Double-angle and half-angle identities are used in trigonometry to rewrite expressions involving twice an angle or half an angle. These formulas are helpful for finding exact trig values, simplifying expressions, solving equations, and verifying identities. This page includes examples with sine, cosine, tangent, quadrant information, and identity proofs.

 

Notes

 

Practice Problems

\(\textbf{1)}\) \(\text{Find exact value of }\sin{165^{\circ}}\)

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The answer is \(\displaystyle\frac{\sqrt{6}-\sqrt{2}}{4}\)

Show Work

\(\,\,\,\,\,165^\circ=\frac{330^\circ}{2}\)

\(\,\,\,\,\,\sin\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos\theta}{2}}\)

\(\,\,\,\,\,\sin(165^\circ)=\sqrt{\frac{1-\cos(330^\circ)}{2}}\)

\(\,\,\,\,\,\sin(165^\circ)=\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}\)

\(\,\,\,\,\,\sin(165^\circ)=\sqrt{\frac{2-\sqrt3}{4}}\)

\(\,\,\,\,\,\sin(165^\circ)=\frac{\sqrt{2-\sqrt3}}{2}\)

\(\,\,\,\,\,\sin(165^\circ)=\frac{\sqrt6-\sqrt2}{4}\)

 

\(\textbf{2)}\) \(\text{Find exact value of }\cos{75^{\circ}}\)

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The answer is \(\displaystyle\frac{\sqrt{2-\sqrt{3}}}{2}\)

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\(\,\,\,\,\,75^\circ=\frac{150^\circ}{2}\)

\(\,\,\,\,\,\cos\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1+\cos\theta}{2}}\)

\(\,\,\,\,\,\cos(75^\circ)=\sqrt{\frac{1+\cos(150^\circ)}{2}}\)

\(\,\,\,\,\,\cos(75^\circ)=\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}\)

\(\,\,\,\,\,\cos(75^\circ)=\sqrt{\frac{2-\sqrt3}{4}}\)

\(\,\,\,\,\,\cos(75^\circ)=\frac{\sqrt{2-\sqrt3}}{2}\)

 

\(\textbf{3)}\) \(\text{Find exact value of }\tan{67.5^{\circ}}\)

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The answer is \(\sqrt{2}+1\)

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\(\,\,\,\,\,67.5^\circ=\frac{135^\circ}{2}\)

\(\,\,\,\,\,\tan\left(\frac{\theta}{2}\right)=\frac{\sin\theta}{1+\cos\theta}\)

\(\,\,\,\,\,\tan(67.5^\circ)=\frac{\sin(135^\circ)}{1+\cos(135^\circ)}\)

\(\,\,\,\,\,\tan(67.5^\circ)=\frac{\frac{\sqrt2}{2}}{1-\frac{\sqrt2}{2}}\)

\(\,\,\,\,\,\tan(67.5^\circ)=\frac{\sqrt2}{2-\sqrt2}\)

\(\,\,\,\,\,\tan(67.5^\circ)=\frac{\sqrt2(2+\sqrt2)}{(2-\sqrt2)(2+\sqrt2)}\)

\(\,\,\,\,\,\tan(67.5^\circ)=\sqrt2+1\)

 

\(\textbf{4)}\) \(\text{Find exact value of }\cos{15^{\circ}}\)

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The answer is \(\displaystyle\frac{\sqrt{2+\sqrt{3}}}{2}\)

Show Work

\(\,\,\,\,\,15^\circ=\frac{30^\circ}{2}\)

\(\,\,\,\,\,\cos\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1+\cos\theta}{2}}\)

\(\,\,\,\,\,\cos(15^\circ)=\sqrt{\frac{1+\cos(30^\circ)}{2}}\)

\(\,\,\,\,\,\cos(15^\circ)=\sqrt{\frac{1+\frac{\sqrt3}{2}}{2}}\)

\(\,\,\,\,\,\cos(15^\circ)=\sqrt{\frac{2+\sqrt3}{4}}\)

