
If $ \sin 3\theta =\cos \left( \theta -{{6}^{\circ }} \right) $ , where $ 3\theta $ and $ \left( \theta -{{6}^{\circ }} \right) $ both are acute angles , then the value of $ \theta $ is:
(A) $ {{18}^{\circ }} $
(B) $ {{24}^{\circ }} $
(C) $ {{36}^{\circ }} $
(D) $ {{30}^{\circ }} $
Answer
507.3k+ views
Hint: For answering this question we will use the transformation of sine angle to cosine angle can be done using the formulae $ \sin \left( {{90}^{\circ }}-x \right)=\cos x $ where $ x $ is an acute angle and then simplify the expression and obtain the value of $ \theta $ .
Complete step by step answer:
Now considering from the question we have $ \sin \left( 3\theta \right)=\cos \left( \theta -{{6}^{\circ }} \right) $ , where $ 3\theta $ and $ \left( \theta -{{6}^{\circ }} \right) $ both are acute angles.
From the basic concept of trigonometry we know that the transformation of sine angle to cosine angle can be done using the formulae $ \sin \left( {{90}^{\circ }}-x \right)=\cos x $ where $ x $ is an acute angle.
After using the transformation formulae we will have $ \sin \left( {{90}^{\circ }}-\left( {{90}^{\circ }}-3\theta \right) \right)=\cos \left( {{90}^{\circ }}-3\theta \right) $ .
By simplifying the expression that we have from the information given in the question $ \sin \left( 3\theta \right)=\cos \left( \theta -{{6}^{\circ }} \right) $ we will get the simplified expression $ \cos \left( {{90}^{\circ }}-3\theta \right)=\cos \left( \theta -{{6}^{\circ }} \right) $ .
Therefore, we can say that $ {{90}^{\circ }}-3\theta =\theta -{{6}^{\circ }} $ .
After simplifying this expression we will have $ {{90}^{\circ }}+{{6}^{\circ }}=3\theta +\theta $ .
By further simplifying this we will get $ {{96}^{\circ }}=4\theta $ .
By performing a transformation from Right-hand side to left-hand side we will have $ \dfrac{{{96}^{\circ }}}{4}=\theta $ .
After calculating we will have $ \theta ={{24}^{\circ }} $ .
Therefore we can conclude that when $ \sin 3\theta =\cos \left( \theta -{{6}^{\circ }} \right) $ , where $ 3\theta $ and $ \left( \theta -{{6}^{\circ }} \right) $ both are acute angles , the value of $ \theta $ is $ {{24}^{\circ }} $ .Hence we will mark option B is correct.
Note:
While answering questions of this type we should be sure with the calculations. The transformation of sine angle to cosine angle can be done using the formulae $ \sin \left( {{90}^{\circ }}-x \right)=\cos x $ where $ x $ is an acute angle. If we had made a mistake while performing calculations and written $ {{96}^{\circ }}=4\theta $ as $ {{120}^{\circ }}=4\theta $ by mistake this will lead us to the conclusion as $ \theta =3{{0}^{\circ }} $ . Then we will mark option D as correct which is wrong.
Complete step by step answer:
Now considering from the question we have $ \sin \left( 3\theta \right)=\cos \left( \theta -{{6}^{\circ }} \right) $ , where $ 3\theta $ and $ \left( \theta -{{6}^{\circ }} \right) $ both are acute angles.
From the basic concept of trigonometry we know that the transformation of sine angle to cosine angle can be done using the formulae $ \sin \left( {{90}^{\circ }}-x \right)=\cos x $ where $ x $ is an acute angle.
After using the transformation formulae we will have $ \sin \left( {{90}^{\circ }}-\left( {{90}^{\circ }}-3\theta \right) \right)=\cos \left( {{90}^{\circ }}-3\theta \right) $ .
By simplifying the expression that we have from the information given in the question $ \sin \left( 3\theta \right)=\cos \left( \theta -{{6}^{\circ }} \right) $ we will get the simplified expression $ \cos \left( {{90}^{\circ }}-3\theta \right)=\cos \left( \theta -{{6}^{\circ }} \right) $ .
Therefore, we can say that $ {{90}^{\circ }}-3\theta =\theta -{{6}^{\circ }} $ .
After simplifying this expression we will have $ {{90}^{\circ }}+{{6}^{\circ }}=3\theta +\theta $ .
By further simplifying this we will get $ {{96}^{\circ }}=4\theta $ .
By performing a transformation from Right-hand side to left-hand side we will have $ \dfrac{{{96}^{\circ }}}{4}=\theta $ .
After calculating we will have $ \theta ={{24}^{\circ }} $ .
Therefore we can conclude that when $ \sin 3\theta =\cos \left( \theta -{{6}^{\circ }} \right) $ , where $ 3\theta $ and $ \left( \theta -{{6}^{\circ }} \right) $ both are acute angles , the value of $ \theta $ is $ {{24}^{\circ }} $ .Hence we will mark option B is correct.
Note:
While answering questions of this type we should be sure with the calculations. The transformation of sine angle to cosine angle can be done using the formulae $ \sin \left( {{90}^{\circ }}-x \right)=\cos x $ where $ x $ is an acute angle. If we had made a mistake while performing calculations and written $ {{96}^{\circ }}=4\theta $ as $ {{120}^{\circ }}=4\theta $ by mistake this will lead us to the conclusion as $ \theta =3{{0}^{\circ }} $ . Then we will mark option D as correct which is wrong.
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