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https://www.box.com/shared/8quza0w4ks#/s/8quza0w4ks/1/84853366/735382824/1

hi can someone explain question 5)b when you have a squred ?solomon A
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raheem94
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Original post by otrivine
https://www.box.com/shared/8quza0w4ks#/s/8quza0w4ks/1/84853366/735382824/1

hi can someone explain question 5)b when you have a squred ?solomon A


6cos⁡2x+sin⁡(2x)=6cos⁡2x+2sin⁡xcos⁡x=2cos⁡x(3cos⁡x+sin⁡x) 6 \cos^2 x + \sin (2x) = 6 \cos^2 x + 2 \sin x \cos x = 2 \cos x ( 3 \cos x + \sin x )

Now use the result of part (a) for 3cos⁡x+sin⁡x 3 \cos x + \sin x
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otrivine
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Original post by raheem94
6cos⁡2x+sin⁡(2x)=6cos⁡2x+2sin⁡xcos⁡x=2cos⁡x(3cos⁡x+sin⁡x) 6 \cos^2 x + \sin (2x) = 6 \cos^2 x + 2 \sin x \cos x = 2 \cos x ( 3 \cos x + \sin x )

Now use the result of part (a) for 3cos⁡x+sin⁡x 3 \cos x + \sin x


got it thanks :wink:
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otrivine
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Original post by raheem94
6cos⁡2x+sin⁡(2x)=6cos⁡2x+2sin⁡xcos⁡x=2cos⁡x(3cos⁡x+sin⁡x) 6 \cos^2 x + \sin (2x) = 6 \cos^2 x + 2 \sin x \cos x = 2 \cos x ( 3 \cos x + \sin x )

Now use the result of part (a) for 3cos⁡x+sin⁡x 3 \cos x + \sin x


and one thing raheem http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/Maths/C3_Jan_2007_Paper.pdf
question 1)a) how do u tackle the question
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raheem94
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Original post by otrivine


sin⁡(3θ)=sin⁡(2θ+θ)=sin⁡(2θ)cos⁡(θ)+cos⁡(2θ)sin⁡(θ) \sin (3 \theta ) = \sin ( 2 \theta + \theta ) = \sin (2 \theta ) \cos ( \theta ) + \cos (2 \theta) \sin ( \theta )

Write sin⁡(2θ)=2sin⁡θcos⁡θ and cos⁡(2θ)=1−2sin⁡2(θ) \sin ( 2 \theta ) = 2 \sin \theta \cos \theta \ and \ \cos ( 2 \theta ) = 1 - 2 \sin^2 ( \theta )
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otrivine
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Original post by raheem94
sin⁡(3θ)=sin⁡(2θ+θ)=sin⁡(2θ)cos⁡(θ)+cos⁡(2θ)sin⁡(θ) \sin (3 \theta ) = \sin ( 2 \theta + \theta ) = \sin (2 \theta ) \cos ( \theta ) + \cos (2 \theta) \sin ( \theta )

Write sin⁡(2θ)=2sin⁡θcos⁡θ and cos⁡(2θ)=1−2sin⁡2(θ) \sin ( 2 \theta ) = 2 \sin \theta \cos \theta \ and \ \cos ( 2 \theta ) = 1 - 2 \sin^2 ( \theta )


thanks and last one http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/January%202009/6665_01_que_20090115.pdf
6)a)ii i did the first part now for the second part do i use the equation and rearrange so i sub the sin3X?
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raheem94
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Original post by otrivine
thanks and last one http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/January%202009/6665_01_que_20090115.pdf
6)a)ii i did the first part now for the second part do i use the equation and rearrange so i sub the sin3X?


Multiply both sides of the equation given in part (i) by −2 -2 , and you should be able to see what to do next.
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otrivine
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Original post by raheem94
Multiply both sides of the equation given in part (i) by −2 -2 , and you should be able to see what to do next.


i am not getting ur point?
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raheem94
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Original post by otrivine
i am not getting ur point?


8sin⁡3θ−6sin⁡θ+1=0  ⟹  −2(−4sin⁡3θ+3sin⁡θ)+1=0  ⟹  −2sin⁡(3θ)+1=0 8 \sin^3 \theta -6 \sin \theta +1 = 0 \implies -2( -4 \sin^3 \theta +3 \sin \theta ) +1 = 0 \\ \implies -2 \sin ( 3 \theta ) +1 =0

Does it make sense?
Reply 9
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otrivine
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Original post by raheem94
8sin⁡3θ−6sin⁡θ+1=0  ⟹  −2(−4sin⁡3θ+3sin⁡θ)+1=0  ⟹  −2sin⁡(3θ)+1=0 8 \sin^3 \theta -6 \sin \theta +1 = 0 \implies -2( -4 \sin^3 \theta +3 \sin \theta ) +1 = 0 \\ \implies -2 \sin ( 3 \theta ) +1 =0

Does it make sense?


yep yep thanks

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