\(\,\,\,\,\,\cos(15^\circ)=\frac{\sqrt{2+\sqrt3}}{2}\)

 

\(\textbf{5)}\) \(\sin\theta=\frac{3}{5},\) and \(90^{\circ}\lt\theta\lt180^{\circ},\) find \(\sin ⁡2\theta\)

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The answer is \(\frac{-24}{25}\)

Show Work

\(\,\,\,\,\,\sin\theta=\frac{3}{5}\)

\(\,\,\,\,\,90^\circ\lt\theta\lt180^\circ,\text{ so }\theta\text{ is in Quadrant II}\)

\(\,\,\,\,\,\cos\theta=-\frac{4}{5}\)

\(\,\,\,\,\,\sin(2\theta)=2\sin\theta\cos\theta\)

\(\,\,\,\,\,\sin(2\theta)=2\left(\frac{3}{5}\right)\left(-\frac{4}{5}\right)\)

\(\,\,\,\,\,\sin(2\theta)=-\frac{24}{25}\)

 

\(\textbf{6)}\) \(\sin\theta=\frac{3}{5},\) and \(90^{\circ}\lt\theta\lt180^{\circ},\) find \(\cos ⁡2\theta\)

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The answer is \(\frac{7}{25}\)

Show Work

\(\,\,\,\,\,\sin\theta=\frac{3}{5}\)

\(\,\,\,\,\,90^\circ\lt\theta\lt180^\circ,\text{ so }\theta\text{ is in Quadrant II}\)

\(\,\,\,\,\,\cos\theta=-\frac{4}{5}\)

\(\,\,\,\,\,\cos(2\theta)=\cos^2\theta-\sin^2\theta\)

\(\,\,\,\,\,\cos(2\theta)=\left(-\frac{4}{5}\right)^2-\left(\frac{3}{5}\right)^2\)

\(\,\,\,\,\,\cos(2\theta)=\frac{16}{25}-\frac{9}{25}\)

\(\,\,\,\,\,\cos(2\theta)=\frac{7}{25}\)

 

\(\textbf{7)}\) \(\sin\theta=\frac{3}{5},\) and \(90^{\circ}\lt\theta\lt180^{\circ},\) find \(\tan ⁡2\theta\)

Show Answer

The answer is \(\frac{-24}{7}\)

Show Work

\(\,\,\,\,\,\sin(2\theta)=-\frac{24}{25}\)

\(\,\,\,\,\,\cos(2\theta)=\frac{7}{25}\)

\(\,\,\,\,\,\tan(2\theta)=\frac{\sin(2\theta)}{\cos(2\theta)}\)

\(\,\,\,\,\,\tan(2\theta)=\frac{-\frac{24}{25}}{\frac{7}{25}}\)

\(\,\,\,\,\,\tan(2\theta)=-\frac{24}{7}\)

 

\(\textbf{8)}\) \(\sin\theta=\frac{3}{5},\) and \(90^{\circ}\lt\theta\lt180^{\circ},\) find \(\sin \frac{\theta}{2}\)

Show Answer

The answer is \(\frac{3\sqrt{10}}{10}\)

Show Work

\(\,\,\,\,\,90^\circ\lt\theta\lt180^\circ,\text{ so }45^\circ\lt\frac{\theta}{2}\lt90^\circ\)

\(\,\,\,\,\,\frac{\theta}{2}\text{ is in Quadrant I, so sine is positive.}\)

\(\,\,\,\,\,\cos\theta=-\frac{4}{5}\)

\(\,\,\,\,\,\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\cos\theta}{2}}\)

\(\,\,\,\,\,\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\left(-\frac{4}{5}\right)}{2}}\)

\(\,\,\,\,\,\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{\frac{9}{5}}{2}}\)

\(\,\,\,\,\,\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{9}{10}}\)

\(\,\,\,\,\,\sin\left(\frac{\theta}{2}\right)=\frac{3\sqrt{10}}{10}\)

 

\(\textbf{9)}\) \(\sin\theta=\frac{3}{5},\) and \(90^{\circ}\lt\theta\lt180^{\circ},\) find \(\cos \frac{\theta}{2}\)

