Question

Write true or false.
If \[\sin x = \cos y\] then \[x + y = {45^ \circ }\]

Answer 1

Hint: Try to remember a principal value of x and see the complimentary value of y then add them and see whether we are getting \[{45^ \circ }\] or not.
Complete Step by Step Solution:
Let us do it by taking some examples first we know that \[\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}\] and the same value of cosine exists when \[\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}\]
So in this case \[\sin \dfrac{\pi }{3} = \cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}\]
So if I add them i will get the required result
\[\begin{array}{l}
\therefore x + y\\
 = \dfrac{\pi }{3} + \dfrac{\pi }{6}\\
 = \dfrac{{2\pi + \pi }}{6}\\
 = \dfrac{{3\pi }}{6}\\
 = \dfrac{\pi }{2}
\end{array}\]
 And we also know that \[\dfrac{\pi }{2} = {90^ \circ }\] so from here we can clearly see that the statement written is false

Note: we also know that \[\sin x = \cos (90 - x)\] so this also settles it because we can also write \[\cos y = \sin (90 - y)\] therefore \[\sin x = \sin (90 - y)\] and now if we remove sin from both the sides we are left with \[x = {90^ \circ } - y\] i.e., \[x + y = {90^ \circ }\]
So in both the cases the answer remains to be false.