Half and Double Angle Identities We have already derived formulas for the sum and difference of angles which allowed us to precisely calculate the sine, cosine, and tangent of many new angles using the special angles. We can take this a step further by deriving formulas for the double and half angles of sine, cosine, and tangent which will gives us even more tools to manipulate the angles inside of trigonometric functions.
Double Angle To derive the double angle formulas, we can use the sum formulas by letting θ = α \theta = \alpha θ = α θ = β \theta = \beta θ = β θ + θ = 2 θ \theta + \theta = 2\theta θ + θ = 2 θ
The double angle formula for sin ( 2 θ ) \sin(2\theta) sin ( 2 θ )
sin ( α + β ) = sin ( α ) cos ( β ) + cos ( α ) sin ( β ) sin ( θ + θ ) = sin ( θ ) cos ( θ ) + cos ( θ ) sin ( θ ) sin ( 2 θ ) = sin ( θ ) cos ( θ ) + cos ( θ ) sin ( θ ) = 2 sin ( θ ) cos ( θ ) \begin{array}{lllllllll} \sin(\alpha + \beta) &=& \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \\[0.5em] \sin(\theta + \theta) &=& \sin(\theta)\cos(\theta) + \cos(\theta)\sin(\theta) \\[0.5em] \sin(2\theta) &=& \sin(\theta)\cos(\theta) + \cos(\theta)\sin(\theta) \\[0.5em] &=& 2\sin(\theta)\cos(\theta) \\[0.5em] \end{array} sin ( α + β ) sin ( θ + θ ) sin ( 2 θ ) = = = = sin ( α ) cos ( β ) + cos ( α ) sin ( β ) sin ( θ ) cos ( θ ) + cos ( θ ) sin ( θ ) sin ( θ ) cos ( θ ) + cos ( θ ) sin ( θ ) 2 sin ( θ ) cos ( θ ) The double angle formula for cos ( 2 θ ) \cos(2\theta) cos ( 2 θ )
cos ( α + β ) = cos ( α ) cos ( β ) − sin ( α ) sin ( β ) cos ( θ + θ ) = cos ( θ ) cos ( θ ) − sin ( θ ) sin ( θ ) cos ( 2 θ ) = cos ( θ ) cos ( θ ) − sin ( θ ) sin ( θ ) = cos 2 ( θ ) − sin 2 ( θ ) \begin{array}{lllllllll} \cos(\alpha + \beta) &=& \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) \\[0.5em] \cos(\theta + \theta) &=& \cos(\theta)\cos(\theta) - \sin(\theta)\sin(\theta) \\[0.5em] \cos(2\theta) &=& \cos(\theta)\cos(\theta) - \sin(\theta)\sin(\theta) \\[0.5em] &=& \cos^2(\theta) - \sin^2(\theta) \\[0.5em] \end{array} cos ( α + β ) cos ( θ + θ ) cos ( 2 θ ) = = = = cos ( α ) cos ( β ) − sin ( α ) sin ( β ) cos ( θ ) cos ( θ ) − sin ( θ ) sin ( θ ) cos ( θ ) cos ( θ ) − sin ( θ ) sin ( θ ) cos 2 ( θ ) − sin 2 ( θ ) We can create two more double angles formulas for cos ( 2 θ ) \cos(2\theta) cos ( 2 θ ) sin 2 ( θ ) + cos 2 ( θ ) = 1 \sin^2(\theta) + \cos^2(\theta) = 1 sin 2 ( θ ) + cos 2 ( θ ) = 1 sin 2 ( θ ) \sin^2(\theta) sin 2 ( θ ) cos 2 ( θ ) \cos^2(\theta) cos 2 ( θ )
