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If $\sin 5\theta = \cos 4\theta$ , where $5\theta$ and $4\theta$ are acute angles , find the values of $\theta$ ?

Last updated date: 17th Jun 2024
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Hint: Start by writing the given equation and all the conditions given. Use the trigonometric identities used for the conversion of one trigonometric ratio to another one. After conversion substitute the value and compare the values and solve the relation formed then , in order to find out the value of $\theta$ . Do a check on the value found by checking the range it can fall into.

Given,
$\sin 5\theta = \cos 4\theta$
Also , $5\theta$ and $4\theta$ are acute angles.
Now, we know that $\sin ({90^ \circ } - \theta ) = \cos \theta$
So , applying this identity for $4\theta$ , we will get
$\sin ({90^ \circ } - 5\theta ) = \cos 4\theta$
Substituting this values , we get
$\sin 5\theta = \sin ({90^ \circ } - 4\theta )$
Now , we know that if $\sin \theta = \sin \phi$ then $\theta = \phi$
$\therefore 5\theta = {90^ \circ } - 4\theta \\ \Rightarrow 5\theta + 4\theta = {90^ \circ } \\ \Rightarrow 9\theta = {90^ \circ } \\ \Rightarrow \theta = \dfrac{{{{90}^ \circ }}}{9} \\ \therefore \theta = {10^ \circ } \\$
So , The value of $\theta$ is ${10^ \circ }$ .
We can do a final check by using the data given in the question that is $5\theta$ and $4\theta$ are acute angles. So which means if we divide $5\theta$ and $4\theta$ by 5 and 4 respectively, we will get $\theta$ also to acute , as
$0 \leqslant 5\theta \leqslant {90^ \circ } \\ 0 \leqslant \dfrac{{5\theta }}{5} \leqslant \dfrac{{{{90}^ \circ }}}{5} \\ 0 \leqslant \theta \leqslant {18^ \circ } \\$
Which is also acute.

Note: Similar problems which involve different angles can be solved by following the same procedure. Attention must be given while converting one trigonometric ratio to another , by following the rules of conversion. A final check is must for such problems as sometimes we might get vague values, this can be done by either putting back the value in any given equation and look for expected value or by using the same procedure as in the solution above. In any case it must hold true.