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The answer is \(\frac{\sqrt{10}}{10}\)

Show Work

\(\,\,\,\,\,90^\circ\lt\theta\lt180^\circ,\text{ so }45^\circ\lt\frac{\theta}{2}\lt90^\circ\)

\(\,\,\,\,\,\frac{\theta}{2}\text{ is in Quadrant I, so cosine is positive.}\)

\(\,\,\,\,\,\cos\theta=-\frac{4}{5}\)

\(\,\,\,\,\,\cos\left(\frac{\theta}{2}\right)=\sqrt{\frac{1+\cos\theta}{2}}\)

\(\,\,\,\,\,\cos\left(\frac{\theta}{2}\right)=\sqrt{\frac{1+\left(-\frac{4}{5}\right)}{2}}\)

\(\,\,\,\,\,\cos\left(\frac{\theta}{2}\right)=\sqrt{\frac{\frac{1}{5}}{2}}\)

\(\,\,\,\,\,\cos\left(\frac{\theta}{2}\right)=\sqrt{\frac{1}{10}}\)

\(\,\,\,\,\,\cos\left(\frac{\theta}{2}\right)=\frac{\sqrt{10}}{10}\)

 

\(\textbf{10)}\) \(\sin\theta=\frac{3}{5},\) and \(90^{\circ}\lt\theta\lt180^{\circ},\) find \(\tan \frac{\theta}{2}\)

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The answer is \(3\)

Show Work

\(\,\,\,\,\,\tan\left(\frac{\theta}{2}\right)=\frac{\sin\theta}{1+\cos\theta}\)

\(\,\,\,\,\,\tan\left(\frac{\theta}{2}\right)=\frac{\frac{3}{5}}{1+\left(-\frac{4}{5}\right)}\)

\(\,\,\,\,\,\tan\left(\frac{\theta}{2}\right)=\frac{\frac{3}{5}}{\frac{1}{5}}\)

\(\,\,\,\,\,\tan\left(\frac{\theta}{2}\right)=3\)

 

\(\textbf{11)}\) \(\sin{A}=\frac{4}{5},\) and \(0^{\circ}\lt A \lt90^{\circ},\) find \(\tan{2A}\)

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The answer is \(-\frac{24}{7}\)

Show Work

\(\,\,\,\,\, \sin{A}=\frac{4}{5}=\frac{\text{opposite}}{\text{hypotenuse}}\)

\(\,\,\,\,\, \text{opposite}=4, \text{hypotenuse}=5\)

\(\,\,\,\,\, \text{(adjacent)}^2+4^2=5^2\)

\(\,\,\,\,\, \text{adjacent}=3\)

\(\,\,\,\,\, \tan{A}=\frac{\text{opposite}}{\text{adjacent}}=\frac{4}{3}\)

\(\,\,\,\,\, \tan{2A}=\frac{2\tan{A}}{1-\tan^2{A}}\)

\(\,\,\,\,\, \tan{2A}=\frac{2(4/3)}{1-(4/3)^2}\)

\(\,\,\,\,\, \tan{2A}=\frac{8/3}{1-16/9}\)

\(\,\,\,\,\, \tan{2A}=\frac{8/3}{9/9-16/9}\)

\(\,\,\,\,\, \tan{2A}=\frac{8/3}{-7/9}\)

\(\,\,\,\,\, \tan{2A}=\frac{8}{3}*\left(-\frac{9}{7}\right)\)

\(\,\,\,\,\, \tan{2A}=-\frac{24}{7}\)

 

\(\textbf{12)}\) \(\sin{A}=\frac{4}{5},\) and \(90^{\circ}\lt A \lt180^{\circ},\) find \(\tan{2A}\)

Show Answer

The answer is \(\frac{24}{7}\)

Show Work

\(\,\,\,\,\, \sin{A}=\frac{4}{5}=\frac{\text{opposite}}{\text{hypotenuse}}\)

\(\,\,\,\,\, \text{opposite}=4, \text{hypotenuse}=5\)

