You are Here: Home >< Maths

# Edexcel C2 Trig help please

Announcements Posted on
TSR's new app is coming! Sign up here to try it first >> 17-10-2016
1. I have been doing: Trigonometry CORE 2 Chapter 10 Exercise 2B question 2 and I am confused as to how to get to some of the solutions.

I have been using the rules :
Sin x = k
First solution is calculator solution: inverse sine (k)= α
Second solution is (180° – α)
Other solutions are found by adding or subtracting multiples of 360°

Cos x = k
First solution is calculator solution: inverse cosine(k) = α
Second solution is (360° – α)
Other solutions are found by adding or subtracting multiples of 360°

Tan x = k
First solution is calculator solution: inverse tangent (k) = α
Second solution is (180° + α)
Other solutions are found by adding or subtracting multiples of 360°

2a)sin x° = -root 3/2 Interval-180< x<540
Inverse sine (-root3/2) =-60°
As sine x is negative it is in the 3rd and 4th quadrants
180° - (-60°) = 240°360 + (-60°) = 300
How do you get to this?

2b) 2sin x° = -0.3 Interval -180<x<180
Sin x° = -0.15°
Inverse sine (-0.15) = -8.63°to 3 s.f.
As sine x is negative it is in the 3rd and 4th quadrants
360° - 8.63 = 351.37 I then took away 180° so that it was within the interval, giving 171.37°
However, there is an additional solution of -171.37° but why?

2c) cos x° = -0.809 Interval -180<x<180
Inverse cosine (-0.809) =144° to 3 s.f.
As cos x° is negative it is in the 2nd and 3rd quadrants
360° - 36° = 216° I then subtracted 360° and got -144°

2d) cos x° = 0.84 Interval -360<x<0
Inverse cosine (0.84) = 32.9°
Cos x is positive so is in the 1st and 3rd quadrants
360° - 32.9° = 327.1° I then subtracted 360° to get -32.9°
However, there is another solution of -327° but I am not sure why?

Any help would be warmly welcomed and very much appreciated, thanks in advance
2. (Original post by IdentityWithheld)
I have been doing: Trigonometry CORE 2 Chapter 10 Exercise 2B question 2 and I am confused as to how to get to some of the solutions.

I have been using the rules :
Sin x = k
First solution is calculator solution: inverse sine (k)= α
Second solution is (180° – α)
Other solutions are found by adding or subtracting multiples of 360°

Cos x = k
First solution is calculator solution: inverse cosine(k) = α
Second solution is (360° – α)
Other solutions are found by adding or subtracting multiples of 360°
Sounds good. The reason you can add or subtract multiples of 360 degrees is because those two functions are 360-periodic functions, i.e: they repeat themselves every 360 degrees.

Tan x = k
First solution is calculator solution: inverse tangent (k) = α
Second solution is (180° + α)
Other solutions are found by adding or subtracting multiples of 360°
This bit, not so much. Tangent is a 180-degree periodic function, you you need to add and subtract multiples of 180 to find other solutions as it repeats every 180 degrees, not 360 degrees like sine and cosine.

2a)sin x° = -root 3/2 Interval-180< x<540
Inverse sine (-root3/2) =-60°
As sine x is negative it is in the 3rd and 4th quadrants
180° - (-60°) = 240°360 + (-60°) = 300
How do you get to this?

inverse sine (+whatever value). The value inside the inverse sine needs to be positive.

So you do inverse sine (+sqrt(3)/2) = 60.

Then solutions are:
180 + 60 = 240 (third quadrant is +180)
360 - 60 = 300 (fourth quadrant is 360-)
240 - 360 = -120 (add/subtract multiples of 360)
300 - 360 = -60 (add/subtract multiples of 360)

2b) 2sin x° = -0.3 Interval -180<x<180
Sin x° = -0.15°
Inverse sine (-0.15) = -8.63°to 3 s.f.
As sine x is negative it is in the 3rd and 4th quadrants
360° - 8.63 = 351.37 I then took away 180° so that it was within the interval, giving 171.37°
However, there is an additional solution of -171.37° but why?
Same mistake, you need to always do inverse sine (positive).

So inverse sine (+0.15) = 8.63.

Then:
360 - 8.63 = 351.37 (fourth quadrant is 360-)
180 + 8.63 = 188.63 (third quadrant is 180 +)
351.37 - 360 = -8.63 (add/subtract multiples)
188.63 - 360 = -171.37 (add/subtract multiples)

2c) cos x° = -0.809 Interval -180<x<180
Inverse cosine (-0.809) =144° to 3 s.f.
As cos x° is negative it is in the 2nd and 3rd quadrants
360° - 36° = 216° I then subtracted 360° and got -144°
Same mistake here, you need to do inverse cosine (positive).

2d) cos x° = 0.84 Interval -360<x<0
Inverse cosine (0.84) = 32.9°
Cos x is positive so is in the 1st and 3rd quadrants
360° - 32.9° = 327.1° I then subtracted 360° to get -32.9°
However, there is another solution of -327° but I am not sure why?
Cos x is positive in the 1st and fourth quadrants.

Any help would be warmly welcomed and very much appreciated, thanks in advance
3. (Original post by IdentityWithheld)
Any help would be warmly welcomed and very much appreciated, thanks in advance
Several things you need to learn:

1. Using the method that you're using, you always need to do inverse trig (positive)

2. Learn what the quadrants are:

3. Know what you need to do for each quadrant.

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: May 16, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### How does exam reform affect you?

From GCSE to A level, it's all changing

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams