Sum and difference of angles formulas are used to find exact trig values for angles that can be written as the sum or difference of familiar unit-circle angles. These identities work for sine, cosine, and tangent, and they are especially helpful for angles like \(15^\circ\), \(75^\circ\), and \(195^\circ\). This page includes exact value problems, tangent sum and difference problems, and identity verification using angle formulas.
Notes
Practice Problems
Find the exact value.
\(\textbf{1)}\) \( \sin{\left(15°\right)} \) The exact value is \( \displaystyle \frac{\sqrt{6}-\sqrt{2}}{4} \)
\(\text{Step 1: Rewrite } \sin\left(15°\right) \text{ using familiar angles.} \)
\(\,\,\,\,\,\,\sin\left(15°\right)=\sin\left(60°-45°\right)\)
\(\text{Step 2: Use the difference formula.} \)
\(\,\,\,\,\,\,\sin\left(A-B\right)=\sin\left(A\right)\cos\left(B\right)-\cos\left(A\right)\sin\left(B\right)\)
\(\,\,\,\,\,\,\sin\left(60°-45°\right)=\sin\left(60°\right)\cos\left(45°\right)-\cos\left(60°\right)\sin\left(45°\right)\)
\(\text{Step 3: Evaluate.} \)
\(\,\,\,\,\,\,\sin\left(60°-45°\right)=\displaystyle\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right)\)
\(\text{Step 4: Simplify.} \)
\(\,\,\,\,\,\,\sin\left(60°-45°\right)=\displaystyle\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}\)
\(\,\,\,\,\,\,\sin\left(15°\right)=\displaystyle\frac{\sqrt{6}-\sqrt{2}}{4}\)
\(\text{Step 1: Rewrite } \sin\left(15°\right) \text{ using familiar angles.} \)
\(\,\,\,\,\,\,\sin\left(15°\right)=\sin\left(60°-45°\right)\)
\(\text{Step 2: Use the difference formula.} \)
\(\,\,\,\,\,\,\sin\left(A-B\right)=\sin\left(A\right)\cos\left(B\right)-\cos\left(A\right)\sin\left(B\right)\)
\(\,\,\,\,\,\,\sin\left(60°-45°\right)=\sin\left(60°\right)\cos\left(45°\right)-\cos\left(60°\right)\sin\left(45°\right)\)
\(\text{Step 3: Evaluate.} \)
\(\,\,\,\,\,\,\sin\left(60°-45°\right)=\displaystyle\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right)\)
\(\text{Step 4: Simplify.} \)
\(\,\,\,\,\,\,\sin\left(60°-45°\right)=\displaystyle\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}\)
\(\,\,\,\,\,\,\sin\left(15°\right)=\displaystyle\frac{\sqrt{6}-\sqrt{2}}{4}\)
\(\textbf{2)}\) \( \cos{\left(195°\right)} \) The exact value is \( \displaystyle -\frac{\sqrt{6}+\sqrt{2}}{4} \)
\(\text{Step 1: Rewrite } \cos\left(195°\right) \text{ using familiar angles.} \)
\(\,\,\,\,\,\,\cos\left(195°\right)=\cos\left(150°+45°\right)\)
\(\text{Step 2: Use the sum formula.} \)
\(\,\,\,\,\,\,\cos\left(A+B\right)=\cos\left(A\right)\cos\left(B\right)-\sin\left(A\right)\sin\left(B\right)\)
