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Question

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If \[\sin x = \cos y\] then \[x + y = {45^ \circ }\]

Answer
Verified

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.

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.

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