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Harold’s Trig ProofsCheat Sheet26 April 2016Proof of Pythagorean IdentitiesProof1320165825500 Givenx2+y2=r2r=1sinθ=oppositehypotenuse=yr=y1=ycosθ=adjacenthypotenuse=xr=x1=x Substitute and Simplifysin2θ+cos2θ=12 Formulasin2θ+cos2θ=1 [1]Proof Givensin2θ+cos2θ=1 [1] Divide by cos2θ, then Simplifysin2θcos2θ+cos2θcos2θ=1cos2θ Formulatan2θ+1=sec2θ [2]Proof Givensin2θ+cos2θ=1 [1] Divide by sin2θ, then Simplifysin2θsin2θ+cos2θsin2θ=1sin2θ Formula1+cot2θ=csc2θ [3]Proof of Sum and Difference FormulasTrig Sum and Difference Formulassinα±β=sinαcosβ±cosαsinβcosα±β=cosαcosβ?sinαsinβProof DiagramProof of sinα±βProve Sum Givensinα+β=EDDA=oppositehypotenuse Alternate interior angles are congruentα=∠CAB=∠HFA=∠HDF Tallest vertical lineED=GF+HD Substitute, then divide and multiply by AF & FDsinα+β=EDAD=GFAD+HDAD=GFAF AFAD+HDFD FDAD Convert back to trig formulassinα+β=sinαcosβ+cosαsinβ [4]Prove Difference Replace +β with -βcos-β=cosβsin-β=-sinβ Simplifysinα-β=sinαcosβ-cosαsinβ [5] General Formula [4+5] sinα±β=sinαcosβ±cosαsinβ [6]Proof of cosα±βProve Sum Givencosα+β=AEAD=adjacenthypotenuse Longest horizontal lineEA=GA-FH Substitute, then divide and multiply by AF & DFcosα+β=EAAD=GAAD-FHAD=GAAF AFAD+FHDF DFAD Convert back to trig formulascosα+β=cosαcosβ-sinαsinβ [7]Prove Difference Replace +β with -βcos-β=cosβsin-β=-sinβ Simplifycosα-β=cosαcosβ+sinαsinβ [8] General Formula [7+8]cosα±β=cosαcosβ?sinαsinβ [9]Proof of tanα±βProve Sum and Difference Giventanα±β=sinα±βcosα±β Substitute sinα±β=sinαcosβ±cosαsinβ [6]cosα±β=cosαcosβ?sinαsinβ [9] Divide by (cosαcosβ), then Simplifytanα±β=sinαcosβ±cosαsinβcosαcosβ?sinαsinβ General Formulatanα±β=tanα± tanβ1? tanα tanβ [10]Proof of Double Angle Formulas (2θ)Proof Givensinα+β=sinαcosβ+cosαsinβ [4] Substituteθ=α=β Simplifysinθ+θ=sinθcosθ+cosθsinθ Formulasin2θ=2sinθcosθ [14]Proof Givencosα+β=cosαcosβ-sinαsinβ [7] Substituteθ=α=β Simplifycosθ+θ=cosθcosθ-sinθsinθ Formulacos2θ=cos2θ-sin2θ [15]Proof Givencos2θ=cos2θ-sin2θ [15]sin2θ+cos2θ=1 [1] Substitutesin2θ=1-cos2θ Simplifycos2θ=cos2θ-1-cos2θ Formulacos2θ=2 cos2θ-1 [16]Proof Givencos2θ=cos2θ-sin2θ [15]sin2θ+cos2θ=1 [1] Substitutecos2θ=1-sin2θ Simplifycos2θ=1-sin2θ-sin2θ Formulacos2θ=1-2sin2θ [17]Proof Giventan2θ=sin2θcos2θ Substitutesin2θ=2sinθcosθ [14]cos2θ=cos2θ-sin2θ [15] Divide by cos2θtan2θ= 2sinθcosθcos2θ-sin2θ Simplifytan2θ= 2sinθcosθcos2θcos2θ-sin2θcos2θ Formulatan2θ=2 tanθ1-tan2θ [18]Proof of Half Angle Formulas (θ/2)Proof Givencos2θ=1-2sin2θ [17] Solve for sin2θsin2θ=1-cos2θ2 [19a] Substituteθ=θ2 Solvesin2θ2=1-cosθ2 Formulasinθ2=±1-cosθ2 [19b]Proof Givencos2θ=2 cos2θ-1 [16] Solve for cos2θcos2θ=1+cos2θ2 [20a] Substituteθ=θ2 Solvecos2θ2=1+cosθ2 Formulacosθ2=±1+cosθ2 [20b]Proof Giventan2θ=sin2θcos2θ Substitutesin2θ=1-cos2θ2 [19a]cos2θ=1+cos2θ2 [20a] Simplifytan2θ=1-cos2θ21+cos2θ2=1-cos2θ1+cos2θ Substituteθ=θ2 Solvetan2θ2=1-cosθ1+cosθ [21a] Formulatanθ2=±1-cosθ1+cosθ [21b]Proof of Cofunction FormulasProof Givensinα-β=sinαcosβ-cosαsinβ [5] Substituteα=π2, β=θ Simplifysinπ2-θ=sinπ2cosθ+cosπ2sinθ Formulasinπ2-θ=cosθ [22]Proof Givencosα-β=cosαcosβ+sinαsinβ [8] Substituteα=π2, β=θ Simplifycosπ2-θ=cosπ2cosθ-sinπ2sinθ Formulacosπ2-θ=sinθ [23]Proof Giventanθ=sinθcosθ Substitutesinπ2-θ=cosθ [22]cosπ2-θ=sinθ [23] Simplifytanπ2-θ=sinπ2-θcosπ2-θ=cosθsinθ=cotθ Formulatanπ2-θ=cotθ [24]Proof Givensecθ=1cosθ Substitutecosπ2-θ=sinθ [23] Simplifysecπ2-θ=1cosπ2-θ=1sinθ=cscθ Formulasecπ2-θ=cscθ [25]Proof Givencscθ=1sinθ Substitutesinπ2-θ=cosθ [22] Simplifycscπ2-θ=1sinπ2-θ=1cosθ=secθ Formulacscπ2-θ=secθ [26]Proof Givencotθ=1tanθ Substitutetanπ2-θ=cotθ [24] Simplifycotπ2-θ=1tanπ2-θ=1cotθ=tanθ Formulacotπ2-θ=tanθ [27] ................
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