Product-sum identities, also called product-to-sum identities, rewrite products of trigonometric functions as sums or differences. They are especially useful for finding exact values, simplifying trig expressions, and proving other identities. These problems include products involving sine, cosine, tangent-related expressions, degrees, radians, and exact unit-circle values.
Notes

Practice Problems
\(\textbf{1)}\) Find the exact value of \(\cos\left(75^{\circ}\right)\cos\left(15^{\circ}\right)\)
The answer is \( \frac{1}{4} \)
\(\,\,\,\,\,\cos\left(\alpha\right)\cos\left(\beta\right)=\left(\displaystyle\frac{\cos\left(\alpha + \beta\right)+\cos\left(\alpha – \beta\right)}{2}\right)\)
\(\,\,\,\,\,\cos\left(75^{\circ}\right)\cos\left(15^{\circ}\right)=\displaystyle\frac{\cos\left(75^{\circ} + 15^{\circ}\right)+\cos\left(75^{\circ} – 15^{\circ}\right)}{2}\)
\(\,\,\,\,\,\cos\left(75^{\circ}\right)\cos\left(15^{\circ}\right)=\displaystyle\frac{\cos\left(90^{\circ}\right)+\cos\left(60^{\circ}\right)}{2}\)
\(\,\,\,\,\,\cos\left(75^{\circ}\right)\cos\left(15^{\circ}\right)=\displaystyle\frac{0+\frac{1}{2}}{2}\)
\(\,\,\,\,\,\cos\left(75^{\circ}\right)\cos\left(15^{\circ}\right)=\frac{1}{4}\)
\(\textbf{2)}\) Find the exact value of \(\sin\left(135^{\circ}\right)\cos\left(75^{\circ}\right)\)
The answer is \( \frac{\sqrt{3}-1}{4} \)
\(\,\,\,\,\,\sin\left(\alpha\right)\cos\left(\beta\right)=\left(\displaystyle\frac{\sin\left(\alpha + \beta\right)+\sin\left(\alpha – \beta\right)}{2}\right)\)
\(\,\,\,\,\,\sin\left(135^{\circ}\right)\cos\left(75^{\circ}\right)=\displaystyle\frac{\sin\left(135^{\circ} + 75^{\circ}\right)+\sin\left(135^{\circ} – 75^{\circ}\right)}{2}\)
\(\,\,\,\,\,\sin\left(135^{\circ}\right)\cos\left(75^{\circ}\right)=\displaystyle\frac{\sin\left(210^{\circ}\right)+\sin\left(60^{\circ}\right)}{2}\)
\(\,\,\,\,\,\sin\left(135^{\circ}\right)\cos\left(75^{\circ}\right)=\displaystyle\frac{-\frac{1}{2}+\frac{\sqrt{3}}{2}}{2}\)
\(\,\,\,\,\,\sin\left(135^{\circ}\right)\cos\left(75^{\circ}\right)=\displaystyle\frac{\sqrt{3}-1}{4}\)
\(\textbf{3)}\) Find the exact value of \(\sin\left(75^{\circ}\right)\sin\left(15^{\circ}\right)\)
The answer is \(\frac{1}{4}\)
\(\,\,\,\,\,\sin\left(\alpha\right)\sin\left(\beta\right)=\displaystyle\frac{\cos\left(\alpha-\beta\right)-\cos\left(\alpha+\beta\right)}{2}\)
\(\,\,\,\,\,\sin\left(75^\circ\right)\sin\left(15^\circ\right)=\displaystyle\frac{\cos\left(75^\circ-15^\circ\right)-\cos\left(75^\circ+15^\circ\right)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\cos\left(60^\circ\right)-\cos\left(90^\circ\right)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\frac{1}{2}-0}{2}\)
\(\,\,\,\,\,=\frac{1}{4}\)
\(\textbf{4)}\) Find the exact value of \(\cos\left(105^{\circ}\right)\cos\left(15^{\circ}\right)\)
The answer is \(-\frac{1}{4}\)
\(\,\,\,\,\,\cos\left(\alpha\right)\cos\left(\beta\right)=\displaystyle\frac{\cos\left(\alpha+\beta\right)+\cos\left(\alpha-\beta\right)}{2}\)
\(\,\,\,\,\,\cos\left(105^\circ\right)\cos\left(15^\circ\right)=\displaystyle\frac{\cos\left(120^\circ\right)+\cos\left(90^\circ\right)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{-\frac{1}{2}+0}{2}\)
\(\,\,\,\,\,=-\frac{1}{4}\)
\(\textbf{5)}\) Find the exact value of \(\sin\left(105^{\circ}\right)\cos\left(15^{\circ}\right)\)
The answer is \(\frac{\sqrt{3}+1}{4}\)
