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## Multiple Angle Formula

by: MaKena Betler

24

0

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# Multiple Angle Formula MTH 151 Cr.4

MaKena Betler
UW - L
Precalculus
No professor available

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Chapter 7 Section 4: Multiple Angle Formula
COURSE
Precalculus
PROF.
No professor available
TYPE
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PAGES
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KARMA
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## Popular in Mathematics (M)

This 0 page Class Notes was uploaded by MaKena Betler on Wednesday November 4, 2015. The Class Notes belongs to MTH 151 Cr.4 at University of Wisconsin - La Crosse taught by a professor in Fall 2015. Since its upload, it has received 24 views. For similar materials see Precalculus in Mathematics (M) at University of Wisconsin - La Crosse.

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Date Created: 11/04/15
Chapter 7 Section 4 Multiple Angle Formula 0 Double Angle Formula 0 sin2xsinxx 39 Zsinxcosxcosxsinx 39 2sinxcosx 0 cos2xcosxx 39 Loosxcosx sinxsinx acoszx Sinzx Here is a second equation to replace cos 2x 39 1 sin2x sin2x21 2sin2x And a third way 39 cos2xcost 1 coszx 39 Ecoszx 1coszx 39 2coszx 1 0 tan2xtanxx x0 1 tanxtan Z tanxtan x 6 a 2tanx o 1 tan2x Examples 0 sinx x isEQuad II F ind all trig values ofcx 39 sin2x2sinxcosx 3 5 4 5 24 I 25 4 lt Negative because on is in 39 cos2x1 2sin2x 2 1 23 21 22 7 5 25 25 25 24 sin oc 25 24 25 600 24 39 tanx x cosx 7 25 7175 7 25 I sin2x2 254 I cos2x 5 I tan2x27 4 I csc20 2 5 I sec2x725 I cot2x27 4 0 Verify or Disprove csc2xcscxsecx 1 1 sin x 23inxcos We 1 l 1 23inxcos W T 23inxcos We ve e 0 Verify or Disprove cos4x sin4xcos 2 x 2 o 2 2 o 2 2 2 cos x s1n xcos xs1n x cos x s1n x cos x sin x1 2 2 2 2 cos x s1n x cos x s1n x ve e 0 Verify or Disprove 0 Verify or Disprove ZCSC P2X sin x cos x cos x sin x 2 2sin xicos W 1 sinx cos ve e 1 sin2xcos f2x 1 in 17 Y1 nns P7 Y1 12SinXCOSX2COSZX 1 12 Sinxcosx 1 1 sinzx 2 Sinxcosx 2 COS x I 7 cinrrnor 7 Qi n2 Y 2 cosx Sinx 605x 2Sinx cosxsinx I cosx Sinx Sinx 0V 605x27 Answer xO2nH 4H 2 H x 3 n o Solve for x 39 cosx cos2x0 39 cosx 2coszx 10 39 2cos2xcosx1 2cosx1 Z cosx1 I cosx 1Vcosc7 Answer xH2nH 2H x272n o Solve for x 39 sinx cosx 1 0 2 2 39 smxcosx 1 39 2 sinxcosx0 39 sin 2 x 0 2 2 8111 x 2 smxcosx cos x 1 I 2x02nH2xH2nH Answer xnHVx o Solve for x HnHH 2 cnt PX COSX Sim Solve for x sinxsin 2x 0 sinx2 sinxcosx0 sinx12cosx0 coxx Sinx VCI IIIC 2H xH2nH x 2nH 4H 2 H x 3 n 1111 2 3 I smx xlsEQuadI Whatiscos4x 39 cos4x2cosz2x 1 2 5 Z 39 1 2ZZ2l Z 3 32 625 Z 98 625 Z 527 625 625 625 O Whatissin3x 39 sinx2xsin xcos2xcos xsin2x 2x cos Lcosx 22 5 2 5 5 5 5 21 120141 125 125 125

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