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Trigonometry Identities

This page will include all common trigonometry identites that the author know; with Derivations. Those are really important, and the author highly recommende to memorize all of those.

Pythagorean Identities​

1​

sin⁡2x+cos⁡2x=1\sin^2{x}+\cos^2{x}=1
Derivation

2​

cot⁡2x+1=csc⁡2x\cot^2{x}+1=\csc^2{x}
Derivation

3​

tan⁡2x+1=sec⁡2x\tan^2{x}+1=\sec^2{x}
Derivation

Angle Sum and Difference Identities​

4​

sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin{\left(\alpha+\beta\right)}=\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta}
Derivation
sin⁡β=BCAB=BC1  ⟹  BC=sin⁡βcos⁡β=ACAB=AC1  ⟹  AC=cos⁡βsin⁡α=CDAC=CDcos⁡β  ⟹  CD=sin⁡αcos⁡βcos⁡α=CFBC=CFsin⁡β  ⟹  CF=cos⁡αsin⁡β \begin{align*} \blue{\sin{\beta}}=\frac{BC}{AB}=\frac{BC}{1}\implies BC&=\blue{\sin{\beta}}\\ \blue{\cos{\beta}}=\frac{AC}{AB}=\frac{AC}{1}\implies AC&=\blue{\cos{\beta}}\\ \green{\sin{\alpha}}=\frac{CD}{AC}=\frac{CD}{\blue{\cos{\beta}}}\implies CD&=\green{\sin{\alpha}}\blue{\cos{\beta}}\\ \green{\cos{\alpha}}=\frac{CF}{BC}=\frac{CF}{\blue{\sin{\beta}}}\implies CF&=\green{\cos{\alpha}}\blue{\sin{\beta}}\\ \end{align*}sin⁡(α+β)=BEAB=BE1=DF=CD+CF=sin⁡αcos⁡β+cos⁡αsin⁡β \begin{align*} \sin{(\green{\alpha}+\blue{\beta})}&=\frac{BE}{AB}=\frac{BE}{1}=DF\\ &=CD+CF\\ &=\green{\sin{\alpha}}\blue{\cos{\beta}}+\green{\cos{\alpha}}\blue{\sin{\beta}} \tag*{$\blacksquare$} \end{align*}

5​

sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β\sin{\left(\alpha-\beta\right)}=\sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}
Derivation
sin⁡β=BCAB=BC1  ⟹  BC=sin⁡βcos⁡β=ACAB=AC1  ⟹  AC=cos⁡βsin⁡α=ADAC=ADcos⁡β  ⟹  AD=sin⁡αcos⁡βcos⁡α=BFBC=BFsin⁡β  ⟹  BF=cos⁡αsin⁡β \begin{align*} \blue{\sin{\beta}}=\frac{BC}{AB}=\frac{BC}{1}\implies BC&=\blue{\sin{\beta}}\\ \blue{\cos{\beta}}=\frac{AC}{AB}=\frac{AC}{1}\implies AC&=\blue{\cos{\beta}}\\ \green{\sin{\alpha}}=\frac{AD}{AC}=\frac{AD}{\blue{\cos{\beta}}}\implies AD&=\green{\sin{\alpha}}\blue{\cos{\beta}}\\ \green{\cos{\alpha}}=\frac{BF}{BC}=\frac{BF}{\blue{\sin{\beta}}}\implies BF&=\green{\cos{\alpha}}\blue{\sin{\beta}}\\ \end{align*}sin⁡(α−β)=BGAB=BG1=AE=AD−ED=AD−BF=sin⁡αcos⁡β−cos⁡αsin⁡β \begin{align*} \sin{(\green{\alpha}-\blue{\beta})}&=\frac{BG}{AB}=\frac{BG}{1}=AE\\ &=AD-ED\\ &=AD-BF\\ &=\green{\sin{\alpha}}\blue{\cos{\beta}}-\green{\cos{\alpha}}\blue{\sin{\beta}} \tag*{$\blacksquare$} \end{align*}

