• YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page

Trigonometric results help, am I using the correct method?? (1 Viewer)

BabzBrah

Member
Joined
May 12, 2012
Messages
38
Gender
Male
HSC
N/A
Hey guys. At the moment i'm trying to teach my self some trigonometric functions. It's just that there is some extension 1 maths in the 2unit section and I was wondering if using it is the correct method or if I should be using another method. Examples of the questions:

Simplify

1) sin(pi-x)

For this question this is my current method of solving it.

sin(pi)cos(x)-cos(pi)sin(x) = sin(180)cos(x)-cos(180)sin(x)
= 0.(cos(x)) - (-1.(sin(x))
= -(-sin(x))
= sin(x)

I understand how to do it and I do the same type of process using the tana and cos rules when needed, it's just that i'm wondering if this is right since in my text book this is a advanced topic and they i'm using extension solving? However in the examples they use a similar type of method
 

Nws m8

Banned
Joined
Oct 21, 2012
Messages
494
Gender
Male
HSC
2012
Uni Grad
2018
ASTC ... Sin (pi - x) is the second quadrant , therefore it equals sin x.

Your method is correct but ^^^^ is shortcut
 

Nws m8

Banned
Joined
Oct 21, 2012
Messages
494
Gender
Male
HSC
2012
Uni Grad
2018
Also the method you are using is 3u, the one I showed you is 2u :)
 

BabzBrah

Member
Joined
May 12, 2012
Messages
38
Gender
Male
HSC
N/A
ASTC ... Sin (pi - x) is the second quadrant , therefore it equals sin x.

Your method is correct but ^^^^ is shortcut
Thanks for the feedback. The only thing is what about questions like

Expand and simplify

c) tan(pi/4 - x) is there any method to do these with 2u. I can solve it just using 3u:

tan(45) - tan(x) / 1 + tan(45).tan(x) = 1 - tan(x) / 1 + tan(x) which is the answer.

So basically im confused because most of the questions in this exercise are using 3u method and this question said "expand and simplify" implying 3u?

Also just noticed one question 2pheta :S
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top