milan5baros
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#1
Report Thread starter 11 years ago
#1
i thought cos 165 would be cos 120+cos 45

which would be -1/2+ 1/root2


the mark scheme says -root3+1/2root2


can anyone spot where I have gone wrong here.
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Chewwy
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#2
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i reckon it was when you assumed cosA + cosB = cos(A+B)
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gyrase
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#3
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(Original post by milan5baros)
i thought cos 165 would be cos 120+cos 45

which would be -1/2+ 1/root2


the mark scheme says -root3+1/2root2


can anyone spot where I have gone wrong here.
cos(165) = cos(120+45) = cos120cos45-sin120sin45
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Lukins
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#4
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#4
use the double angle formula
cos(a+b)=cosacosb-sinacosb
that would probably give the right answer, too lazy to check
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Kolya
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#5
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#5
It's easiest to just see that \cos 165 = -\cos 15. (If that isn't obvious, look at the graph of cosine)
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milan5baros
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#6
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can I just ask something...why Do I have to use the double angle formula?

I mean I don't understand when and when not too use it.
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RightSaidJames
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#7
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#7
(Original post by gyrase)
cos(165) = cos(120+45) = cos120cos45-sin120sin45
Yeah, ditto that. You can't separate what's inside the brackets into two cos terms, but you can use the double angle formulae to split things into a cos term and a sin term:
cos(A  \pm B) = cosA cosB \mp sinA sinB
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RightSaidJames
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#8
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(Original post by milan5baros)
can I just ask something...why Do I have to use the double angle formula?

I mean I don't understand when and when not too use it.
Because cos(A \pm B) \neq cosA \pm cosB
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milan5baros
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#9
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#9
when does this formula HAVE to be used because when I do say

sin 15= sin 60- sin 45

if I just normally subtract the answers I get the correct answer
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Chaoslord
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#10
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(Original post by milan5baros)
when does this formula HAVE to be used because when I do say

sin 15= sin 60- sin 45

if I just normally subtract the answers I get the correct answer
thats cause sin15 doesnt = sin60 - sin45
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RightSaidJames
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(Original post by milan5baros)
when does this formula HAVE to be used because when I do say

sin 15= sin 60- sin 45

if I just normally subtract the answers I get the correct answer
sin 15 = \frac{\sqrt{6} - \sqrt{2}}{4} \\

sin 60 = \frac{\sqrt{3}}{2} \\

sin 45 = \frac{\sqrt{2}}{2} \\

sin 60 - sin 45 = \frac{\sqrt{3} - \sqrt{2}}{2} \\

\frac{\sqrt{6} - \sqrt{2}}{4} \neq \frac{\sqrt{3} - \sqrt{2}}{2} \\
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milan5baros
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#12
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do I always use the rules then?
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Dadeyemi
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#13
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O.o

...you cant assume cosA+cosB=cos(A+B)
where did you get that idea from?

anyway:
remember the curve cos(theta)
cos(165)=cos(180-15)= -cos(15)

if you dont know cos(15) then using cos(165) = cos(120+45) = cos120cos45-sin120sin45 is you safest bet.
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Dadeyemi
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#14
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(Original post by milan5baros)
do I always use the rules then?
no sometimes you can just make stuff up.

/sarcasm

yes, that kinda the basic idea of how maths works
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RightSaidJames
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#15
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There's normally a question in GCSE papers that asks you to prove this kind of thing wrong, no idea how you've managed to get this far without knowing this.
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milan5baros
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#16
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well if someone would clear it up properly it would have helped!
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RightSaidJames
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#17
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(Original post by RightSaidJames)
Because cos(A \pm B) \neq cosA \pm cosB
How is that not clearing it up properly?
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chr15chr15
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#18
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#18
wait you do double angle formulae in gcse???
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milan5baros
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#19
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#19
thats just stating a formula.....
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RightSaidJames
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#20
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(Original post by chr15chr15)
wait you do double angle formulae in gcse???
noooooooooo lol... but they might ask you a question where someone says "Rachel says:
cos(this) + cos(that) is the same as cos(this + that), therefore cos(A) + cos(B) must always equal cos(A + B).
Prove rachel is wrong in her assumption"

All you do is choose two numbers where that doesn't work.
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