C3 trig help Watch

kendellex
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#1
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#1
I've tried this question twice and it should be fairly easy but I keep getting it wrong :/

cos2x = 1 - cosx

I originally put all the cosx on one side then factorised by taking out cos. that didn't work so I replaced cos2x with 2cos^2x-1 to form a quadratic which also didn't end in the correct answer :/




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krisshP
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#2
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Cos2x=1-cosx
Let x=45
Cos90=1-cos45
0=1-1/√2
LHS≠RHS
So cos2x≠1-cosx
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m4ths/maths247
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#3
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(Original post by kendellex)
I've tried this question twice and it should be fairly easy but I keep getting it wrong :/

cos2x = 1 - cosx

I originally put all the cosx on one side then factorised by taking out cos. that didn't work so I replaced cos2x with 2cos^2x-1 to form a quadratic which also didn't end in the correct answer :/




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What was your quadratic?
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kendellex
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(Original post by m4ths/maths247)
What was your quadratic?
I thought it was 2y^2 + y - 2 = 0 were y = cosx. It doesn't produce the correct answer though so must be wrong


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m4ths/maths247
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(Original post by krisshP)
Cos2x=1-cosx
Let x=45
Cos90=1-cos45
0=1-1/√2
LHS≠RHS
So cos2x≠1-cosx
I'm not sure what you are trying to do here?
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m4ths/maths247
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(Original post by kendellex)
I thought it was 2y^2 + y - 2 = 0 were y = cosx. It doesn't produce the correct answer though so must be wrong


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Answer or answers?
What is your interval?
Have you solved for x and not cosx?
There are certainly 2 solutions between 0 and 2pi
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tsumannai
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2 cos^2(x) - 1 =1 - cosx

2 cos^2(x) + cosx - 2 =0

cosx= (1
+/− (1-4(2)(-2))^1/2)4 (quadratic equation)
= (1
+/−3)/4
cosx= 1 OR -1/2

Therefore x= 0.54 rad or 0.88 rad

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m4ths/maths247
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(Original post by tsumannai)
2 cos^2(x) - 1 =1 - cosx

2 cos^2(x) + cosx - 2 =0

cosx= (1
+/− (1-4(2)(-2))^1/2)4 (quadratic equation)
= (1
+/−3)/4
cosx= 1 OR -1/2

Therefore x= 0.54 rad or 0.88 rad

The board etiquette is such that full solutions are not posted.
This has not answered the question anyhow....
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krisshP
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(Original post by m4ths/maths247)
I'm not sure what you are trying to do here?
Oh no, sorry, I thought that we were supposed to prove that cos2x is identically equal to 1-cosx, now I realise we're we're solving cos2x=1-cosx after looking at post 7. After all there are many trig questions based on proving.
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