\(\,\,\,\,\, \text{(adjacent)}^2+4^2=5^2\)

\(\,\,\,\,\, \text{adjacent}=-3\)

\(\,\,\,\,\, \tan{A}=\frac{\text{opposite}}{\text{adjacent}}=-\frac{4}{3}\)

\(\,\,\,\,\, \tan{2A}=\frac{2\tan{A}}{1-\tan^2{A}}\)

\(\,\,\,\,\, \tan{2A}=\frac{2(-4/3)}{1-(-4/3)^2}\)

\(\,\,\,\,\, \tan{2A}=\frac{-8/3}{1-16/9}\)

\(\,\,\,\,\, \tan{2A}=\frac{-8/3}{9/9-16/9}\)

\(\,\,\,\,\, \tan{2A}=\frac{-8/3}{-7/9}\)

\(\,\,\,\,\, \tan{2A}=-\frac{8}{3}*\left(-\frac{9}{7}\right)\)

\(\,\,\,\,\, \tan{2A}=\frac{24}{7}\)

 

\(\textbf{13)}\) \(2 \sin (75) \cos (75)⁡ \)

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The answer is \(\frac{1}{2}\)

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\(\,\,\,\,\,2\sin{x}\cos{x}=\sin(2x)\)

\(\,\,\,\,\,2\sin(75^\circ)\cos(75^\circ)=\sin(150^\circ)\)

\(\,\,\,\,\,\sin(150^\circ)=\frac{1}{2}\)

 

\(\textbf{14)}\) \(\cos^2(22.5)-\sin^2⁡(22.5)\)

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The answer is \(\frac{\sqrt{2}}{2}\)

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\(\,\,\,\,\,\cos^2{x}-\sin^2{x}=\cos(2x)\)

\(\,\,\,\,\,\cos^2(22.5^\circ)-\sin^2(22.5^\circ)=\cos(45^\circ)\)

\(\,\,\,\,\,\cos(45^\circ)=\frac{\sqrt2}{2}\)

 

\(\textbf{15)}\) \(\displaystyle\frac{\sin{2x}}{\sin{x}}-\frac{\cos{2x}}{\cos{x}}=\sec{x}\)

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Verified

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\(\,\,\,\,\,\displaystyle\frac{\sin{2x}}{\sin{x}}-\frac{\cos{2x}}{\cos{x}}=\sec{x}\,\,\,\,\, \left(\text{Given}\right) \)

\(\,\,\,\,\,\displaystyle\frac{2\sin{x}\cos{x}}{\sin{x}}-\frac{2\cos^2{x}-1}{\cos{x}}=\sec{x}\,\,\,\,\, \left(\text{Double Angle Formulas}\right) \)

\(\,\,\,\,\,\displaystyle\frac{2\cos{x}}{1}-\frac{2\cos^2{x}-1}{\cos{x}}=\sec{x}\,\,\,\,\, \left(\text{Simplify}\right) \)

\(\,\,\,\,\,\displaystyle\frac{2\cos^2{x}}{\cos{x}}-\frac{2\cos^2{x}-1}{\cos{x}}=\sec{x}\,\,\,\,\, \left(\text{Common Denominator}\right) \)

\(\,\,\,\,\,\displaystyle\frac{2\cos^2{x}-\left(2\cos^2{x}-1\right)}{\cos{x}}=\sec{x}\,\,\,\,\, \left(\text{Combine into 1 fraction}\right) \)

\(\,\,\,\,\,\displaystyle\frac{2\cos^2{x}-2\cos^2{x}+1}{\cos{x}}=\sec{x}\,\,\,\,\, \left(\text{Distribute the negative}\right) \)

\(\,\,\,\,\,\displaystyle\frac{1}{\cos{x}}=\sec{x}\,\,\,\,\, \left(\text{Simplify}\right) \)

\(\,\,\,\,\,\sec{x}=\sec{x}\,\,\,\,\, \left(\text{Reciprocal Identities}\right) \)

 

