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What is the trigonometric function sec?
The function sec(x) is the secant function. It is related to the other functions by the expression 1/cos(x). It is not the inverse cosine or arccosine, it is one over the cosine function. Ex. cos(pi/4)= sqrt(2)/2 therefore secant is sec(pi/4)= 1/sqrt(2)/2 or 2/sqrt(2).
What is the exact value of the expression cos 7pi over 12 cos pi over 6 -sin 7pi over 12 sin pi over 6?
cos(a)cos(b)-sin(a)sin(b)=cos(a+b) a=7pi/12 and b=pi/6 a+b = 7pi/12 + pi/6 = 7pi/12 + 2pi/12 = 9pi/12 We want to find cos(9pi/12) cos(9pi/12) = cos(3pi/4) cos(3pi/4)= cos(pi-pi/4) cos(pi)cos(pi/4)-sin(pi)sin(pi/4) cos(pi)=-1 sin(pi)=0 cos(pi/4) = √2/2 sin(pi/4) =√2/2 cos(pi)cos(pi/4)-sin(pi)sin(pi/4) = - cos(pi/4) = -√2/2
Why is the period of secant 2 pi?
It is the same period as cosine function which is 2 pi because sec x = 1/cos x
What is the reciprocal of 4 - pi?
pi-4 is the opposite and 1/4-pi is the reciprocal
What is cot of pi - pi over 4 given that tan of pi over 4 equals 1?
First: note 3 things about cot and tan, and note the given statement:cot = 1/tantan is cyclic with a period of π, that is tan(nπ + x) = tan(x)tan is an odd function, that is tan(-x) = -tan(x)tan(π/4) = 1Now apply them to the problem:cot(π - π/4) = 1/tan(π - π/4)= 1/tan(-π/4)= 1/-tan(π/4)= 1/-1 = -1Thus:cot(π - π/4) = -1.
