#sin^2 63^@ +sin^2 27^@# eqas to why ?

1 Answer
Jun 22, 2018

#sin^2 63^@ + sin^2 27^@ = 1#

Explanation:

Note that:

#cos theta = sin (90^@ - theta)#

#cos^2 theta + sin^2 theta = 1#

So we find:

#sin^2 63^@ + sin^2 27^@ = sin^2 (90^@-27^@) + sin^2 27^@#

#color(white)(sin^2 63^@ + sin^2 27^@) = cos^2 27^@ + sin^2 27^@ = 1#

As to why, consider a right-angled triangle with other angles #A# and #B#...

enter image source here

Note that:

  • #A+B = 90^@#

  • The leg which is #"opposite"# for #A# is #"adjacent"# for #B# and vice versa.

So:

#sin A = "opposite"_A/"hyptotenuse" = "adjacent"_B/"hypotenuse" = cos B = cos (90^@ - A)#