This page covers the double-angle and half-angle identities used in trigonometry to simplify expressions and solve equations. You’ll find clear formulas, and a variety of practice problems to help reinforce how and when to apply each identity.

 

Notes

 

Questions

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

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

 

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

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

 

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

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

 

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

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

 

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

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

 

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

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

 

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

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

 

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

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

 

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

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

 

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

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The answer is \(3\)

 

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

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

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\(\,\,\,\,\, \sin{A}=\frac{4}{5}=\frac{\text{opposite}}{\text{hypotenuse}}\)
\(\,\,\,\,\, \text{opposite}=4, \text{hypotenuse}=5\)
\(\,\,\,\,\, \text{(adjacent)}^2+4^2=5^2\)
\(\,\,\,\,\, \text{adjacent}=3\)
\(\,\,\,\,\, \tan{A}=\frac{\text{opposite}}{\text{adjacent}}=\frac{4}{3}\)
\(\,\,\,\,\, \tan{2A}=\frac{2\tan{A}}{1-\tan^2{A}}\)
\(\,\,\,\,\, \tan{2A}=\frac{2(4/3)}{1-(4/3)^2}\)
\(\,\,\,\,\, \tan{2A}=\frac{8/3}{1-16/9}\)
\(\,\,\,\,\, \tan{2A}=\frac{8/3}{9/9-16/9}\)
\(\,\,\,\,\, \tan{2A}=\frac{8/3}{-7/9}\)
\(\,\,\,\,\, \tan{2A}=\frac{8}{3}*\left(-\frac{9}{7}\right)\)
\(\,\,\,\,\, \tan{2A}=-\frac{24}{7}\)

 

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

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

Show Work

\(\,\,\,\,\, \sin{A}=\frac{4}{5}=\frac{\text{opposite}}{\text{hypotenuse}}\)
\(\,\,\,\,\, \text{opposite}=4, \text{hypotenuse}=5\)
\(\,\,\,\,\, \text{(adjacent)}^2+4^2=5^2\)
\(\,\,\,\,\, \text{adjacent}=-3\)
\(\,\,\,\,\, \tan{A}=\frac{\text{opposite}}{\text{adjacent}}=-\frac{4}{3}\)
\(\,\,\,\,\, \tan{2A}=\frac{2\tan{A}}{1-\tan^2{A}}\)
\(\,\,\,\,\, \tan{2A}=\frac{2(-4/3)}{1-(-4/3)^2}\)
\(\,\,\,\,\, \tan{2A}=\frac{-8/3}{1-16/9}\)
\(\,\,\,\,\, \tan{2A}=\frac{-8/3}{9/9-16/9}\)
\(\,\,\,\,\, \tan{2A}=\frac{-8/3}{-7/9}\)
\(\,\,\,\,\, \tan{2A}=-\frac{8}{3}*\left(-\frac{9}{7}\right)\)
\(\,\,\,\,\, \tan{2A}=\frac{24}{7}\)

 

Find each value.

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

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

 

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

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

 

Verify the Identities using Double Angle Formulas

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

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\(\,\,\,\,\,\displaystyle\frac{\sin{2x}}{\sin{x}}-\frac{\cos{2x}}{\cos{x}}=\sec{x}\,\,\,\,\, \left(\text{Given}\right) \)
\(\,\,\,\,\,\displaystyle\frac{2\sin{x}\cos{x}}{\sin{x}}-\frac{2\cos^2{x}-1}{\cos{x}}=\sec{x}\,\,\,\,\, \left(\text{Double Angle Formulas}\right) \)
\(\,\,\,\,\,\displaystyle\frac{2\cos{x}}{1}-\frac{2\cos^2{x}-1}{\cos{x}}=\sec{x}\,\,\,\,\, \left(\text{Simplify}\right) \)
\(\,\,\,\,\,\displaystyle\frac{2\cos^2{x}}{\cos{x}}-\frac{2\cos^2{x}-1}{\cos{x}}=\sec{x}\,\,\,\,\, \left(\text{Common Denominator}\right) \)
\(\,\,\,\,\,\displaystyle\frac{2\cos^2{x}-\left(2\cos^2{x}-1\right)}{\cos{x}}=\sec{x}\,\,\,\,\, \left(\text{Combine into 1 fraction}\right) \)
\(\,\,\,\,\,\displaystyle\frac{2\cos^2{x}-2\cos^2{x}+1}{\cos{x}}=\sec{x}\,\,\,\,\, \left(\text{Distribute the negative}\right) \)
\(\,\,\,\,\,\displaystyle\frac{1}{\cos{x}}=\sec{x}\,\,\,\,\, \left(\text{Simplify}\right) \)
\(\,\,\,\,\,\sec{x}=\sec{x}\,\,\,\,\, \left(\text{Reciprocal Identities}\right) \)