cos ( 2 θ ) = cos 2 ( θ ) − sin 2 ( θ ) = ( 1 − sin 2 ( θ ) ) − sin 2 ( θ ) = 1 − 2 sin 2 ( θ ) \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = (1 - \sin^2(\theta)) - \sin^2(\theta) = 1 - 2\sin^2(\theta) cos ( 2 θ ) = cos 2 ( θ ) − sin 2 ( θ ) = ( 1 − sin 2 ( θ )) − sin 2 ( θ ) = 1 − 2 sin 2 ( θ ) cos ( 2 θ ) = cos 2 ( θ ) − sin 2 ( θ ) = cos 2 ( θ ) − ( 1 − cos 2 ( θ ) ) = 2 cos 2 ( θ ) − 1 \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = \cos^2(\theta) - (1 - \cos^2(\theta)) = 2\cos^2(\theta) - 1 cos ( 2 θ ) = cos 2 ( θ ) − sin 2 ( θ ) = cos 2 ( θ ) − ( 1 − cos 2 ( θ )) = 2 cos 2 ( θ ) − 1 Finally, the double angle formula for tan ( 2 θ ) \tan(2\theta) tan ( 2 θ )
tan ( α + β ) = tan ( α ) + tan ( β ) 1 − tan ( α ) tan ( β ) tan ( θ + θ ) = tan ( θ ) + tan ( θ ) 1 − tan ( θ ) tan ( θ ) tan ( 2 θ ) = tan ( θ ) + tan ( θ ) 1 − tan 2 ( θ ) = 2 tan ( θ ) 1 − tan 2 ( θ ) \begin{array}{lllllllll} \tan(\alpha + \beta) &=& \dfrac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)} \\[1.5em] \tan(\theta + \theta) &=& \dfrac{\tan(\theta) + \tan(\theta)}{1 - \tan(\theta)\tan(\theta)} \\[1.5em] \tan(2\theta) &=& \dfrac{\tan(\theta) + \tan(\theta)}{1 - \tan^2(\theta)} \\[1.5em] &=& \dfrac{2\tan(\theta)}{1 - \tan^2(\theta)} \\[1.5em] \end{array} tan ( α + β ) tan ( θ + θ ) tan ( 2 θ ) = = = = 1 − tan ( α ) tan ( β ) tan ( α ) + tan ( β ) 1 − tan ( θ ) tan ( θ ) tan ( θ ) + tan ( θ ) 1 − tan 2 ( θ ) tan ( θ ) + tan ( θ ) 1 − tan 2 ( θ ) 2 tan ( θ ) Verify Identities As we grow our toolbox of identities, we are able to manipulate trigonometric expressions in more ways. We can use these identities like the double angle formulas to verify other identities...
E x a m p l e ‾ \underline{Example} E x am pl e
Verify the identity cos 4 ( θ ) − sin 4 ( θ ) = cos ( 2 θ ) \cos^4(\theta) - \sin^4(\theta) = \cos(2\theta) cos 4 ( θ ) − sin 4 ( θ ) = cos ( 2 θ )
We can manipulate cos 4 θ − sin 4 θ \cos^{4}\theta - \sin^{4}\theta cos 4 θ − sin 4 θ cos ( 2 θ ) \cos(2\theta) cos ( 2 θ )
cos ( 2 θ ) = ( cos 2 θ ) 2 − ( sin 2 θ ) 2 = ( cos 2 θ − sin 2 θ ) ( cos 2 θ + sin 2 θ ) = ( cos 2 θ − sin 2 θ ) ( 1 ) = cos 2 θ − sin 2 θ = cos ( 2 θ ) \begin{array}{lllllllll} \cos(2\theta) &=& (\cos^{2}\theta)^2 - (\sin^{2}\theta)^2 \\[1.5em] &=& (\cos^{2}\theta - \sin^{2}\theta)(\cos^{2}\theta + \sin^{2}\theta) \\[1.5em] &=& (\cos^{2}\theta - \sin^{2}\theta)(1) \\[1.5em] &=& \cos^{2}\theta - \sin^{2}\theta \\[1.5em] &=& \cos(2\theta) \\[1.5em] \end{array} cos ( 2 θ ) = = = = = ( cos 2 θ ) 2 − ( sin 2 θ ) 2 ( cos 2 θ − sin 2 θ ) ( cos 2 θ + sin 2 θ ) ( cos 2 θ − sin 2 θ ) ( 1 ) cos 2 θ − sin 2 θ cos ( 2 θ ) This verifies the identity.
The reduction formulas are formulas which allow us to reduce the power of a trigonometric function to a lower power. This can be useful when we are trying to integrate trigonometric functions and we can derive them from the double angles formulas.
