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Mathematics 16 Online
OpenStudy (anonymous):

If sin x = .21,what is cos x?

OpenStudy (solomonzelman):

I would do it like this. \[\sin x = .21\]\[\sin ^2x = (.21)^2\] \[We~~know~~that:~~Cos^2x=1-Sin^2x~~~~~~S O,~...\]\[\sin ^2x = (.21)^2~~~~~~->~~~~~-\sin^2x=-(.21)^2\] \[1-\sin^2x=\cos^2x~~~~~~~Sub,~~i n~~~~-(.21)^2~~~for~~~-\sin^2x\]\[1-(.21)^2=\cos^2x\] I think you got it from here...This isn't a good approach.. I am tired -:(

OpenStudy (anonymous):

.98

OpenStudy (solomonzelman):

How do you get it ?

OpenStudy (anonymous):

21^2 + a^2 = 100^2 a^2 = 100^2 - 21^2 a^2 = (100 - 21)(100 + 21) a^2 = 79 x 121 a^2 = 9559 a = sqrt (9559) = 97.77... Cos x = a/h = 97.77.. / 100 Cos x = 0.977777... Cos x = 0.98

OpenStudy (solomonzelman):

Okay !

OpenStudy (anonymous):

i did good right?

OpenStudy (solomonzelman):

Yes

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