Challenge Problems

\(\textbf{16)}\) \(\displaystyle\frac{1-\tan^2{x}}{1+\tan^2{x}}=\cos{2x}\)

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Verified

Show Work

\(\,\,\,\,\,\displaystyle\frac{1-\tan^2{x}}{1+\tan^2{x}}=\cos{2x}\,\,\,\,\, \left(\text{Given}\right) \)

\(\,\,\,\,\,\displaystyle\frac{1-\tan^2{x}}{\sec^2{x}}=\cos{2x}\,\,\,\,\, \left(\text{Pythagorean Identities}\right) \)

\(\,\,\,\,\,\displaystyle\frac{\cos^2{x}\left(1-\tan^2{x}\right)}{1}=\cos{2x}\,\,\,\,\, \left(\text{Reciprocal Identities}\right) \)

\(\,\,\,\,\,\cos^2{x}-\cos^2{x}\tan^2{x}=\cos{2x}\,\,\,\,\, \left(\text{Distribution}\right) \)

\(\,\,\,\,\,\cos^2{x}-\cos^2{x}\frac{\sin^2{x}}{\cos^2{x}}=\cos{2x}\,\,\,\,\, \left(\text{Quotient Identities}\right) \)

\(\,\,\,\,\,\cos^2{x}-\sin^2{x}=\cos{2x}\,\,\,\,\, \left(\text{Simplify}\right)\)

\(\,\,\,\,\,\cos{2x}=\cos{2x}\,\,\,\,\, \left(\text{Double Angle Formulas}\right)\)

 

\(\textbf{17)}\) \(\sin{A}=\frac{5}{13},\) and \(0^{\circ}\lt A \lt90^{\circ},\) find \(\sin{2A}\)

Show Answer

The answer is \(\frac{120}{169}\)

Show Work

\(\,\,\,\,\,\sin{A}=\frac{5}{13}\)

\(\,\,\,\,\,0^\circ\lt A\lt90^\circ,\text{ so }A\text{ is in Quadrant I}\)

\(\,\,\,\,\,\cos{A}=\frac{12}{13}\)

\(\,\,\,\,\,\sin(2A)=2\sin A\cos A\)

\(\,\,\,\,\,\sin(2A)=2\left(\frac{5}{13}\right)\left(\frac{12}{13}\right)\)

\(\,\,\,\,\,\sin(2A)=\frac{120}{169}\)

 

\(\textbf{18)}\) \(\cos{A}=-\frac{5}{13},\) and \(90^{\circ}\lt A \lt180^{\circ},\) find \(\cos{2A}\)

Show Answer

The answer is \(-\frac{119}{169}\)

Show Work

\(\,\,\,\,\,\cos{A}=-\frac{5}{13}\)

\(\,\,\,\,\,90^\circ\lt A \lt180^\circ,\text{ so }A\text{ is in Quadrant II}\)

\(\,\,\,\,\,\sin{A}=\frac{12}{13}\)

\(\,\,\,\,\,\cos(2A)=\cos^2 A-\sin^2 A\)

\(\,\,\,\,\,\cos(2A)=\left(-\frac{5}{13}\right)^2-\left(\frac{12}{13}\right)^2\)

\(\,\,\,\,\,\cos(2A)=\frac{25}{169}-\frac{144}{169}\)

\(\,\,\,\,\,\cos(2A)=-\frac{119}{169}\)

 

\(\textbf{19)}\) \(\cos\theta=\frac{7}{25},\) and \(270^{\circ}\lt\theta\lt360^{\circ},\) find \(\sin\frac{\theta}{2}\)

Show Answer

The answer is \(\frac{3}{5}\)

Show Work

\(\,\,\,\,\,270^\circ\lt\theta\lt360^\circ,\text{ so }135^\circ\lt\frac{\theta}{2}\lt180^\circ\)

\(\,\,\,\,\,\frac{\theta}{2}\text{ is in Quadrant II, so sine is positive.}\)

\(\,\,\,\,\,\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\cos\theta}{2}}\)

\(\,\,\,\,\,\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\frac{7}{25}}{2}}\)