6​

cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos{\left(\alpha+\beta\right)}=\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}
Derivation
sin⁡β=BCAB=BC1  ⟹  BC=sin⁡βcos⁡β=ACAB=AC1  ⟹  AC=cos⁡βsin⁡α=BFAC=BFsin⁡β  ⟹  BF=sin⁡αsin⁡βcos⁡α=ADBC=ADcos⁡β  ⟹  AD=cos⁡αcos⁡β \begin{align*} \blue{\sin{\beta}}=\frac{BC}{AB}=\frac{BC}{1}\implies BC&=\blue{\sin{\beta}}\\ \blue{\cos{\beta}}=\frac{AC}{AB}=\frac{AC}{1}\implies AC&=\blue{\cos{\beta}}\\ \green{\sin{\alpha}}=\frac{BF}{AC}=\frac{BF}{\blue{\sin{\beta}}}\implies BF&=\green{\sin{\alpha}}\blue{\sin{\beta}}\\ \green{\cos{\alpha}}=\frac{AD}{BC}=\frac{AD}{\blue{\cos{\beta}}}\implies AD&=\green{\cos{\alpha}}\blue{\cos{\beta}}\\ \end{align*}cos⁡(α+β)=AEAB=AE1=AD−ED=AD−BF=cos⁡αcos⁡β−sin⁡αsin⁡β \begin{align*} \cos{(\green{\alpha}+\blue{\beta})}&=\frac{AE}{AB}=\frac{AE}{1}\\ &=AD-ED\\ &=AD-BF\\ &=\green{\cos{\alpha}}\blue{\cos{\beta}}-\green{\sin{\alpha}}\blue{\sin{\beta}} \tag*{$\blacksquare$} \end{align*}

7​

cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos{\left(\alpha-\beta\right)}=\cos{\alpha}\cos{\beta}+\sin{\alpha}\sin{\beta}
Derivation
sin⁡β=BCAB=BC1  ⟹  BC=sin⁡βcos⁡β=ACAB=AC1  ⟹  AC=cos⁡βsin⁡α=CFBC=CFsin⁡β  ⟹  CF=sin⁡αsin⁡βcos⁡α=CDAC=CDcos⁡β  ⟹  CD=cos⁡αcos⁡β \begin{align*} \blue{\sin{\beta}}=\frac{BC}{AB}=\frac{BC}{1}\implies BC&=\blue{\sin{\beta}}\\ \blue{\cos{\beta}}=\frac{AC}{AB}=\frac{AC}{1}\implies AC&=\blue{\cos{\beta}}\\ \green{\sin{\alpha}}=\frac{CF}{BC}=\frac{CF}{\blue{\sin{\beta}}}\implies CF&=\green{\sin{\alpha}}\blue{\sin{\beta}}\\ \green{\cos{\alpha}}=\frac{CD}{AC}=\frac{CD}{\blue{\cos{\beta}}}\implies CD&=\green{\cos{\alpha}}\blue{\cos{\beta}}\\ \end{align*}cos⁡(α−β)=AGAB=AG1=DF=CD+CF=cos⁡αcos⁡β+sin⁡αsin⁡β \begin{align*} \cos{(\green{\alpha}-\blue{\beta})}&=\frac{AG}{AB}=\frac{AG}{1}=DF\\ &=CD+CF\\ &=\green{\cos{\alpha}}\blue{\cos{\beta}}+\green{\sin{\alpha}}\blue{\sin{\beta}} \tag*{$\blacksquare$} \end{align*}

8​

tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β\tan{\left(\alpha+\beta\right)}=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}
Derivation
tan⁡α=CDAD=CD1  ⟹  CD=tan⁡αsec⁡α=ACAD=AC1  ⟹  AC=sec⁡αtan⁡β=BCAC=BCsec⁡α  ⟹  BC=sec⁡αtan⁡βsec⁡α=BCCF=sec⁡αtan⁡βCF  ⟹  CF=tan⁡βtan⁡α=BFCF=BFtan⁡β  ⟹  BF=tan⁡αtan⁡β \begin{align*} \green{\tan{\alpha}}=\frac{CD}{AD}=\frac{CD}{1}\implies CD&=\green{\tan{\alpha}}\\ \green{\sec{\alpha}}=\frac{AC}{AD}=\frac{AC}{1}\implies AC&=\green{\sec{\alpha}}\\ \blue{\tan{\beta}}=\frac{BC}{AC}=\frac{BC}{\green{\sec{\alpha}}}\implies BC&=\green{\sec{\alpha}}\blue{\tan{\beta}}\\ \green{\sec{\alpha}}=\frac{BC}{CF}=\frac{\green{\sec{\alpha}}\blue{\tan{\beta}}}{CF}\implies CF&=\blue{\tan{\beta}}\\ \green{\tan{\alpha}}=\frac{BF}{CF}=\frac{BF}{\blue{\tan{\beta}}}\implies BF&=\green{\tan{\alpha}}\blue{\tan{\beta}} \end{align*}tan⁡(α+β)=BEAE=DFAE=CD+CFAD−BF=tan⁡α+tan⁡β1−tan⁡αtan⁡β \begin{align*} \tan{(\green{\alpha}+\blue{\beta})}&=\frac{BE}{AE}=\frac{DF}{AE}=\frac{CD+CF}{AD-BF}\\ &=\frac{\green{\tan{\alpha}}+\blue{\tan{\beta}}}{1-\green{\tan{\alpha}}\blue{\tan{\beta}}} \tag*{$\blacksquare$} \end{align*}

