Half Angle Formula

Trigonometry equalities, inequalities and expressions - sin, cos, tan, cot

Half Angle Formula

Postby Guest » Fri Sep 28, 2018 12:50 am

Trying to find the number that goes in the red box. Please help
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Re: Half Angle Formula

Postby Math Tutor » Fri Sep 28, 2018 6:17 am

Please, look at the formulas here:
Power-Reducing Formulas

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Re: Half Angle Formula

Postby Guest » Fri Sep 28, 2018 10:53 pm

Formulas of degrees lowering : [tex]sin^{2}[/tex][tex]\alpha[/tex]=[tex]\frac{1-cos(2\alpha)}{2}[/tex] ; [tex]cos^{2}[/tex][tex]\alpha[/tex]=[tex]\frac{1+cos(2\alpha)}{2}[/tex]

[tex]sin^{4}[/tex](2x)=[tex](sin^{2}(2x) )^{2}[/tex]=[tex][\frac{1-cos(2.2x)}{2}]^{2}[/tex]=[[tex]\frac{1-cos(4x)}{2}]^{2}[/tex]=

=[tex]\frac{1-2cos(4x)+cos^{2}(4x)}{4}[/tex]=[tex]\frac{1}{4}[/tex]-[tex]\frac{2cos(4x)}{4}[/tex]+[tex]\frac{cos^{2}(4x)}{4}[/tex]=

=[tex]\frac{1}{4}[/tex]-[tex]\frac{1}{2}[/tex]cos(4x)+[tex]\frac{\frac{1+cos(2.4x)}{2}}{4}[/tex]=[tex]\frac{1}{4}[/tex]-[tex]\frac{1}{2}[/tex]cos(4x)+[tex]\frac{1+cos(8x)}{8}[/tex]=[tex]\frac{1}{4}[/tex]-[tex]\frac{1}{2}[/tex]cos(4x)+[tex]\frac{1}{8}[/tex]+[tex]\frac{cos(8x)}{8}[/tex]=

=[tex]\frac{3}{8}[/tex]-[tex]\frac{1}{2}[/tex]cos(4x)+[tex]\frac{1}{8}[/tex]cos(8x)
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