is only simple trig anyways