 

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

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\(\,\,\,\,\,\displaystyle\frac{1-\tan^2{x}}{1+\tan^2{x}}=\cos{2x}\,\,\,\,\, \left(\text{Given}\right) \)
\(\,\,\,\,\,\displaystyle\frac{1-\tan^2{x}}{\sec^2{x}}=\cos{2x}\,\,\,\,\, \left(\text{Pythagorean Identities}\right) \)
\(\,\,\,\,\,\displaystyle\frac{\cos^2{x}\left(1-\tan^2{x}\right)}{1}=\cos{2x}\,\,\,\,\, \left(\text{Reciprocal Identities}\right) \)
\(\,\,\,\,\,\cos^2{x}-\cos^2{x}\tan^2{x}=\cos{2x}\,\,\,\,\, \left(\text{Distribution}\right) \)
\(\,\,\,\,\,\cos^2{x}-\cos^2{x}\frac{\sin^2{x}}{\cos^2{x}}=\cos{2x}\,\,\,\,\, \left(\text{Quotient Identities}\right) \)
\(\,\,\,\,\,\cos^2{x}-\sin^2{x}=\cos{2x}\,\,\,\,\, \left(\text{Simplify}\right)\)
\(\,\,\,\,\,\cos{2x}=\cos{2x}\,\,\,\,\, \left(\text{Double Angle Formulas}\right)\)

 

 

See Related Pages\(\)

\(\bullet\text{ Trig Calculator }\)
\(\,\,\,\,\,\,\,\,\text{(Symbolab.com)}\)

\(\bullet\text{ Right Triangle Trigonometry}\)
\(\,\,\,\,\,\,\,\,\sin{(x)}=\displaystyle\frac{\text{opp}}{\text{hyp}}…\)

\(\bullet\text{ Angle of Depression and Elevation}\)
\(\,\,\,\,\,\,\,\,\text{Angle of Depression}=\text{Angle of Elevation}…\)

\(\bullet\text{ Convert to Radians and to Degrees}\)
\(\,\,\,\,\,\,\,\,\text{Radians} \rightarrow \text{Degrees}, \times \displaystyle \frac{180^{\circ}}{\pi}…\)

\(\bullet\text{ Degrees, Minutes and Seconds}\)
\(\,\,\,\,\,\,\,\,48^{\circ}34’21”…\)

\(\bullet\text{ Coterminal Angles}\)
\(\,\,\,\,\,\,\,\,\pm 360^{\circ} \text { or } \pm 2\pi n…\)

\(\bullet\text{ Reference Angles}\)
\(\,\,\,\,\,\,\,\,\)\(…\)

\(\bullet\text{ Find All 6 Trig Functions}\)
\(\,\,\,\,\,\,\,\,\)\(…\)

\(\bullet\text{ Unit Circle}\)
\(\,\,\,\,\,\,\,\,\sin{(60^{\circ})}=\displaystyle\frac{\sqrt{3}}{2}…\)

\(\bullet\text{ Law of Sines}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{\sin{A}}{a}=\frac{\sin{B}}{b}=\frac{\sin{C}}{c}\) \(…\)

\(\bullet\text{ Area of SAS Triangles}\)
\(\,\,\,\,\,\,\,\,\text{Area}=\frac{1}{2}ab \sin{C}\) \(…\)

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

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

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

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

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

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

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

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

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

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

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

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

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