How do you prove #sin^-1x+cos^-1x=pi/2#?

1 Answer
Oct 7, 2016

See the Explanation.

Explanation:

Let #sin^-1x=theta", where, "|x|le1.#

Then, by the Defn. of #sin^-1# fun.,

#sintheta=x, &, theta in [-pi/2,pi/2].#

#theta in [-pi/2,pi/2] rArr -pi/2le thetalepi/2.#

Multiplying the inequality by #-1#, hence, reversing its order,

#pi/2ge-thetage-pi/2.# Now, adding, #pi/2#,

#pigepi/2-thetage0, i.e., (pi/2-theta) in [0,pi].#

Further, #cos(pi/2-theta)=sintheta=x.#

Thus, #cos(pi/2-theta)=x, where, (pi/2-theta) in [0,pi].#

This, together with the Defn. of #cos^-1# fun., allow us to say that,

#(pi/2-theta)=cos^-1x.#

Replacing #theta# by #sin^-1x,# we finally arrive at,

#pi/2-sin^-1x=cos^-1x, or, sin^-1x+cos^-1x=pi/2.#

Enjoy maths.!