Question

If we have a trigonometric expression as \[\sin (36+\theta )=cos(16+\theta )\]; then find the value of \[\theta \] where \[{{(36+\theta )}^{\circ }}\] and \[{{(16+\theta )}^{\circ }}\] are both acute angles.

Answer 1

Hint: As \[{{(36+\theta )}^{\circ }}\] and \[{{(16+\theta )}^{\circ }}\] are acute angles then represent cosine in the form of sine. Compare them and find the value of \[\theta \].

Complete step-by-step solution
The given trigonometric equation is \[\sin (36+\theta )=cos(16+\theta )\]
Now, it is given that both \[{{(36+\theta )}^{\circ }}\] and \[{{(16+\theta )}^{\circ }}\] are acute angles. So that means \[\theta \] will also be an acute angle in order to both of them be an acute one.
We know that an acute angle is an angle which lies between \[{{0}^{\circ }}\] and \[{{90}^{\circ }}\].
Now, from trigonometric identities we know that \[\cos x=\sin ({{90}^{\circ }}-x)\] where x is an acute angle measured in degrees.
So, applying this identity to our given equation we get,\[\sin (36+\theta )=cos(16+\theta )\]
\[\Rightarrow \sin (36+\theta )=\sin ({{90}^{\circ }}-(16+\theta ))\]
\[\Rightarrow \sin (36+\theta )=\sin ({{90}^{\circ }}-16-\theta )\] [All angles inside sine are measured in degrees]
\[\Rightarrow \sin (36+\theta )=\sin (74-\theta )\]
 Now, we have arrived in a form where both the left-hand side and right-hand side contain the sine function. Therefore, we can compare them, as they are comparable. We can tell that the angles inside the sine function are the same and equivalent.
Hence, we can write \[36+\theta =74-\theta \]
Or,\[2\theta =74-36\]
Or,\[\theta =37-18=19\]
Hence, we obtain our required value of \[\theta \]. This is the only acute angle, which satisfies the given equation.
Hence, the required value of \[\theta \] asked in this question is \[{{19}^{\circ }}\].

Note: We have written two same values in LHS and RHS and compared them. This is only possible as it is given that it is an acute angle. Otherwise, it can have multiple solutions, when we have to use the general solution of sine to find such ones.