How do you calculate #Cos^-1(5/13) - Cos^-1(8/17)#?

2 Answers
Truong-Son N.
Jun 15, 2015

With a calculator. A TI-8# can do this with 2nd+COS. 13 and 17 are prime numbers, so 5/13 and 8/17 will give you irrational values. This isn't easily doable enough by hand to be something you need to do by hand.

#arccos(5/13) - arccos(8/17) = 67.38^o - 61.93^o = 5.45^o ~~ 0.095 "rad"#

Antoine
Jul 3, 2015

#=cos^(-1)(220/221)#

Explanation:

Another way would be to let #A=cos^(-1)(5/13)# and #B=cos^(-1)(8/17)#

#=>cosA=5/13#

And , #cosB=8/17#

From the identity #cos^2x+sin^2x=1#
We can obtain that,

#sinx=sqrt(1-cos^2x)#

#=>sinA=sqrt(1-(5/13)^2)=sqrt(144/169)=12/13#

Similarly , #=>sinB=sqrt(1-(8/17)^2)=sqrt(225/289)=15/17#

Now, say #" "C=A-B#

#=>cosC=cos(A-B)=cosAcosB+sinAsinB=5/13*8/17+12/13*15/17=40/221+180/221=220/221#

#=>cosC=220/221#

#=>C=cos^(-1)(220/221)#

#=>A-B=cos^(-1)(220/221)#

#=cos^(-1)(5/13)-cos^(-1)(8/17)=cos^(-1)(220/221)#

Related questions
See all questions in Inverse Trigonometric Properties
Impact of this question
8286 views around the world
You can reuse this answer
Creative Commons License