How can i solve this sin,cos,tan Mathematical equation? #sin(cos^-1(63/65)+2tan^-1(3/4))# Help me guys...

#sin(cos^-1(63/65)+2tan^-1(3/4))#

2 Answers
Nghi N
May 16, 2017

0.999

Explanation:

The easiest way is to use calculator:
#arctan (3/4) = 36^@87#
#2arctan (3/4) = 73^@75#
#arccos (63/65) = 14^@25#
#(arccos (63/65) + 2arctan (3/4)) = 88^@#
sin (88^@) = 0.999

Remind: #sin 90^@ = 1#

NickTheTurtle
May 16, 2017

#1624/1625#

Explanation:

Remember the compound-angle identities #sin(a+b)=sin(a)cos(b)+sin(b)cos(a)# and #cos(a+b)=cos(a)cos(b)-sin(a)sin(b)#.

Then, #sin(arccos(63/65)+2arctan(3/4))#
#=sin(arccos(63/65))cos(2arctan(3/4))+sin(2arctan(3/4))cos(arccos(63/65))#.

Now, #cos(arccos(63/65))=63/65#, and, since #sin^2(x)+cos^2(x)=1#, #sin(arccos(63/65))=sqrt(1-cos^2(arccos(63/65)))=16/65#.

Thus, the above equals #16/65cos(2arctan(3/4))+63/65sin(2arctan(3/4))#.

Apply the compound-angle identities again: #16/65(cos^2(arctan(3/4))-sin^2(arctan(3/4)))+63/65(2sin(arctan(3/4))cos(arctan(3/4)))#.

Now, since #tan^2(x)+1=1/cos^2(x)# and #1+1/tan^2(x)=1/sin^2(x)#, #cos(arctan(x))=sqrt(1/(tan^2(arctan(x))+1))=1/sqrt(x^2+1)# and #sin(arctan(x))=sqrt(1/(1+1/tan^2(arctan(x))))=sqrt(1/(1+1/x^2))=x/sqrt(x^2+1)#.

Thus, the above becomes #16/65((1/sqrt((3/4)^2+1))^2-((3/4)/sqrt((3/4)^2+1))^2)+63/65(2(3/4)/sqrt((3/4)^2+1)1/(sqrt((3/4)^2+1)))#.

Simplify to get #1624/1625#.

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