\(\,\,\,\,\,\,\cos\left(150°+45°\right)=\cos\left(150°\right)\cos\left(45°\right)-\sin\left(150°\right)\sin\left(45°\right)\)
\(\text{Step 3: Evaluate.} \)
\(\,\,\,\,\,\,\cos\left(150°+45°\right)=\displaystyle\left(-\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right)\)
\(\text{Step 4: Simplify.} \)
\(\,\,\,\,\,\,\cos\left(150°+45°\right)=\displaystyle-\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}\)
\(\,\,\,\,\,\,\cos\left(195°\right)=\displaystyle -\frac{\sqrt{6}+\sqrt{2}}{4}\)
\(\text{Step 1: Rewrite } \cos\left(195°\right) \text{ using familiar angles.} \)
\(\,\,\,\,\,\,\cos\left(195°\right)=\cos\left(150°+45°\right)\)
\(\text{Step 2: Use the sum formula.} \)
\(\,\,\,\,\,\,\cos\left(A+B\right)=\cos\left(A\right)\cos\left(B\right)-\sin\left(A\right)\sin\left(B\right)\)
\(\,\,\,\,\,\,\cos\left(150°+45°\right)=\cos\left(150°\right)\cos\left(45°\right)-\sin\left(150°\right)\sin\left(45°\right)\)
\(\text{Step 3: Evaluate.} \)
\(\,\,\,\,\,\,\cos\left(150°+45°\right)=\displaystyle\left(-\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right)\)
\(\text{Step 4: Simplify.} \)
\(\,\,\,\,\,\,\cos\left(150°+45°\right)=\displaystyle-\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}\)
\(\,\,\,\,\,\,\cos\left(195°\right)=\displaystyle -\frac{\sqrt{6}+\sqrt{2}}{4}\)
\(\textbf{3)}\) \( \cos{\left(75°\right)} \) The exact value is \( \displaystyle \frac{\sqrt{6}-\sqrt{2}}{4} \)
\(\text{Step 1: Rewrite } \cos{\left(75°\right)} \text{ using familiar angles.} \)
\(\,\,\,\,\,\,\cos{\left(75°\right)} = \cos{\left(30°+45°\right)}\)
\(\text{Step 2: Use the sum formula.} \)
\(\,\,\,\,\,\,\cos{\left(A+B\right)} = \cos{\left(A\right)}\cos{\left(B\right)} – \sin{\left(A\right)}\sin{\left(B\right)}\)
\(\,\,\,\,\,\,\cos{\left(30°+45°\right)} = \cos{\left(30°\right)}\cos{\left(45°\right)} – \sin{\left(30°\right)}\sin{\left(45°\right)}\)
\(\text{Step 3: Evaluate.} \)
\(\,\,\,\,\,\,\cos{\left(30°+45°\right)} = \displaystyle\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) – \displaystyle\left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right)\)
\(\text{Step 4: Simplify.} \)
\(\,\,\,\,\,\,\cos{\left(30°+45°\right)} = \displaystyle\frac{\sqrt{6}}{4} – \frac{\sqrt{2}}{4}\)
\(\,\,\,\,\,\,\cos{\left(75°\right)} = \displaystyle\frac{\sqrt{6}-\sqrt{2}}{4}\)
\(\text{Step 1: Rewrite } \cos{\left(75°\right)} \text{ using familiar angles.} \)
\(\,\,\,\,\,\,\cos{\left(75°\right)} = \cos{\left(30°+45°\right)}\)
\(\text{Step 2: Use the sum formula.} \)
