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What is the exact value of cos (67.5°)?
OA.
OB.
OC. -√2+√2
√2+√2
2
-
√2-√2
2
OD. √2+ √2
4


What Is The Exact Value Of Cos 675 OA OB OC 22 22 2 22 2 OD 2 2 4 class=

Sagot :

first off, make sure you have a Unit Circle, if you don't do get one, you'll need it, you can find many online.

let's double up 67.5°, that way we can use the half-angle identity for the cosine of it, so hmmm twice 67.5 is simply 135°, keeping in mind that 135° is really 90° + 45°, and that whilst 135° is on the 2nd Quadrant and its cosine is negative 67.5° is on the 1st Quadrant where cosine is positive, so

[tex]cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta) \\\\\\ cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}} \\\\[-0.35em] ~\dotfill\\\\ cos(135^o)\implies cos(90^o+45^o)\implies cos(90^o)cos(45^o)~~ - ~~sin(90^o)sin(45^o) \\\\\\ \left( 0 \right)\left( \cfrac{\sqrt{2}}{2} \right)~~ - ~~\left( 1\right)\left( \cfrac{\sqrt{2}}{2} \right)\implies -\cfrac{\sqrt{2}}{2} \\\\[-0.35em] ~\dotfill[/tex]

[tex]cos(67.5^o)\implies cos\left( \frac{135^o}{2} \right)\implies \pm \sqrt{\cfrac{ ~~ 1-\frac{\sqrt{2} ~~ }{2}}{2}}\implies \stackrel{I~Quadrant}{+\sqrt{\cfrac{ ~~ 1-\frac{\sqrt{2} ~~ }{2}}{2}}} \\\\\\ \sqrt{\cfrac{ ~~ \frac{2-\sqrt{2}}{2} ~~ }{2}}\implies \sqrt{\cfrac{2-\sqrt{2}}{4}}\implies \cfrac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}\implies \cfrac{\sqrt{2-\sqrt{2}}}{2}[/tex]

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