For right ∆ABC with right ∠C, prove each of the following. a. sin A < 1 b. cos A < 1
both sin and cos have maximum values at 1, what angles does that occure at for each
um sorry im not to sure
since the radius is 1, all the hypotenuse are 1, so sin(a) = y and cos(a) = x each point on the circle (x , y) = (cos(a), sin(a))
i have to write a proof
start by stating givens , i havent done a formal one in awhile
try and get through it before doing the formal proof maybe
given sin A< 1
the given triangles are right the sum of the 3 angles in a triangle are 180, so the angles A and B add to 90
the range of values for cos(a) and sin(a), are from -1 to 1 cos(0) = 1 cos(90) = -1, for example
ok
angles A + B + C = 180, and C is said to be 90, so A + B = 90
ok
we are talking about a distance, lengths of the sides, forget about the negative part on the domain of sin and cos, so sin(a) < 1 cos(b) < 1
ok
that restricts all the angles to the first quadrant on that circle, and not including 90, or 0
to prove those, say A+B=90 so angles A and B must both be some value between 0 and 90, not including 0 or 90
ok thanks!!
then apply that to the definition of sin and cos sin(0) = 0 sin(90 = 1 sin has to be less than 1
ok thanks
welcome, i think most of the ideas are there, just organize em,
ok i will thanks so mcuh
overall the remaining unknown angles A and B have to total 90, so A can be 89.99999 and B the rest 0.00001 deg, and that works for example that statement will tell you what the possible values for sin and cos will be
0 < A < 90 0 < B < 90 sin(0) < sin(A) < Sin(90) 0 < sin (A) < 1
sme for cos(A) 0 < A < 90 sin(0) < sin ( A) < sin(90) 0 < sin (A) < 1
the hard part is remembering all the 'official' names for properties and stuff for the reasons
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