\(\,\,\,\,\,\,\cos{\left(A+B\right)} = \cos{\left(A\right)}\cos{\left(B\right)} – \sin{\left(A\right)}\sin{\left(B\right)}\)
\(\,\,\,\,\,\,\cos{\left(30°+45°\right)} = \cos{\left(30°\right)}\cos{\left(45°\right)} – \sin{\left(30°\right)}\sin{\left(45°\right)}\)
\(\text{Step 3: Evaluate.} \)
\(\,\,\,\,\,\,\cos{\left(30°+45°\right)} = \displaystyle\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) – \displaystyle\left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right)\)
\(\text{Step 4: Simplify.} \)
\(\,\,\,\,\,\,\cos{\left(30°+45°\right)} = \displaystyle\frac{\sqrt{6}}{4} – \frac{\sqrt{2}}{4}\)
\(\,\,\,\,\,\,\cos{\left(75°\right)} = \displaystyle\frac{\sqrt{6}-\sqrt{2}}{4}\)
\(\textbf{4)}\) \( \sin{\left(285°\right)} \) The exact value is \( \displaystyle -\frac{\sqrt{6}+\sqrt{2}}{4} \)
\(\sin\left(285^\circ\right)=\sin\left(240^\circ+45^\circ\right)\)
\(\sin(A+B)=\sin A\cos B+\cos A\sin B\)
\(\sin\left(240^\circ+45^\circ\right)=\sin(240^\circ)\cos(45^\circ)+\cos(240^\circ)\sin(45^\circ)\)
\(=\left(-\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)+\left(-\frac{1}{2}\right)\left(\frac{\sqrt2}{2}\right)\)
\(=-\frac{\sqrt6}{4}-\frac{\sqrt2}{4}\)
\(=-\frac{\sqrt6+\sqrt2}{4}\)
\(\sin\left(285^\circ\right)=\sin\left(240^\circ+45^\circ\right)\)
\(\sin(A+B)=\sin A\cos B+\cos A\sin B\)
\(\sin\left(240^\circ+45^\circ\right)=\sin(240^\circ)\cos(45^\circ)+\cos(240^\circ)\sin(45^\circ)\)
\(=\left(-\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)+\left(-\frac{1}{2}\right)\left(\frac{\sqrt2}{2}\right)\)
\(=-\frac{\sqrt6}{4}-\frac{\sqrt2}{4}\)
\(=-\frac{\sqrt6+\sqrt2}{4}\)
\(\textbf{5)}\) \( \cos{\left(-15°\right)} \) The exact value is \( \displaystyle \frac{\sqrt{6}+\sqrt{2}}{4} \)
\(\cos(-15^\circ)=\cos(30^\circ-45^\circ)\)
\(\cos(A-B)=\cos A\cos B+\sin A\sin B\)
\(\cos(30^\circ-45^\circ)=\cos(30^\circ)\cos(45^\circ)+\sin(30^\circ)\sin(45^\circ)\)
\(=\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{\sqrt2}{2}\right)\)
\(=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}\)
\(=\frac{\sqrt6+\sqrt2}{4}\)
\(\cos(-15^\circ)=\cos(30^\circ-45^\circ)\)
\(\cos(A-B)=\cos A\cos B+\sin A\sin B\)
\(\cos(30^\circ-45^\circ)=\cos(30^\circ)\cos(45^\circ)+\sin(30^\circ)\sin(45^\circ)\)
\(=\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{\sqrt2}{2}\right)\)
\(=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}\)
\(=\frac{\sqrt6+\sqrt2}{4}\)
\(\textbf{6)}\) Simplify \( \displaystyle \frac{\tan{\frac{11\pi}{8}}-\tan{\frac{3\pi}{8}}}{1+\tan{\frac{11\pi}{8}}\tan{\frac{3\pi}{8}}} \) The answer is \( 0 \)
