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Sum and Product Identities

Adding on to the previous identities, we can also express the sum and product of trigonometric functions in terms of each other. These identities are useful for simplifying expressions and solving equations.

Product as Sum​

We can derive the product-to-sum identities by adding or subtracting the sum-to-product identities together. The product of two cosine functions can be derived by adding cos⁡(α−β)\cos(\alpha - \beta) and cos⁡(α+β)\cos(\alpha + \beta) together...

+ cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β+ cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡βcos⁡(α−β)+cos⁡(α+β)=2cos⁡αcos⁡β12[cos⁡(α−β)+cos⁡(α+β)]=cos⁡αcos⁡β\begin{array}{ccccc} \phantom{+\:} \cos(\alpha - \beta) &=& \cos \alpha \cos \beta + \sin \alpha \sin \beta \\[0.5em] +\: \cos(\alpha + \beta) &=& \cos \alpha \cos \beta - \sin \alpha \sin \beta \\[1em] \hline \\[0em] \cos(\alpha - \beta) + \cos(\alpha + \beta) &=& 2 \cos \alpha \cos \beta \\[0.5em] \dfrac{1}{2} \Bigl[\cos(\alpha - \beta) + \cos(\alpha + \beta)\Bigr] &=& \cos \alpha \cos \beta \\[0.5em] \end{array}

The product of a sine and a cosine function can be derived by adding sin⁡(α+β)\sin(\alpha + \beta) and sin⁡(α−β)\sin(\alpha - \beta) together...

+ sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β+ sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡βsin⁡(α+β)+sin⁡(α−β)=2sin⁡αcos⁡β12[sin⁡(α+β)+sin⁡(α−β)]=sin⁡αcos⁡β\begin{array}{ccccc} \phantom{+\:} \sin(\alpha + \beta) &=& \sin \alpha \cos \beta + \cos \alpha \sin \beta \\[0.5em] +\: \sin(\alpha - \beta) &=& \sin \alpha \cos \beta - \cos \alpha \sin \beta \\[1em] \hline \\[0em] \sin(\alpha + \beta) + \sin(\alpha - \beta) &=& 2 \sin \alpha \cos \beta \\[0.5em] \dfrac{1}{2} \Bigl[\sin(\alpha + \beta) + \sin(\alpha - \beta)\Bigr] &=& \sin \alpha \cos \beta \\[0.5em] \end{array}

The product of two sine functions can be derived by subtracting cos⁡(α+β)\cos(\alpha + \beta) from cos⁡(α−β)\cos(\alpha - \beta)...

− cos⁡(α−β)=sin⁡αsin⁡β+cos⁡αcos⁡β− cos⁡(α+β)=sin⁡αsin⁡β−cos⁡αcos⁡βcos⁡(α−β)−cos⁡(α+β)=2sin⁡αsin⁡β12[cos⁡(α−β)−cos⁡(α+β)]=sin⁡αsin⁡β\begin{array}{ccccc} \phantom{-\:} \cos(\alpha - \beta) &=& \sin \alpha \sin \beta + \cos \alpha \cos \beta \\[0.5em] -\: \cos(\alpha + \beta) &=& \sin \alpha \sin \beta - \cos \alpha \cos \beta \\[1em] \hline \\[0em] \cos(\alpha - \beta) - \cos(\alpha + \beta) &=& 2 \sin \alpha \sin \beta \\[0.5em] \dfrac{1}{2} \Bigl[\cos(\alpha - \beta) - \cos(\alpha + \beta)\Bigr] &=& \sin \alpha \sin \beta \\[0.5em] \end{array}

Finally, the product of cosine and a sine function can be derived by subtracting sin⁡(α−β)\sin(\alpha - \beta) from sin⁡(α+β)\sin(\alpha + \beta)...

− sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β− sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡βsin⁡(α+β)−sin⁡(α−β)=2cos⁡αsin⁡β12[sin⁡(α+β)−sin⁡(α−β)]=cos⁡αsin⁡β\begin{array}{ccccc} \phantom{-\:} \sin(\alpha + \beta) &=& \sin \alpha \cos \beta + \cos \alpha \sin \beta \\[0.5em] -\: \sin(\alpha - \beta) &=& \sin \alpha \cos \beta - \cos \alpha \sin \beta \\[1em] \hline \\[0em] \sin(\alpha + \beta) - \sin(\alpha - \beta) &=& 2 \cos \alpha \sin \beta \\[0.5em] \dfrac{1}{2} \Bigl[\sin(\alpha + \beta) - \sin(\alpha - \beta)\Bigr] &=& \cos \alpha \sin \beta \\[0.5em] \end{array}

Sum as Product​

Sometimes it is useful to reverse the process and express the sum of two trigonometric functions in terms of their product. These identities can be derived through the use of product-to-sum identities and substitute. We can express both α\alpha and β\beta in terms of uu and vv where α=u+v2\alpha = \dfrac{u + v}{2} and β=u−v2\beta = \dfrac{u - v}{2}. Substituting these values into α+β\alpha + \beta and α−β\alpha - \beta gives us the following equations...

α+β=u+v2+u−v2        α−β=u+v2−u−v2=u+v+u−v2=u+v−u+v2=2u2=2v2=u=v\begin{array}{cccccccccc} \alpha + \beta &=& \dfrac{u + v}{2} + \dfrac{u - v}{2} \:\:\:\:&&\:\:\:\: \alpha - \beta &=& \dfrac{u + v}{2} - \dfrac{u - v}{2} \\[1em] &=& \dfrac{u + v + u - v}{2} && &=& \dfrac{u + v - u + v}{2} \\[1em] &=& \dfrac{2u}{2} && &=& \dfrac{2v}{2} \\[1em] &=& u && &=& v \\[1em] \end{array}

We can use the facts that α=u+v2\alpha = \dfrac{u + v}{2}, β=u−v2\beta = \dfrac{u - v}{2}, α+β=u\alpha + \beta = u, and α−β=v\alpha - \beta = v to derive all the sum-to-product identities. Substituting into cos⁡(α)cos⁡(β)\cos(\alpha)\cos(\beta) gives us the following equation...

cos⁡(α)cos⁡(β)=12[cos⁡(α+β)+cos⁡(α−β)]2cos⁡(α)cos⁡(β)=cos⁡(α+β)+cos⁡(α−β)2cos⁡(u+v2)cos⁡(u−v2)=cos⁡(u)+cos⁡(v)\begin{array}{ccccc} \cos(\alpha)\cos(\beta) &=& \dfrac{1}{2} \Bigl[\cos(\alpha + \beta) + \cos(\alpha - \beta)\Bigr] \\[1em] 2\cos(\alpha)\cos(\beta) &=& \cos(\alpha + \beta) + \cos(\alpha - \beta) \\[1em] 2\cos(\dfrac{u + v}{2})\cos(\dfrac{u - v}{2}) &=& \cos(u) + \cos(v) \\[1em] \end{array}

Substituting into sin⁡(α)cos⁡(β)\sin(\alpha)\cos(\beta) gives us the following equation...

sin⁡(α)cos⁡(β)=12[sin⁡(α+β)+sin⁡(α−β)]2sin⁡(α)cos⁡(β)=sin⁡(α+β)+sin⁡(α−β)2sin⁡(u+v2)cos⁡(u−v2)=sin⁡(u)+sin⁡(v)\begin{array}{ccccc} \sin(\alpha)\cos(\beta) &=& \dfrac{1}{2} \Bigl[\sin(\alpha + \beta) + \sin(\alpha - \beta)\Bigr] \\[1em] 2\sin(\alpha)\cos(\beta) &=& \sin(\alpha + \beta) + \sin(\alpha - \beta) \\[1em] 2\sin(\dfrac{u + v}{2})\cos(\dfrac{u - v}{2}) &=& \sin(u) + \sin(v) \\[1em] \end{array}

Substituting into sin⁡(α)sin⁡(β)\sin(\alpha)\sin(\beta) gives us the following equation...