\(\,\,\,\,\,\sin\left(\alpha\right)\cos\left(\beta\right)=\displaystyle\frac{\sin\left(\alpha+\beta\right)+\sin\left(\alpha-\beta\right)}{2}\)
\(\,\,\,\,\,\sin\left(105^\circ\right)\cos\left(15^\circ\right)=\displaystyle\frac{\sin\left(120^\circ\right)+\sin\left(90^\circ\right)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\frac{\sqrt{3}}{2}+1}{2}\)
\(\,\,\,\,\,=\frac{\sqrt{3}+2}{4}\)
\(\textbf{6)}\) Find the exact value of \(\cos\left(150^{\circ}\right)\cos\left(30^{\circ}\right)\)
The answer is \(-\frac{3}{4}\)
\(\,\,\,\,\,\cos\left(\alpha\right)\cos\left(\beta\right)=\displaystyle\frac{\cos\left(\alpha+\beta\right)+\cos\left(\alpha-\beta\right)}{2}\)
\(\,\,\,\,\,\cos\left(150^\circ\right)\cos\left(30^\circ\right)=\displaystyle\frac{\cos\left(180^\circ\right)+\cos\left(120^\circ\right)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{-1-\frac{1}{2}}{2}\)
\(\,\,\,\,\,=-\frac{3}{4}\)
\(\textbf{7)}\) Find the exact value of \(\sin\left(150^{\circ}\right)\sin\left(30^{\circ}\right)\)
The answer is \(\frac{1}{4}\)
\(\,\,\,\,\,\sin\left(\alpha\right)\sin\left(\beta\right)=\displaystyle\frac{\cos\left(\alpha-\beta\right)-\cos\left(\alpha+\beta\right)}{2}\)
\(\,\,\,\,\,\sin\left(150^\circ\right)\sin\left(30^\circ\right)=\displaystyle\frac{\cos\left(120^\circ\right)-\cos\left(180^\circ\right)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{-\frac{1}{2}-(-1)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\frac{1}{2}}{2}\)
\(\,\,\,\,\,=\frac{1}{4}\)
\(\textbf{8)}\) Rewrite \(\cos(5x)\cos(2x)\) as a sum.
The answer is \(\frac{\cos(7x)+\cos(3x)}{2}\)
\(\,\,\,\,\,\cos(\alpha)\cos(\beta)=\displaystyle\frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}\)
\(\,\,\,\,\,\cos(5x)\cos(2x)=\displaystyle\frac{\cos(5x+2x)+\cos(5x-2x)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\cos(7x)+\cos(3x)}{2}\)
\(\textbf{9)}\) Rewrite \(\sin(6x)\sin(4x)\) as a sum or difference.
The answer is \(\frac{\cos(2x)-\cos(10x)}{2}\)
\(\,\,\,\,\,\sin(\alpha)\sin(\beta)=\displaystyle\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2}\)
\(\,\,\,\,\,\sin(6x)\sin(4x)=\displaystyle\frac{\cos(6x-4x)-\cos(6x+4x)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\cos(2x)-\cos(10x)}{2}\)
\(\textbf{10)}\) Rewrite \(\sin(3x)\cos(8x)\) as a sum.
The answer is \(\frac{\sin(11x)-\sin(5x)}{2}\)
\(\,\,\,\,\,\sin(\alpha)\cos(\beta)=\displaystyle\frac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}{2}\)
\(\,\,\,\,\,\sin(3x)\cos(8x)=\displaystyle\frac{\sin(3x+8x)+\sin(3x-8x)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\sin(11x)+\sin(-5x)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\sin(11x)-\sin(5x)}{2}\)
\(\textbf{11)}\) Rewrite \(\cos(4x)\sin(x)\) as a sum.
The answer is \(\frac{\sin(5x)-\sin(3x)}{2}\)
\(\,\,\,\,\,\cos(\alpha)\sin(\beta)=\displaystyle\frac{\sin(\alpha+\beta)-\sin(\alpha-\beta)}{2}\)
\(\,\,\,\,\,\cos(4x)\sin(x)=\displaystyle\frac{\sin(4x+x)-\sin(4x-x)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\sin(5x)-\sin(3x)}{2}\)
\(\textbf{12)}\) Rewrite \(\cos(9x)\cos(4x)\) as a sum.
The answer is \(\frac{\cos(13x)+\cos(5x)}{2}\)
\(\,\,\,\,\,\cos(\alpha)\cos(\beta)=\displaystyle\frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}\)
\(\,\,\,\,\,\cos(9x)\cos(4x)=\displaystyle\frac{\cos(9x+4x)+\cos(9x-4x)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\cos(13x)+\cos(5x)}{2}\)
\(\textbf{13)}\) Rewrite \(\sin(7x)\sin(2x)\) as a sum or difference.