\(\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\)
\(\displaystyle \frac{\tan{\frac{11\pi}{8}}-\tan{\frac{3\pi}{8}}}{1+\tan{\frac{11\pi}{8}}\tan{\frac{3\pi}{8}}}=\tan\left(\frac{11\pi}{8}-\frac{3\pi}{8}\right)\)
\(=\tan\left(\frac{8\pi}{8}\right)\)
\(=\tan(\pi)\)
\(=0\)
\(\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\)
\(\displaystyle \frac{\tan{\frac{11\pi}{8}}-\tan{\frac{3\pi}{8}}}{1+\tan{\frac{11\pi}{8}}\tan{\frac{3\pi}{8}}}=\tan\left(\frac{11\pi}{8}-\frac{3\pi}{8}\right)\)
\(=\tan\left(\frac{8\pi}{8}\right)\)
\(=\tan(\pi)\)
\(=0\)
\(\textbf{7)}\) \( \cos{\left(90°+\theta\right)}=-\sin{\theta} \) Verified
\(\,\,\,\,\, \cos{\left(90°+\theta\right)}=-\sin{\theta} \,\,\,\,\, \left(\text{Given}\right) \)
\(\,\,\,\,\, \cos{90^\circ}\cos{\theta}-\sin{90^\circ}\sin{\theta}=-\sin{\theta} \,\,\,\,\, \left(\text{Sum Angle Formula}\right) \)
\(\,\,\,\,\, (0)\cos{\theta}-(1)\sin{\theta}=-\sin{\theta} \,\,\,\,\, \left(\sin{90^\circ}=1 \text{ and } \cos(90^\circ)=0\right) \)
\(\,\,\,\,\, -\sin{\theta}=-\sin{\theta} \,\,\,\,\, \left(\text{Simplify}\right) \)
\(\,\,\,\,\, \cos{\left(90°+\theta\right)}=-\sin{\theta} \,\,\,\,\, \left(\text{Given}\right) \)
\(\,\,\,\,\, \cos{90^\circ}\cos{\theta}-\sin{90^\circ}\sin{\theta}=-\sin{\theta} \,\,\,\,\, \left(\text{Sum Angle Formula}\right) \)
\(\,\,\,\,\, (0)\cos{\theta}-(1)\sin{\theta}=-\sin{\theta} \,\,\,\,\, \left(\sin{90^\circ}=1 \text{ and } \cos(90^\circ)=0\right) \)
\(\,\,\,\,\, -\sin{\theta}=-\sin{\theta} \,\,\,\,\, \left(\text{Simplify}\right) \)
\(\textbf{8)}\) \(\sin\left(75^\circ\right)\) The exact value is \(\displaystyle\frac{\sqrt6+\sqrt2}{4}\)
\(\sin(75^\circ)=\sin(30^\circ+45^\circ)\)
\(\sin(A+B)=\sin A\cos B+\cos A\sin B\)
\(\sin(30^\circ+45^\circ)=\sin(30^\circ)\cos(45^\circ)+\cos(30^\circ)\sin(45^\circ)\)
\(=\left(\frac12\right)\left(\frac{\sqrt2}{2}\right)+\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)\)
\(=\frac{\sqrt2}{4}+\frac{\sqrt6}{4}\)
\(=\frac{\sqrt6+\sqrt2}{4}\)
\(\sin(75^\circ)=\sin(30^\circ+45^\circ)\)
\(\sin(A+B)=\sin A\cos B+\cos A\sin B\)
\(\sin(30^\circ+45^\circ)=\sin(30^\circ)\cos(45^\circ)+\cos(30^\circ)\sin(45^\circ)\)
\(=\left(\frac12\right)\left(\frac{\sqrt2}{2}\right)+\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)\)
\(=\frac{\sqrt2}{4}+\frac{\sqrt6}{4}\)
\(=\frac{\sqrt6+\sqrt2}{4}\)
\(\textbf{9)}\) \(\cos\left(15^\circ\right)\) The exact value is \(\displaystyle\frac{\sqrt6+\sqrt2}{4}\)
\(\cos(15^\circ)=\cos(60^\circ-45^\circ)\)
\(\cos(A-B)=\cos A\cos B+\sin A\sin B\)
