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cos(9pi/4+5pi/6)=
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\[\cos(u+v)=\cos(u)\cos(v)-\sin(u)\sin(v)\] is what you need for this one
what does u and v stand for?
I wonder why they just didn't say \(\cos\left(\dfrac{37\pi}{12} \right)\)
set it up like this cos(9pi/4 + 5pi/6)=cos9pi/4(cos5pi/6)-sin9pi/4(sin5pi/6)=
@ParthKohli because you have no idea what that is
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@satellite73 ?
OK well, Wolfram does.
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