The reduction formula for sin 2 θ \sin^{2}\theta sin 2 θ
cos ( 2 θ ) = 1 − 2 sin 2 θ cos ( 2 θ ) − 1 = − 2 sin 2 θ 1 − cos ( 2 θ ) = 2 sin 2 θ 1 − cos ( 2 θ ) 2 = sin 2 θ \begin{array}{lllllllll} \cos(2\theta) &=& 1 - 2\sin^{2}\theta \\[1.5em] \cos(2\theta) - 1 &=& -2\sin^{2}\theta \\[1.5em] 1 - \cos(2\theta) &=& 2\sin^{2}\theta \\[1.5em] \dfrac{1 - \cos(2\theta)}{2} &=& \sin^{2}\theta \\[1.5em] \end{array} cos ( 2 θ ) cos ( 2 θ ) − 1 1 − cos ( 2 θ ) 2 1 − cos ( 2 θ ) = = = = 1 − 2 sin 2 θ − 2 sin 2 θ 2 sin 2 θ sin 2 θ The reduction formula for cos 2 θ \cos^{2}\theta cos 2 θ
cos ( 2 θ ) = 2 cos 2 θ − 1 1 + cos ( 2 θ ) = 2 cos 2 θ 1 + cos ( 2 θ ) 2 = cos 2 θ \begin{array}{lllllllll} \cos(2\theta) &=& 2\cos^{2}\theta - 1 \\[1.5em] 1 + \cos(2\theta) &=& 2\cos^{2}\theta \\[1.5em] \dfrac{1 + \cos(2\theta)}{2} &=& \cos^{2}\theta \\[1.5em] \end{array} cos ( 2 θ ) 1 + cos ( 2 θ ) 2 1 + cos ( 2 θ ) = = = 2 cos 2 θ − 1 2 cos 2 θ cos 2 θ Finally, the reduction formula for tan 2 θ \tan^{2}\theta tan 2 θ
tan 2 θ = sin 2 θ cos 2 θ = ( 1 − cos ( 2 θ ) 2 ) ÷ ( 1 + cos ( 2 θ ) 2 ) = ( 1 − cos ( 2 θ ) 2 ) ( 2 1 + cos ( 2 θ ) ) = 1 − cos ( 2 θ ) 1 + cos ( 2 θ ) \begin{array}{lllllllll} \tan^{2}\theta &=& \dfrac{\sin^{2}\theta}{\cos^{2}\theta} \\[1.5em] &=& (\dfrac{1 - \cos(2\theta)}{2}) \div (\dfrac{1 + \cos(2\theta)}{2}) \\[1.5em] &=& (\dfrac{1 - \cos(2\theta)}{2})(\dfrac{2}{1 + \cos(2\theta)}) \\[1.5em] &=& \dfrac{1 - \cos(2\theta)}{1 + \cos(2\theta)} \\[1.5em] \end{array} tan 2 θ = = = = cos 2 θ sin 2 θ ( 2 1 − cos ( 2 θ ) ) ÷ ( 2 1 + cos ( 2 θ ) ) ( 2 1 − cos ( 2 θ ) ) ( 1 + cos ( 2 θ ) 2 ) 1 + cos ( 2 θ ) 1 − cos ( 2 θ ) Half Angle The half angle formulas are formulas which allow us to find the sine, cosine, and tangent of half angles. These formulas are useful for finding the exact value of angles that are half of special angles and we can derive them by substituting α 2 \dfrac{\alpha}{2} 2 α θ \theta θ
The half angle formula for sin ( α 2 ) \sin(\dfrac{\alpha}{2}) sin ( 2 α )
sin 2 θ = 1 − cos ( 2 θ ) 2 sin 2 ( α 2 ) = 1 − cos ( 2 ( α / 2 ) ) 2 = 1 − cos ( α ) 2 sin ( α 2 ) = ± 1 − cos α 2 \begin{array}{lllllllll} \sin^{2}\theta &=& \dfrac{1 - \cos(2\theta)}{2} \\[1.5em] \sin^{2}(\dfrac{\alpha}{2}) &=& \dfrac{1 - \cos(2 (\alpha / 2))}{2} \\[1.5em] &=& \dfrac{1 - \cos(\alpha)}{2} \\[1.5em] \sin(\dfrac{\alpha}{2}) &=& \pm\sqrt{\dfrac{1 - \cos \alpha}{2}} \\[1.5em] \end{array} sin 2 θ sin 2 ( 2 α ) sin ( 2 α ) = = = = 2 1 − cos ( 2 θ ) 2 1 − cos ( 2 ( α /2 )) 2 1 − cos ( α ) ± 2 1 − cos α The half angle formula for cos ( α 2 ) \cos(\dfrac{\alpha}{2}) cos ( 2 α )
