How would I go on about solving:
cos(2⋅arctan(12)) ?
My attempt was to find v in sin(v)cos(v)=12 and substitute that with arctan(12), but I end up with sin(v)=√15 which I cant solve.
Is there any other way to approach the problem?
How to find limh→0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
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