9​

tan⁡(α−β)=tan⁡α−tan⁡β1+tan⁡αtan⁡β\tan{\left(\alpha-\beta\right)}=\frac{\tan{\alpha}-\tan{\beta}}{1+\tan{\alpha}\tan{\beta}}
Derivation
tan⁡α=ADCD=AD1  ⟹  AD=tan⁡αsec⁡α=ACCD=AC1  ⟹  AC=sec⁡αtan⁡β=BCAC=BCsec⁡α  ⟹  BC=sec⁡αtan⁡βsec⁡α=BCBF=sec⁡αtan⁡βBF  ⟹  BF=tan⁡βtan⁡α=CFBF=CFtan⁡β  ⟹  CF=tan⁡αtan⁡β \begin{align*} \green{\tan{\alpha}}=\frac{AD}{CD}=\frac{AD}{1}\implies AD&=\green{\tan{\alpha}}\\ \green{\sec{\alpha}}=\frac{AC}{CD}=\frac{AC}{1}\implies AC&=\green{\sec{\alpha}}\\ \blue{\tan{\beta}}=\frac{BC}{AC}=\frac{BC}{\green{\sec{\alpha}}}\implies BC&=\green{\sec{\alpha}}\blue{\tan{\beta}}\\ \green{\sec{\alpha}}=\frac{BC}{BF}=\frac{\green{\sec{\alpha}}\blue{\tan{\beta}}}{BF}\implies BF&=\blue{\tan{\beta}}\\ \green{\tan{\alpha}}=\frac{CF}{BF}=\frac{CF}{\blue{\tan{\beta}}}\implies CF&=\green{\tan{\alpha}}\blue{\tan{\beta}} \end{align*}tan⁡(α−β)=BGAG=AEDF=AD−BFCD+CF=tan⁡α−tan⁡β1+tan⁡αtan⁡β \begin{align*} \tan{(\green{\alpha}-\blue{\beta})}&=\frac{BG}{AG}=\frac{AE}{DF}=\frac{AD-BF}{CD+CF}\\ &=\frac{\green{\tan{\alpha}}-\blue{\tan{\beta}}}{1+\green{\tan{\alpha}}\blue{\tan{\beta}}} \tag*{$\blacksquare$} \end{align*}

Multiple Angle Formulas​

10​

sin⁡2α=2sin⁡αcos⁡α\sin{2\alpha}=2\sin{\alpha}\cos{\alpha}
Derivation
sin⁡2α=sin⁡(α+α)=sin⁡αcos⁡α+cos⁡αsin⁡α(Trig Identity #4)=2sin⁡αcos⁡α\begin{align*} \sin{2\alpha}&=\sin{({\alpha}+{\alpha})}\\ &=\sin{{\alpha}}\cos{{\alpha}}+\cos{{\alpha}}\sin{{\alpha}} &&({\text{Trig Identity \#4}})\\ &=2\sin{\alpha}\cos{\alpha} \tag*{$\blacksquare$} \end{align*}

Trig Identity #4

11​

cos⁡2α=cos⁡2α−sin⁡2α=2cos⁡2α−1=1−2sin⁡2α\cos{2\alpha}=\cos^2{\alpha}-\sin^2{\alpha}=2\cos^2{\alpha}-1=1-2\sin^2{\alpha}
Derivation
cos⁡2α=cos⁡(α+α)=cos⁡αcos⁡α−sin⁡αsin⁡α(Trig Identity #6)=cos⁡2α−sin⁡2α=cos⁡2α−(1−cos⁡2α)=2cos⁡2α−1(Trig Identity #1)=(1−sin⁡2α)−sin⁡2α=1−2sin⁡2α(Trig Identity #1)\begin{align*} \cos{2\alpha}&=\cos{({\alpha}+{\alpha})}\\ &=\cos{{\alpha}}\cos{{\alpha}}-\sin{{\alpha}}\sin{{\alpha}} &&({\text{Trig Identity \#6}})\\ &=\cos^2{\alpha}-\sin^2{\alpha}\\ &=\cos^2{\alpha}-(1-\cos^2{\alpha})=2\cos^2{\alpha}-1 &&({\text{Trig Identity \#1}})\\ &=(1-\sin^2{\alpha})-\sin^2{\alpha}=1-2\sin^2{\alpha}&&({\text{Trig Identity \#1}}) \tag*{$\blacksquare$} \end{align*}

