How do you use a half-angle formula to find the exact value of #cos22.5#?

2 Answers
Aug 10, 2018

#color(maroon)(cos 22.5^@ = + sqrt((sqrt2 -1)/(2 sqrt2)) ~~ + 0.3827#

Explanation:

http://www2.clarku.edu/~djoyce/trig/identities.html

#cos (theta/2) = +- sqrt((1-cos theta) / 2)#

Let #hat (theta/2) = 22.5^@#

#hat theta = 2 * 22.5 = 45 ^@#

cos (theta/2) = cos 22.5^@ = + sqrt((1 - cos 45) / 2)#

We know #cos 45 = 1 / sqrt2#

#:. color(maroon)(cos 22.5^@ = + sqrt((1 - 1/sqrt2)/2) = + sqrt((sqrt2 -1)/(2 sqrt2)) ~~ + 0.3827#

Aug 10, 2018

#cos22.5^@=1/2(sqrt(2+sqrt2))#

Explanation:

#"using the "color(blue)"half-angle formula for cos"#

#•color(white)(x)cos(x/2)=+-sqrt((1+cosx)/2)#

#cos22.5^@=+sqrt((1+cos45^@)/2)#

#color(white)(xxxxxx)=sqrt((1+sqrt2/2)/2)#

#color(white)(xxxxxx)=sqrt((2+sqrt2)/4)#

#color(white)(xxxxxx)=1/2(sqrt(2+sqrt2))#