Trig Identities – Addition and Subtraction

Deriving the addition and subtraction trigonometric identities.

We will start with cosine, and use the result to derive the remaining identities.

Proving cos(A-B)=cos(A)cos(B)+sin(A)(sin(B).

A and B are represented in the unit circle below.

Remember P(x_1,y_1)=(cos(A), sin(A)) and Q(x_2, y_2)=(cos(B), sin(B))

Using the cosine rule and triangle OPQ, find PQ

    \begin{equation*}(PQ)^2=1^2+1^2-2(1)(1)cos(A-B)\end{equation}

    \begin{equation*}(PQ)^2=2-2cos(A-B)\end{equation}

Using the distance between points, find PQ

    \begin{equation*}(PQ)^2=(x_1-x_2)^2+(y_1-y_2)^2\end{equation}

    \begin{equation*}(PQ)^2=(cosA-cosB)^2+(sinA-sinB)^2\end{equation}

(PQ)^2=cos^2A-2cosAcosB+cos^2B+sin^2A-2sinAsinB+sin^2B

Remember the Pythagorean identity

    \begin{equation*}cos^2\theta+sin^2\theta=1\end{equation}

    \begin{equation*}(PQ)^2=2-2cosAcosB-2sinAsinB\end{equation}

Hence

    \begin{equation*}2-2cos(A-B)=2-2cosAcosB-2sinAsinB\end{equation}

(1)   \begin{equation*}cos(A-B)=cosAcosB+sinAsinB\end{equation*}

We can then use this identity to find cos(A+B).

    \begin{equation*}cos(A+B)=cos(A-(-B))\end{equation}

    \begin{equation*}cos(A-(-B))=cos(A)cos(-B)+sinAsin(-B)\end{equation}

Remember cos(-B)=cos(B) and sin(-B)=-sin(B)

    \begin{equation*}cos(A-(-B))=cosAcosB-sinAsinB\end{equation}

(2)   \begin{equation*}cos(A+B)=cosAcosB-sinAsinB\end{equation*}

We can also find sin(A+B)

Remember, sin\theta=cos(\frac{\pi}{2}-\theta)

    \begin{equation*}sin(A+B)=cos(\frac{\pi}{2}-(A+B))\end{equation}

    \begin{equation*}sin(A+B)=cos((\frac{\pi}{2}-A)-B)\end{equation}

    \begin{equation*}sin(A+B)=cos(\frac{\pi}{2}-A)cosB+sin(\frac{\pi}{2}-A)sinB\end{equation}

(3)   \begin{equation*}sin(A+B)=sinAcosB+cosAsinB\end{equation*}

We can use equation 2 to find sin(A-B)

    \begin{equation*}sin(A-B)=sin(A+(-B))\end{equation}

    \begin{equation*}sin(A+(-B))=sinAcos(-B)+cosAsin(-B)\end{equation}

    \begin{equation*}sin(A+(-B))=sinAcos(B)-cosAsin(B)\end{equation}

(4)   \begin{equation*}sin(A-B)=sinAcos(B)-cosAsin(B)\end{equation*}

And we can use both the sine and cosine identities to find tan(A+B)

Remember tan\theta=\frac{sin\theta}{cos\theta}

    \begin{equation*}tan(A+B)=\frac{sin(A+B)}{cos(A+B)}\end{equation}

    \begin{equation*}=\frac{sinAcosB+cosAsinB}{cosAcosB-sinAsinB}\end{equation}

    \begin{equation*}=\frac{sinAcosB+cosAsinB}{cosAcosB-sinAsinB}\times \frac{cosAcosB}{cosAcosB}\end{equation}

    \begin{equation*}=\frac{\frac{sinAcosB}{cosAcosB}+\frac{cosAsinB}{cosAcosB}}{\frac{cosAcosB}{cosAcosB}+\frac{sinAsinB}{cosAcosB}}\end{equation}

    \begin{equation*}=\frac{tanA+tanB}{1-tanAtanB}\end{equation}

(5)   \begin{equation*}tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}\end{equation*}

and

(6)   \begin{equation*}tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}\end{equation*}

    \begin{equation*}cos(A\pm B)=cosAcosB\mp sinAsinB\end{equation}


    \begin{equation*}sin(A\pm B)=sinAcosB\pm cosAsinB\end{equation}


    \begin{equation*}tan(A \pm B)=\frac{tan A \pm tanB}{1 \mp tanAtanB}\end{equation}

Leave a Comment

Filed under Addition and Subtraction Identities, Identities, Non-Right Trigonometry, Simplifying fractions, Trigonometry, Year 11 Mathematical Methods, Year 11 Specialist Mathematics

Leave a Reply

Your email address will not be published. Required fields are marked *