Question

If $\theta $ and $3\theta - {30^ \circ }$ are acute angles such that $\sin \theta = \cos \left( {3\theta - {{30}^ \circ }} \right)$ , then find the value of $\tan \theta $.

Answer 1

Hint: Here we use the concept of trigonometric functions sine and cosine having complementary angles to each other and write the given equation in the question in such a form that we have the same trigonometric function on both sides. Then taking the inverse of the trigonometric function on both sides we calculate the angle. We check the value of$\theta $ from the inequality formed by taking $3\theta - {30^ \circ }$ as an acute angle.
* Acute angle is defined as an angle that has a measure of less than \[{90^ \circ }\].
* Two angles are said to be complementary to each other if the sum of angles is \[{90^ \circ }\].
Therefore, \[\sin x = \cos (90 - x),\cos x = \sin ({90^ \circ } - x)\]

Complete step by step solution:
We have two acute angles $\theta $ and $3\theta - {30^ \circ }$
So, from the definition of acute angles \[\theta < {90^ \circ },3\theta - {30^ \circ } < {90^ \circ }\]
We can solve the inequality which will give us an idea about the measure of the angle \[\theta \].
\[\because 3\theta - {30^ \circ } < {90^ \circ }\]
Add \[{30^ \circ }\]on both sides of the inequality
\[
   \Rightarrow 3\theta - {30^ \circ } + {30^ \circ } < {90^ \circ } + {30^ \circ } \\
   \Rightarrow 3\theta < {120^ \circ } \\
 \]
Divide both sides of the inequality by 3
\[ \Rightarrow \dfrac{{3\theta }}{3} < \dfrac{{{{120}^ \circ }}}{3}\]
Cancel out the common terms from numerator and denominator.
\[ \Rightarrow \theta < {40^ \circ }\] … (1)
It is given that $\sin \theta = \cos \left( {3\theta - {{30}^ \circ }} \right)$ … (2)
Since, we know sine and cosine are complimentary angles, we write \[\cos x = \sin ({90^ \circ } - x)\]
Substitute the value of \[x = 3\theta - {30^ \circ }\]
\[ \Rightarrow \cos (3\theta - {30^ \circ }) = \sin ({90^ \circ } - (3\theta - {30^ \circ }))\]
Open the bracket to solve for the value of the angle
\[
   \Rightarrow \cos (3\theta - {30^ \circ }) = \sin ({90^ \circ } - 3\theta + {30^ \circ }) \\
   \Rightarrow \cos (3\theta - {30^ \circ }) = \sin ({120^ \circ } - 3\theta ) \\
 \]
Substitute the value of \[\cos (3\theta - {30^ \circ }) = \sin ({120^ \circ } - 3\theta )\]in equation (2)
$ \Rightarrow \sin \theta = \sin ({120^ \circ } - 3\theta )$
Take the inverse function of sine on both sides of the equation.
$ \Rightarrow {\sin ^{ - 1}}(\sin \theta ) = {\sin ^{ - 1}}(\sin ({120^ \circ } - 3\theta ))$
Since we know inverse and the function cancel each other,
$ \Rightarrow \theta = {120^ \circ } - 3\theta $
Shift the values with the variable on one side of the equation
$
   \Rightarrow \theta + 3\theta = {120^ \circ } \\
   \Rightarrow 4\theta = {120^ \circ } \\
 $
Divide both sides of the equation by 4
$ \Rightarrow \dfrac{{4\theta }}{4} = \dfrac{{{{120}^ \circ }}}{4}$
Cancel out the common terms from numerator and denominator.
$ \Rightarrow \theta = {30^ \circ }$
We check from equation (1)
\[{30^ \circ } < {40^ \circ }\].
Now we have to find the value of $\tan \theta $.
Substitute the value of $\theta = {30^ \circ }$to find the value of $\tan \theta $
$ \Rightarrow \tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$

So, the value of $\tan \theta = \dfrac{1}{{\sqrt 3 }}$.

Note:
Students many times make the mistake of taking complementary functions instead of complementary angles and they write \[\sin \theta + \cos \theta = 90\]which is wrong because by complimentary functions we mean the functions have sum 90 and by complementary angles, we mean sum of angles is 90. Also, many students who don’t remember the values of trigonometric functions at various angles should try making the table first and then write their answers.

Angles (in degrees)	${0^ \circ }$	${30^ \circ }$	${45^ \circ }$	${60^ \circ }$	${90^ \circ }$
sin	0	$\dfrac{1}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{{\sqrt 3 }}{2}$	$1$
cos	1	$\dfrac{{\sqrt 3 }}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{1}{2}$	0
tan	0	$\dfrac{1}{{\sqrt 3 }}$	1	$\sqrt 3 $	Not defined