What does #-csc(arc cot(7))+2csc(arctan(5))# equal?

2 Answers
Jun 10, 2016

-5.03

Explanation:

S = -csc(arccot (7)) + 2csc(arctan (5))
Use calculator -->
a. cot x = 7 --> tan x = 1/7
arccot(7) = arctan(1/7) --> arc #x = 8^@13#
sin x = sin 8^@13 = 0.14 --> #csc x = 1/0.14 = 7.07#

b. tan y = 5 --> arc #y = 78^@69#
sin y = sin 78^@69 = 0.98
#csc y = 1/(sin 78^@69) = 1/0.98 = 1.02#
Finally,
S = - 707 + 2(1.02) = -5.03

Jun 12, 2016

#-sqrt50+(2sqrt26)/5approx-5.03146#

Explanation:

This is solvable without a calculator. It all depends on drawing pictures of the triangles.

For #-csc("arccot"(7))#, this is asking for the cosecant of the triangle where the cotangent is already equal to #7#.

Since cotangent is equal to the adjacent side of the angle in question divided by the opposite side, we can say that #"adjacent"=7# and #"opposite"=1#.

Through the Pythagorean Theorem, #"hypotenuse"=sqrt50#.

Since we want to find cosecant of this triangle, we will take the hypotenuse over the opposite side, so

#csc("arccot"(7))=sqrt50/1#

and

#-csc("arccot"(7))=-sqrt50#

We can find #csc(arctan(5))# through a similar method.

If the tangent of an angle is #5#, then #"opposite"=5# and #"adjacent"=1#. Again, through the Pythagorean Theorem, we see that #"hypotenuse"=sqrt26#.

We want to find the cosecant of this angle as well, which will be #"hypotenuse"/"opposite"=sqrt26/5#.

Thus

#csc(arctan(5))=sqrt26/5#

and

#2csc(arctan(5))=(2sqrt26)/5#

So, combining these, we see that

#-csc("arccot"(7))+2csc(arctan(5))=-sqrt50+(2sqrt26)/5#