\(\cos(60^\circ-45^\circ)=\cos(60^\circ)\cos(45^\circ)+\sin(60^\circ)\sin(45^\circ)\)
\(=\left(\frac12\right)\left(\frac{\sqrt2}{2}\right)+\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)\)
\(=\frac{\sqrt2}{4}+\frac{\sqrt6}{4}\)
\(=\frac{\sqrt6+\sqrt2}{4}\)
\(\cos(15^\circ)=\cos(60^\circ-45^\circ)\)
\(\cos(A-B)=\cos A\cos B+\sin A\sin B\)
\(\cos(60^\circ-45^\circ)=\cos(60^\circ)\cos(45^\circ)+\sin(60^\circ)\sin(45^\circ)\)
\(=\left(\frac12\right)\left(\frac{\sqrt2}{2}\right)+\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)\)
\(=\frac{\sqrt2}{4}+\frac{\sqrt6}{4}\)
\(=\frac{\sqrt6+\sqrt2}{4}\)
\(\textbf{10)}\) \(\sin\left(105^\circ\right)\) The exact value is \(\displaystyle\frac{\sqrt6+\sqrt2}{4}\)
\(\sin(105^\circ)=\sin(60^\circ+45^\circ)\)
\(\sin(A+B)=\sin A\cos B+\cos A\sin B\)
\(\sin(60^\circ+45^\circ)=\sin(60^\circ)\cos(45^\circ)+\cos(60^\circ)\sin(45^\circ)\)
\(=\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)+\left(\frac12\right)\left(\frac{\sqrt2}{2}\right)\)
\(=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}\)
\(=\frac{\sqrt6+\sqrt2}{4}\)
\(\sin(105^\circ)=\sin(60^\circ+45^\circ)\)
\(\sin(A+B)=\sin A\cos B+\cos A\sin B\)
\(\sin(60^\circ+45^\circ)=\sin(60^\circ)\cos(45^\circ)+\cos(60^\circ)\sin(45^\circ)\)
\(=\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)+\left(\frac12\right)\left(\frac{\sqrt2}{2}\right)\)
\(=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}\)
\(=\frac{\sqrt6+\sqrt2}{4}\)
\(\textbf{11)}\) \(\tan\left(75^\circ\right)\) The exact value is \(2+\sqrt3\)
\(\tan(75^\circ)=\tan(45^\circ+30^\circ)\)
\(\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\)
\(\tan(45^\circ+30^\circ)=\frac{1+\frac{\sqrt3}{3}}{1-1\cdot\frac{\sqrt3}{3}}\)
\(=\frac{\frac{3+\sqrt3}{3}}{\frac{3-\sqrt3}{3}}\)
\(=\frac{3+\sqrt3}{3-\sqrt3}\)
\(=2+\sqrt3\)
\(\tan(75^\circ)=\tan(45^\circ+30^\circ)\)
\(\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\)
\(\tan(45^\circ+30^\circ)=\frac{1+\frac{\sqrt3}{3}}{1-1\cdot\frac{\sqrt3}{3}}\)
\(=\frac{\frac{3+\sqrt3}{3}}{\frac{3-\sqrt3}{3}}\)
\(=\frac{3+\sqrt3}{3-\sqrt3}\)
\(=2+\sqrt3\)
\(\textbf{12)}\) \(\tan\left(15^\circ\right)\) The exact value is \(2-\sqrt3\)
\(\tan(15^\circ)=\tan(45^\circ-30^\circ)\)
\(\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\)
\(\tan(45^\circ-30^\circ)=\frac{1-\frac{\sqrt3}{3}}{1+1\cdot\frac{\sqrt3}{3}}\)
\(=\frac{3-\sqrt3}{3+\sqrt3}\)
\(=2-\sqrt3\)
\(\tan(15^\circ)=\tan(45^\circ-30^\circ)\)
\(\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\)
\(\tan(45^\circ-30^\circ)=\frac{1-\frac{\sqrt3}{3}}{1+1\cdot\frac{\sqrt3}{3}}\)