cos 2 θ = 1 + cos ( 2 θ ) 2 cos 2 ( α 2 ) = 1 + cos ( 2 ( α / 2 ) ) 2 = 1 + cos ( α ) 2 cos ( α 2 ) = ± 1 + cos α 2 \begin{array}{lllllllll} \cos^{2}\theta &=& \dfrac{1 + \cos(2\theta)}{2} \\[1.5em] \cos^{2}(\dfrac{\alpha}{2}) &=& \dfrac{1 + \cos(2 (\alpha / 2))}{2} \\[1.5em] &=& \dfrac{1 + \cos(\alpha)}{2} \\[1.5em] \cos(\dfrac{\alpha}{2}) &=& \pm\sqrt{\dfrac{1 + \cos \alpha}{2}} \\[1.5em] \end{array} cos 2 θ cos 2 ( 2 α ) cos ( 2 α ) = = = = 2 1 + cos ( 2 θ ) 2 1 + cos ( 2 ( α /2 )) 2 1 + cos ( α ) ± 2 1 + cos α Finally, the half angle formula for tan ( α 2 ) \tan(\dfrac{\alpha}{2}) tan ( 2 α )
tan 2 θ = 1 − cos ( 2 θ ) 1 + cos ( 2 θ ) tan 2 ( α 2 ) = 1 − cos ( 2 ( α / 2 ) ) 1 + cos ( 2 ( α / 2 ) ) = 1 − cos ( α ) 1 + cos ( α ) tan ( α 2 ) = ± 1 − cos α 1 + cos α \begin{array}{lllllllll} \tan^{2}\theta &=& \dfrac{1 - \cos(2\theta)}{1 + \cos(2\theta)} \\[1.5em] \tan^{2}(\dfrac{\alpha}{2}) &=& \dfrac{1 - \cos(2 (\alpha / 2))}{1 + \cos(2 (\alpha / 2))} \\[1.5em] &=& \dfrac{1 - \cos(\alpha)}{1 + \cos(\alpha)} \\[1.5em] \tan(\dfrac{\alpha}{2}) &=& \pm\sqrt{\dfrac{1 - \cos \alpha}{1 + \cos \alpha}} \\[1.5em] \end{array} tan 2 θ tan 2 ( 2 α ) tan ( 2 α ) = = = = 1 + cos ( 2 θ ) 1 − cos ( 2 θ ) 1 + cos ( 2 ( α /2 )) 1 − cos ( 2 ( α /2 )) 1 + cos ( α ) 1 − cos ( α ) ± 1 + cos α 1 − cos α Summary The double angles formulas are...
sin ( 2 θ ) = 2 sin ( θ ) cos ( θ ) cos ( 2 θ ) = cos 2 ( θ ) − sin 2 ( θ ) = 2 cos 2 ( θ ) − 1 = 1 − 2 sin 2 ( θ ) tan ( 2 θ ) = 2 tan ( θ ) 1 − tan 2 ( θ ) \begin{array}{lllllllll} \sin(2\theta) &=& 2\sin(\theta)\cos(\theta) \\[1.5em] \cos(2\theta) &=& \cos^{2}(\theta) - \sin^{2}(\theta) \\[0.5em] &=& 2\cos^{2}(\theta) - 1 \\[0.5em] &=& 1 - 2\sin^{2}(\theta) \\[1.5em] \tan(2\theta) &=& \dfrac{2\tan(\theta)}{1 - \tan^{2}(\theta)} \\[1.5em] \end{array} sin ( 2 θ ) cos ( 2 θ ) tan ( 2 θ ) = = = = = 2 sin ( θ ) cos ( θ ) cos 2 ( θ ) − sin 2 ( θ ) 2 cos 2 ( θ ) − 1 1 − 2 sin 2 ( θ ) 1 − tan 2 ( θ ) 2 tan ( θ ) The half angle formulas are...
sin ( α 2 ) = ± 1 − cos α 2 cos ( α 2 ) = ± 1 + cos α 2 tan ( α 2 ) = ± 1 − cos α 1 + cos α \begin{array}{lllllllll} \sin(\dfrac{\alpha}{2}) &=& \pm\sqrt{\dfrac{1 - \cos \alpha}{2}} \\[1.5em] \cos(\dfrac{\alpha}{2}) &=& \pm\sqrt{\dfrac{1 + \cos \alpha}{2}} \\[1.5em] \tan(\dfrac{\alpha}{2}) &=& \pm\sqrt{\dfrac{1 - \cos \alpha}{1 + \cos \alpha}} \\[1.5em] \end{array} sin ( 2 α ) cos ( 2 α ) tan ( 2 α ) = = = ± 2 1 − cos α ± 2 1 + cos α ± 1 + cos α 1 − cos α Finally, the reduction formulas are...
sin 2 θ = 1 − cos ( 2 θ ) 2 cos 2 θ = 1 + cos ( 2 θ ) 2 tan 2 θ = 1 − cos ( 2 θ ) 1 + cos ( 2 θ ) \begin{array}{lllllllll} \sin^{2}\theta &=& \dfrac{1 - \cos(2\theta)}{2} \\[1.5em] \cos^{2}\theta &=& \dfrac{1 + \cos(2\theta)}{2} \\[1.5em] \tan^{2}\theta &=& \dfrac{1 - \cos(2\theta)}{1 + \cos(2\theta)} \\[1.5em] \end{array} sin 2 θ cos 2 θ tan 2 θ = = = 2 1 − cos ( 2 θ ) 2 1 + cos ( 2 θ ) 1 + cos ( 2 θ ) 1 − cos ( 2 θ )