\(\,\,\,\,\,\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{\frac{18}{25}}{2}}\)

\(\,\,\,\,\,\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{9}{25}}\)

\(\,\,\,\,\,\sin\left(\frac{\theta}{2}\right)=\frac{3}{5}\)

 

\(\textbf{20)}\) \(\tan\theta=-\frac{8}{15},\) and \(90^{\circ}\lt\theta\lt180^{\circ},\) find \(\tan{2\theta}\)

Show Answer

The answer is \(-\frac{240}{161}\)

Show Work

\(\,\,\,\,\,\tan(2\theta)=\frac{2\tan\theta}{1-\tan^2\theta}\)

\(\,\,\,\,\,\tan(2\theta)=\frac{2\left(-\frac{8}{15}\right)}{1-\left(-\frac{8}{15}\right)^2}\)

\(\,\,\,\,\,\tan(2\theta)=\frac{-\frac{16}{15}}{1-\frac{64}{225}}\)

\(\,\,\,\,\,\tan(2\theta)=\frac{-\frac{16}{15}}{\frac{161}{225}}\)

\(\,\,\,\,\,\tan(2\theta)=-\frac{16}{15}\cdot\frac{225}{161}\)

\(\,\,\,\,\,\tan(2\theta)=-\frac{240}{161}\)

 

 

See Related Pages\(\)

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\(\bullet\text{ Right Triangle Trigonometry}\)
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\(\,\,\,\,\,\,\,\,\text{Area}=\frac{1}{2}ab \sin{C}\) \(…\)

\(\bullet\text{ Law of Cosines}\)
\(\,\,\,\,\,\,\,\,a^2=b^2+c^2-2bc \cos{A}\) \(…\)

\(\bullet\text{ Area of SSS Triangles (Heron’s formula)}\)
\(\,\,\,\,\,\,\,\,\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}\) \(…\)

\(\bullet\text{ Geometric Mean}\)
\(\,\,\,\,\,\,\,\,x=\sqrt{ab} \text{ or } \displaystyle\frac{a}{x}=\frac{x}{b}…\)

\(\bullet\text{ Geometric Mean- Similar Right Triangles}\)
\(\,\,\,\,\,\,\,\,\)\(…\)

\(\bullet\text{ Inverse Trigonmetric Functions}\)
\(\,\,\,\,\,\,\,\,\sin {\left(cos^{-1}\left(\frac{3}{5}\right)\right)}…\)

\(\bullet\text{ Sum and Difference of Angles Formulas}\)
\(\,\,\,\,\,\,\,\,\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}…\)

\(\bullet\text{ Double-Angle and Half-Angle Formulas}\)
\(\,\,\,\,\,\,\,\,\sin{(2A)}=2\sin{(A)}\cos{(A)}…\)

\(\bullet\text{ Trigonometry-Pythagorean Identities}\)
\(\,\,\,\,\,\,\,\,\sin^2{(x)}+\cos^2{(x)}=1…\)

\(\bullet\text{ Product-Sum Identities}\)
\(\,\,\,\,\,\,\,\,\cos{\alpha}\cos{\beta}=\left(\displaystyle\frac{\cos{(\alpha+\beta)}+\cos{(\alpha-\beta)}}{2}\right)…\)

\(\bullet\text{ Cofunction Identities}\)
\(\,\,\,\,\,\,\,\,\sin{(x)}=\cos{(\frac{\pi}{2}-x)}…\)

\(\bullet\text{ Proving Trigonometric Identities}\)
\(\,\,\,\,\,\,\,\,\sec{x}-\cos{x}=\displaystyle\frac{\tan^2{x}}{\sec{x}}…\)

\(\bullet\text{ Graphing Trig Functions- sin and cos}\)
\(\,\,\,\,\,\,\,\,f(x)=A \sin{B(x-c)}+D \) \(…\)

\(\bullet\text{ Solving Trigonometric Equations}\)
\(\,\,\,\,\,\,\,\,2\cos{(x)}=\sqrt{3}…\)