\(=\frac{3-\sqrt3}{3+\sqrt3}\)
\(=2-\sqrt3\)
\(\textbf{13)}\) \(\sin\left(\frac{7\pi}{12}\right)\) The exact value is \(\displaystyle\frac{\sqrt6+\sqrt2}{4}\)
\(\frac{7\pi}{12}=\frac{\pi}{3}+\frac{\pi}{4}\)
\(\sin\left(\frac{7\pi}{12}\right)=\sin\left(\frac{\pi}{3}+\frac{\pi}{4}\right)\)
\(\sin(A+B)=\sin A\cos B+\cos A\sin B\)
\(=\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)+\left(\frac12\right)\left(\frac{\sqrt2}{2}\right)\)
\(=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}\)
\(=\frac{\sqrt6+\sqrt2}{4}\)
\(\frac{7\pi}{12}=\frac{\pi}{3}+\frac{\pi}{4}\)
\(\sin\left(\frac{7\pi}{12}\right)=\sin\left(\frac{\pi}{3}+\frac{\pi}{4}\right)\)
\(\sin(A+B)=\sin A\cos B+\cos A\sin B\)
\(=\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)+\left(\frac12\right)\left(\frac{\sqrt2}{2}\right)\)
\(=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}\)
\(=\frac{\sqrt6+\sqrt2}{4}\)
\(\textbf{14)}\) \(\cos\left(\frac{5\pi}{12}\right)\) The exact value is \(\displaystyle\frac{\sqrt6-\sqrt2}{4}\)
\(\frac{5\pi}{12}=\frac{\pi}{3}+\frac{\pi}{12}=\frac{\pi}{3}+\frac{\pi}{4}-\frac{\pi}{6}\text{ is possible, but use }75^\circ.\)
\(\cos\left(\frac{5\pi}{12}\right)=\cos(75^\circ)\)
\(\cos(75^\circ)=\cos(30^\circ+45^\circ)\)
\(\cos(A+B)=\cos A\cos B-\sin A\sin B\)
\(=\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)-\left(\frac12\right)\left(\frac{\sqrt2}{2}\right)\)
\(=\frac{\sqrt6-\sqrt2}{4}\)
\(\frac{5\pi}{12}=\frac{\pi}{3}+\frac{\pi}{12}=\frac{\pi}{3}+\frac{\pi}{4}-\frac{\pi}{6}\text{ is possible, but use }75^\circ.\)
\(\cos\left(\frac{5\pi}{12}\right)=\cos(75^\circ)\)
\(\cos(75^\circ)=\cos(30^\circ+45^\circ)\)
\(\cos(A+B)=\cos A\cos B-\sin A\sin B\)
\(=\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)-\left(\frac12\right)\left(\frac{\sqrt2}{2}\right)\)
\(=\frac{\sqrt6-\sqrt2}{4}\)
\(\textbf{15)}\) \(\tan\left(\frac{5\pi}{12}\right)\) The exact value is \(2+\sqrt3\)
\(\frac{5\pi}{12}=75^\circ=45^\circ+30^\circ\)
\(\tan(75^\circ)=\tan(45^\circ+30^\circ)\)
\(\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\)
\(=\frac{1+\frac{\sqrt3}{3}}{1-\frac{\sqrt3}{3}}\)
\(=\frac{3+\sqrt3}{3-\sqrt3}\)
\(=2+\sqrt3\)
\(\frac{5\pi}{12}=75^\circ=45^\circ+30^\circ\)
\(\tan(75^\circ)=\tan(45^\circ+30^\circ)\)
\(\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\)
\(=\frac{1+\frac{\sqrt3}{3}}{1-\frac{\sqrt3}{3}}\)
\(=\frac{3+\sqrt3}{3-\sqrt3}\)
\(=2+\sqrt3\)
Challenge Problems
\(\textbf{16)}\) Verify \(\sin\left(\frac{\pi}{2}+x\right)=\cos{x}\) Verified
\(\sin\left(\frac{\pi}{2}+x\right)=\cos{x}\)