The answer is \(\frac{\cos(5x)-\cos(9x)}{2}\)
\(\,\,\,\,\,\sin(\alpha)\sin(\beta)=\displaystyle\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2}\)
\(\,\,\,\,\,\sin(7x)\sin(2x)=\displaystyle\frac{\cos(7x-2x)-\cos(7x+2x)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\cos(5x)-\cos(9x)}{2}\)
\(\textbf{14)}\) Rewrite \(\sin(10x)\cos(3x)\) as a sum.
The answer is \(\frac{\sin(13x)+\sin(7x)}{2}\)
\(\,\,\,\,\,\sin(\alpha)\cos(\beta)=\displaystyle\frac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}{2}\)
\(\,\,\,\,\,\sin(10x)\cos(3x)=\displaystyle\frac{\sin(10x+3x)+\sin(10x-3x)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\sin(13x)+\sin(7x)}{2}\)
\(\textbf{15)}\) Find the exact value of \(\cos\left(\frac{5\pi}{12}\right)\cos\left(\frac{\pi}{12}\right)\)
The answer is \(\frac{1}{4}\)
\(\,\,\,\,\,\cos(\alpha)\cos(\beta)=\displaystyle\frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}\)
\(\,\,\,\,\,\cos\left(\frac{5\pi}{12}\right)\cos\left(\frac{\pi}{12}\right)=\displaystyle\frac{\cos\left(\frac{6\pi}{12}\right)+\cos\left(\frac{4\pi}{12}\right)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\cos\left(\frac{\pi}{2}\right)+\cos\left(\frac{\pi}{3}\right)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{0+\frac{1}{2}}{2}\)
\(\,\,\,\,\,=\frac{1}{4}\)
Challenge Problems
\(\textbf{16)}\) Find the exact value of \(\sin\left(\frac{7\pi}{12}\right)\cos\left(\frac{\pi}{12}\right)\)
The answer is \(\frac{\sqrt{3}+1}{4}\)
\(\,\,\,\,\,\sin(\alpha)\cos(\beta)=\displaystyle\frac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}{2}\)
\(\,\,\,\,\,\sin\left(\frac{7\pi}{12}\right)\cos\left(\frac{\pi}{12}\right)=\displaystyle\frac{\sin\left(\frac{8\pi}{12}\right)+\sin\left(\frac{6\pi}{12}\right)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\sin\left(\frac{2\pi}{3}\right)+\sin\left(\frac{\pi}{2}\right)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\frac{\sqrt{3}}{2}+1}{2}\)
\(\,\,\,\,\,=\frac{\sqrt{3}+2}{4}\)
\(\textbf{17)}\) Find the exact value of \(\sin\left(\frac{5\pi}{12}\right)\sin\left(\frac{\pi}{12}\right)\)
The answer is \(\frac{1}{4}\)
\(\,\,\,\,\,\sin(\alpha)\sin(\beta)=\displaystyle\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2}\)
\(\,\,\,\,\,\sin\left(\frac{5\pi}{12}\right)\sin\left(\frac{\pi}{12}\right)=\displaystyle\frac{\cos\left(\frac{4\pi}{12}\right)-\cos\left(\frac{6\pi}{12}\right)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\cos\left(\frac{\pi}{3}\right)-\cos\left(\frac{\pi}{2}\right)}{2}\)
\(\,\,\,\,\,=\displaystyle\frac{\frac{1}{2}-0}{2}\)
\(\,\,\,\,\,=\frac{1}{4}\)
\(\textbf{18)}\) Rewrite \(2\cos(6x)\cos(4x)\) as a sum.
The answer is \(\cos(10x)+\cos(2x)\)
\(\,\,\,\,\,\cos(\alpha)\cos(\beta)=\displaystyle\frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}\)
\(\,\,\,\,\,2\cos(\alpha)\cos(\beta)=\cos(\alpha+\beta)+\cos(\alpha-\beta)\)
\(\,\,\,\,\,2\cos(6x)\cos(4x)=\cos(10x)+\cos(2x)\)
\(\textbf{19)}\) Rewrite \(2\sin(9x)\sin(5x)\) as a sum or difference.
The answer is \(\cos(4x)-\cos(14x)\)
\(\,\,\,\,\,\sin(\alpha)\sin(\beta)=\displaystyle\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2}\)
\(\,\,\,\,\,2\sin(\alpha)\sin(\beta)=\cos(\alpha-\beta)-\cos(\alpha+\beta)\)
\(\,\,\,\,\,2\sin(9x)\sin(5x)=\cos(4x)-\cos(14x)\)
\(\textbf{20)}\) Rewrite \(2\sin(8x)\cos(3x)\) as a sum.
The answer is \(\sin(11x)+\sin(5x)\)
\(\,\,\,\,\,\sin(\alpha)\cos(\beta)=\displaystyle\frac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}{2}\)
\(\,\,\,\,\,2\sin(\alpha)\cos(\beta)=\sin(\alpha+\beta)+\sin(\alpha-\beta)\)
\(\,\,\,\,\,2\sin(8x)\cos(3x)=\sin(11x)+\sin(5x)\)
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