Trig Identity #6

Trig Identity #1

12​

sin⁡3α=−4sin⁡3α+3sin⁡α\sin{3\alpha}=-4\sin^3{\alpha}+3\sin{\alpha}
Derivation
sin⁡3α=sin⁡(2α+α)=sin⁡2αcos⁡α+cos⁡2αsin⁡α(Trig Identity #4)=2sin⁡αcos⁡αcos⁡α+cos⁡2αsin⁡α(Trig Identity #10)=2sin⁡α(cos⁡2α)+(1−2sin⁡2α)sin⁡α(Trig Identity #11)=2sin⁡α(1−sin⁡2α)+(1−2sin⁡2α)sin⁡α(Trig Identity #1)=2sin⁡α−2sin⁡3α+sin⁡α−2sin⁡3α=−4sin⁡3α+3sin⁡α\begin{align*} \sin{3\alpha}&=\sin{({2\alpha}+{\alpha})}\\ &=\sin{{2\alpha}}\cos{{\alpha}}+\cos{{2\alpha}}\sin{{\alpha}} &&({\text{Trig Identity \#4}})\\ &=2\sin{\alpha}\cos{\alpha}\cos{\alpha}+\cos{{2\alpha}}\sin{{\alpha}} &&({\text{Trig Identity \#10}})\\ &=2\sin{\alpha}(\cos^2{\alpha})+(1-2\sin^2{\alpha})\sin{{\alpha}} &&({\text{Trig Identity \#11}})\\ &=2\sin{\alpha}(1-\sin^2{\alpha})+(1-2\sin^2{\alpha})\sin{{\alpha}} &&({\text{Trig Identity \#1}})\\ &=2\sin{\alpha}-2\sin^3{\alpha}+\sin{\alpha}-2\sin^3{\alpha}\\ &=-4\sin^3{\alpha}+3\sin{\alpha} \tag*{$\blacksquare$} \end{align*}

Trig Identity #4 \quad Trig Identity #10

Trig Identity #11 \quad Trig Identity #1

13​

cos⁡3α=4cos⁡3α−3cos⁡α\cos{3\alpha}=4\cos^3{\alpha}-3\cos{\alpha}
Derivation
cos⁡3α=cos⁡(2α+α)=cos⁡2αcos⁡α−sin⁡2αsin⁡α(Trig Identity #6)=(2cos⁡2α−1)cos⁡α−sin⁡2αsin⁡α(Trig Identity #11)=2cos⁡3α−cos⁡α−2sin⁡αcos⁡αsin⁡α(Trig Identity #10)=2cos⁡3α−cos⁡α−2cos⁡α(1−cos⁡2α)(Trig Identity #1)=2cos⁡3α−cos⁡α−2cos⁡α+2cos⁡2α=4cos⁡3α−3cos⁡α\begin{align*} \cos{3\alpha}&=\cos{({2\alpha}+{\alpha})}\\ &=\cos{{2\alpha}}\cos{{\alpha}}-\sin{{2\alpha}}\sin{{\alpha}} &&({\text{Trig Identity \#6}})\\ &=(2\cos^2{\alpha}-1)\cos{\alpha}-\sin{{2\alpha}}\sin{{\alpha}} &&({\text{Trig Identity \#11}})\\ &=2\cos^3{\alpha}-\cos{\alpha}-2\sin{\alpha}\cos{\alpha}\sin{{\alpha}} &&({\text{Trig Identity \#10}})\\ &=2\cos^3{\alpha}-\cos{\alpha}-2\cos{\alpha}(1-\cos^2{\alpha}) &&({\text{Trig Identity \#1}})\\ &=2\cos^3{\alpha}-\cos{\alpha}-2\cos{\alpha}+2\cos^2{\alpha}\\ &=4\cos^3{\alpha}-3\cos{\alpha} \tag*{$\blacksquare$} \end{align*}

Trig Identity #6 \quad Trig Identity #11

Trig Identity #11 \quad Trig Identity #1

Power-Reduction Formulas​

14​

sin⁡2α=1−cos⁡2α2\sin^2{\alpha}=\frac{1-\cos{2\alpha}}{2}
Derivation

15​

cos⁡2α=1+cos⁡2α2\cos^2{\alpha}=\frac{1+\cos{2\alpha}}{2}
Derivation

Half Angle Formulas​

Why ±\pm ?

16​

sin⁡12α=±1−cos⁡α2\sin{\frac{1}{2}\alpha}=\pm \sqrt{\frac{1-\cos{\alpha}}{2}}
Derivation

17​

cos⁡12α=±1+cos⁡α2\cos{\frac{1}{2}\alpha}=\pm \sqrt{\frac{1+\cos{\alpha}}{2}}
Derivation

18​

tan⁡12α=±1−cos⁡α1+cos⁡α=1−cos⁡αsin⁡α=sin⁡α1+cos⁡α\tan{\frac{1}{2}\alpha}=\pm \sqrt{\frac{1-\cos{\alpha}}{1+\cos{\alpha}}}=\frac{1-\cos{\alpha}}{\sin{\alpha}}=\frac{\sin{\alpha}}{1+\cos{\alpha}}
Derivation