Question

Help here please:

Simplify the following expressions:

a. cos(180° + x) + cos(180° - x)

b. cos(126°)cos(36°) + sin(126°)sin(36°)

Answer 1

Can you help me through a? I will solve b myself .

Answer 2

you now cos C + cos D formula??

Answer 3

(a) cos(180)cos(x) - sin(180)sin(x) + cos(180)cos(x) + sin(180)sin(x)

cos(180) = -1

so it simplifies to -cos(x) - cos(x) = -2cos(x)

Answer 4

cos c =cos d?

Answer 5

http://answers.yahoo.com/question/index?qid=20090209193628AAUvpWz

you can use this formula formula or ... use above formula given by campbell_st

Answer 6

(b) will give cos(126 - 36) = cos(90) = 0

Answer 7

above **method

Answer 8

cos(180° + x) = cox x in quadrant III =?
cos(180° - x) = cox x in quadrant II =?
Can you try these 2 first?

Answer 9

so the first one is in Q3 and the second one in Q2, is it right?

Answer 10

yes, but what would it be?
Sorry, typo : cox x -> cos x :S

Answer 11

For example
cos(180° + x) = cos x in quadrant III = -cos x

Answer 12

yeah, so both of them are -cos x right?

Answer 13

Yes~
so, what would you get for cos(180° + x) + cos(180° - x) ?

Answer 14

-2cos x

Answer 15

yup~ first question is done :)

Answer 16

I have a question, the formula that campbell used is the sum difference formula right?

Answer 17

it's compound angle formula

Answer 18

that is weird, I didnt learn that one, I only learned the 6 sum difference formulas, and the double angle formula

Answer 19

question b is the expanded difference for cos... I recognised it straight away

rewrote the question as the difference and got the solution.... its a working backwards question

Answer 20

ok, let me see if I can manage question b

Answer 21

is questino be a first pythagorian identity question?

Answer 22

Try to match this:
cos (A-B) = cosAcosB + sinAsinB

Answer 23

pythagorean*

Answer 24

ok, one sec, I think i figured this out