Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

How do you evaluate # cos (pi/2) cos(pi/7) + sin(pi/2) sin(pi/7)#?

1 Answer

Oct 16, 2016

Inserting #cos(pi/2)=0 and sin(pi/2)=1# in the given expression we get

#cos(pi/2)cos(pi/7)+sin(pi/2)sin(pi/7)#

#=0xxcos(pi/7)+1xxsin(pi/7)#

#=sin(pi/7)=0.434#

Related questions

See all questions in Trigonometric Functions of Any Angle

Impact of this question

2768 views around the world

You can reuse this answer
Creative Commons License