sin⁡(α)sin⁡(β)=12[cos⁡(α−β)−cos⁡(α+β)]2sin⁡(α)sin⁡(β)=cos⁡(α−β)−cos⁡(α+β)2sin⁡(u+v2)sin⁡(u−v2)=cos⁡(u)−cos⁡(v)\begin{array}{ccccc} \sin(\alpha)\sin(\beta) &=& \dfrac{1}{2} \Bigl[\cos(\alpha - \beta) - \cos(\alpha + \beta)\Bigr] \\[1em] 2\sin(\alpha)\sin(\beta) &=& \cos(\alpha - \beta) - \cos(\alpha + \beta) \\[1em] 2\sin(\dfrac{u + v}{2})\sin(\dfrac{u - v}{2}) &=& \cos(u) - \cos(v) \\[1em] \end{array}

Finally, substituting into cos⁡(α)sin⁡(β)\cos(\alpha)\sin(\beta) gives us the following equation...

cos⁡(α)sin⁡(β)=12[sin⁡(α+β)−sin⁡(α−β)]2cos⁡(α)sin⁡(β)=sin⁡(α+β)−sin⁡(α−β)2cos⁡(u+v2)sin⁡(u−v2)=sin⁡(u)−sin⁡(v)\begin{array}{ccccc} \cos(\alpha)\sin(\beta) &=& \dfrac{1}{2} \Bigl[\sin(\alpha + \beta) - \sin(\alpha - \beta)\Bigr] \\[1em] 2\cos(\alpha)\sin(\beta) &=& \sin(\alpha + \beta) - \sin(\alpha - \beta) \\[1em] 2\cos(\dfrac{u + v}{2})\sin(\dfrac{u - v}{2}) &=& \sin(u) - \sin(v) \\[1em] \end{array}

Trigonometric Identities​

All the product-to-sum identities are...

cos⁡(α)cos⁡(β)=12[cos⁡(α+β)+cos⁡(α−β)]sin⁡(α)cos⁡(β)=12[sin⁡(α+β)+sin⁡(α−β)]sin⁡(α)sin⁡(β)=12[cos⁡(α−β)−cos⁡(α+β)]cos⁡(α)sin⁡(β)=12[sin⁡(α+β)−sin⁡(α−β)]\begin{array}{ccccccccc} \cos(\alpha)\cos(\beta) &=& \dfrac{1}{2} \Bigl[\cos(\alpha + \beta) + \cos(\alpha - \beta)\Bigr] \\[1em] \sin(\alpha)\cos(\beta) &=& \dfrac{1}{2} \Bigl[\sin(\alpha + \beta) + \sin(\alpha - \beta)\Bigr] \\[1em] \sin(\alpha)\sin(\beta) &=& \dfrac{1}{2} \Bigl[\cos(\alpha - \beta) - \cos(\alpha + \beta)\Bigr] \\[1em] \cos(\alpha)\sin(\beta) &=& \dfrac{1}{2} \Bigl[\sin(\alpha + \beta) - \sin(\alpha - \beta)\Bigr] \\[1em] \end{array}

All the sum-to-product identities are...

cos⁡(α)+cos⁡(β)=2cos⁡(α+β2)cos⁡(α−β2)sin⁡(α)+sin⁡(β)=2sin⁡(α+β2)cos⁡(α−β2)sin⁡(α)−sin⁡(β)=2cos⁡(α+β2)sin⁡(α−β2)cos⁡(α)−cos⁡(β)=−2sin⁡(α+β2)sin⁡(α−β2)\begin{array}{ccccccccc} \cos(\alpha) + \cos(\beta) &=& 2 \cos (\dfrac{\alpha + \beta}{2}) \cos (\dfrac{\alpha - \beta}{2}) \\[1em] \sin(\alpha) + \sin(\beta) &=& 2 \sin (\dfrac{\alpha + \beta}{2}) \cos (\dfrac{\alpha - \beta}{2}) \\[1em] \sin(\alpha) - \sin(\beta) &=& 2 \cos (\dfrac{\alpha + \beta}{2}) \sin (\dfrac{\alpha - \beta}{2}) \\[1em] \cos(\alpha) - \cos(\beta) &=& -2 \sin (\dfrac{\alpha + \beta}{2}) \sin (\dfrac{\alpha - \beta}{2}) \\[1em] \end{array}