\(\sin(A+B)=\sin A\cos B+\cos A\sin B\)
\(\sin\left(\frac{\pi}{2}+x\right)=\sin\left(\frac{\pi}{2}\right)\cos{x}+\cos\left(\frac{\pi}{2}\right)\sin{x}\)
\(=(1)\cos{x}+(0)\sin{x}\)
\(=\cos{x}\)
\(\sin\left(\frac{\pi}{2}+x\right)=\cos{x}\)
\(\sin(A+B)=\sin A\cos B+\cos A\sin B\)
\(\sin\left(\frac{\pi}{2}+x\right)=\sin\left(\frac{\pi}{2}\right)\cos{x}+\cos\left(\frac{\pi}{2}\right)\sin{x}\)
\(=(1)\cos{x}+(0)\sin{x}\)
\(=\cos{x}\)
\(\textbf{17)}\) Verify \(\cos\left(\frac{\pi}{2}-x\right)=\sin{x}\) Verified
\(\cos\left(\frac{\pi}{2}-x\right)=\sin{x}\)
\(\cos(A-B)=\cos A\cos B+\sin A\sin B\)
\(\cos\left(\frac{\pi}{2}-x\right)=\cos\left(\frac{\pi}{2}\right)\cos{x}+\sin\left(\frac{\pi}{2}\right)\sin{x}\)
\(=(0)\cos{x}+(1)\sin{x}\)
\(=\sin{x}\)
\(\cos\left(\frac{\pi}{2}-x\right)=\sin{x}\)
\(\cos(A-B)=\cos A\cos B+\sin A\sin B\)
\(\cos\left(\frac{\pi}{2}-x\right)=\cos\left(\frac{\pi}{2}\right)\cos{x}+\sin\left(\frac{\pi}{2}\right)\sin{x}\)
\(=(0)\cos{x}+(1)\sin{x}\)
\(=\sin{x}\)
\(\textbf{18)}\) Simplify \(\sin(x+y)-\sin(x-y)\) The answer is \(2\cos{x}\sin{y}\)
\(\sin(x+y)=\sin x\cos y+\cos x\sin y\)
\(\sin(x-y)=\sin x\cos y-\cos x\sin y\)
\(\sin(x+y)-\sin(x-y)\)
\(=(\sin x\cos y+\cos x\sin y)-(\sin x\cos y-\cos x\sin y)\)
\(=2\cos x\sin y\)
\(\sin(x+y)=\sin x\cos y+\cos x\sin y\)
\(\sin(x-y)=\sin x\cos y-\cos x\sin y\)
\(\sin(x+y)-\sin(x-y)\)
\(=(\sin x\cos y+\cos x\sin y)-(\sin x\cos y-\cos x\sin y)\)
\(=2\cos x\sin y\)
\(\textbf{19)}\) Simplify \(\cos(x-y)-\cos(x+y)\) The answer is \(2\sin{x}\sin{y}\)
\(\cos(x-y)=\cos x\cos y+\sin x\sin y\)
\(\cos(x+y)=\cos x\cos y-\sin x\sin y\)
\(\cos(x-y)-\cos(x+y)\)
\(=(\cos x\cos y+\sin x\sin y)-(\cos x\cos y-\sin x\sin y)\)
\(=2\sin x\sin y\)
\(\cos(x-y)=\cos x\cos y+\sin x\sin y\)
\(\cos(x+y)=\cos x\cos y-\sin x\sin y\)
\(\cos(x-y)-\cos(x+y)\)
\(=(\cos x\cos y+\sin x\sin y)-(\cos x\cos y-\sin x\sin y)\)
\(=2\sin x\sin y\)
\(\textbf{20)}\) Verify \(\tan\left(\frac{\pi}{4}+x\right)=\frac{1+\tan{x}}{1-\tan{x}}\) Verified
\(\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\)
\(\tan\left(\frac{\pi}{4}+x\right)=\frac{\tan\left(\frac{\pi}{4}\right)+\tan{x}}{1-\tan\left(\frac{\pi}{4}\right)\tan{x}}\)
\(=\frac{1+\tan{x}}{1-1\cdot\tan{x}}\)
\(=\frac{1+\tan{x}}{1-\tan{x}}\)
\(\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\)
\(\tan\left(\frac{\pi}{4}+x\right)=\frac{\tan\left(\frac{\pi}{4}\right)+\tan{x}}{1-\tan\left(\frac{\pi}{4}\right)\tan{x}}\)
\(=\frac{1+\tan{x}}{1-1\cdot\tan{x}}\)
\(=\frac{1+\tan{x}}{1-\tan{x}}\)
See Related Pages\(\)
