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If sin x = .21,what is cos x?
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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 -:(
.98
How do you get it ?
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
Okay !
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i did good right?
Yes
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