\(\bullet\text{ Right Triangle Trigonometry}\)
\(\,\,\,\,\,\,\,\,\sin{(x)}=\displaystyle\frac{\text{opp}}{\text{hyp}}…\)
\(\bullet\text{ Angle of Depression and Elevation}\)
\(\,\,\,\,\,\,\,\,\text{Angle of Depression}=\text{Angle of Elevation}…\)
\(\bullet\text{ Convert to Radians and to Degrees}\)
\(\,\,\,\,\,\,\,\,\text{Radians} \rightarrow \text{Degrees}, \times \displaystyle \frac{180^{\circ}}{\pi}…\)
\(\bullet\text{ Degrees, Minutes and Seconds}\)
\(\,\,\,\,\,\,\,\,48^{\circ}34’21”…\)
\(\bullet\text{ Coterminal Angles}\)
\(\,\,\,\,\,\,\,\,\pm 360^{\circ} \text { or } \pm 2\pi n…\)
\(\bullet\text{ Reference Angles}\)
\(\,\,\,\,\,\,\,\,\)
\(…\)
\(\bullet\text{ Find All 6 Trig Functions}\)
\(\,\,\,\,\,\,\,\,\)
\(…\)
\(\bullet\text{ Unit Circle}\)
\(\,\,\,\,\,\,\,\,\sin{(60^{\circ})}=\displaystyle\frac{\sqrt{3}}{2}…\)
\(\bullet\text{ Law of Sines}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{\sin{A}}{a}=\frac{\sin{B}}{b}=\frac{\sin{C}}{c}\) 
\(…\)
\(\bullet\text{ Area of SAS Triangles}\)
\(\,\,\,\,\,\,\,\,\text{Area}=\frac{1}{2}ab \sin{C}\) \(…\)
\(\bullet\text{ Law of Cosines}\)
\(\,\,\,\,\,\,\,\,a^2=b^2+c^2-2bc \cos{A}\) \(…\)
\(\bullet\text{ Area of SSS Triangles (Heron’s formula)}\)
\(\,\,\,\,\,\,\,\,\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}\) \(…\)
\(\bullet\text{ Geometric Mean}\)
\(\,\,\,\,\,\,\,\,x=\sqrt{ab} \text{ or } \displaystyle\frac{a}{x}=\frac{x}{b}…\)
\(\bullet\text{ Geometric Mean- Similar Right Triangles}\)
\(\,\,\,\,\,\,\,\,\)
\(…\)
\(\bullet\text{ Inverse Trigonmetric Functions}\)
\(\,\,\,\,\,\,\,\,\sin {\left(cos^{-1}\left(\frac{3}{5}\right)\right)}…\)
\(\bullet\text{ Sum and Difference of Angles Formulas}\)
\(\,\,\,\,\,\,\,\,\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}…\)
\(\bullet\text{ Double-Angle and Half-Angle Formulas}\)
\(\,\,\,\,\,\,\,\,\sin{(2A)}=2\sin{(A)}\cos{(A)}…\)
\(\bullet\text{ Trigonometry-Pythagorean Identities}\)
\(\,\,\,\,\,\,\,\,\sin^2{(x)}+\cos^2{(x)}=1…\)
\(\bullet\text{ Product-Sum Identities}\)
\(\,\,\,\,\,\,\,\,\cos{\alpha}\cos{\beta}=\left(\displaystyle\frac{\cos{(\alpha+\beta)}+\cos{(\alpha-\beta)}}{2}\right)…\)
\(\bullet\text{ Cofunction Identities}\)
\(\,\,\,\,\,\,\,\,\sin{(x)}=\cos{(\frac{\pi}{2}-x)}…\)
\(\bullet\text{ Proving Trigonometric Identities}\)
\(\,\,\,\,\,\,\,\,\sec{x}-\cos{x}=\displaystyle\frac{\tan^2{x}}{\sec{x}}…\)
\(\bullet\text{ Graphing Trig Functions- sin and cos}\)
\(\,\,\,\,\,\,\,\,f(x)=A \sin{B(x-c)}+D \) 
\(…\)
\(\bullet\text{ Solving Trigonometric Equations}\)
\(\,\,\,\,\,\,\,\,2\cos{(x)}=\sqrt{3}…\)