Answer

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Hint: First of all square both sides of the given equation and convert the whole equation in terms of \[\sin \theta \] and find the value of \[\sin \theta \]. Then if \[\sin \theta =\sin \alpha \], then \[\theta =n\pi +{{\left( -1 \right)}^{n}{\alpha }}\] where \[\alpha \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\] and \[n\in I\]. Then choose the acute \[\theta \] among all.

Complete step-by-step answer:

Here, we have to find the values of \[\theta \] if \[\sin \theta =\cos \theta \] given that \[\theta \] is acute. First of all, let us consider the equation given in the question, \[\sin \theta =\cos \theta \]

By squaring both sides of the above equation, we get

\[{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \]

Now, we know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] or \[{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \]. By substituting the value of \[{{\cos }^{2}}\theta \] in terms of \[{{\sin }^{2}}\theta \] in the above equation, we get,

\[{{\sin }^{2}}\theta =1-{{\sin }^{2}}\theta \]

By adding \[{{\sin }^{2}}\theta \] on both sides of the above equation, we get,

\[{{\sin }^{2}}\theta +{{\sin }^{2}}\theta =1\]

Or, \[2{{\sin }^{2}}\theta =1\]

By dividing 2 on both sides of the above equation, we get,

\[\Rightarrow \dfrac{2{{\sin }^{2}}\theta }{2}=\dfrac{1}{2}\]

By canceling the like terms in LHS of the above equation, we get,

\[\Rightarrow {{\sin }^{2}}\theta =\dfrac{1}{2}\]

By taking square root on both sides of the above equation, we get,

\[\sqrt{{{\sin }^{2}}\theta }=\sqrt{\dfrac{1}{2}}\]

We know that \[\sqrt{{{a}^{2}}}=\pm a\] and \[\sqrt{1}=1\]. By applying these in the above equation, we get,

\[\sin \theta =\pm \dfrac{1}{\sqrt{2}}\]

Since we are given that \[\theta \] is acute, so \[\sin \theta >0\].

Therefore, we take \[\sin \theta =\dfrac{1}{\sqrt{2}}\].

We know that \[\sin {{45}^{o}}=\dfrac{1}{\sqrt{2}}\]. So by substituting the value of \[\dfrac{1}{\sqrt{2}}\] in terms of sine in the above equation, we get,

\[\Rightarrow \sin \theta =\sin {{45}^{o}}=\sin \left( 45\times \dfrac{\pi }{180} \right)=\sin \dfrac{\pi }{4}\]

We know that if \[\sin \theta =\sin \alpha \], then

\[\theta =n\pi +{{\left( -1 \right)}^{n}{\alpha }},\alpha \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\text{ and }n\in I\]

By applying this in the above equation, we get,

\[\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}\]

By substituting n = 0, we get

\[\theta =0.\left( \pi \right)+{{\left( -1 \right)}^{0}}\dfrac{\pi }{4}\]

We get, \[\theta =\dfrac{\pi }{4}\]

By substituting n = 1, we get

\[\theta =\pi +{{\left( -1 \right)}^{1}}\dfrac{\pi }{4}\]

\[=\pi -\dfrac{\pi }{4}\]

We get, \[\theta =\dfrac{3\pi }{4}\]

By substituting n = 2, we get,

\[\theta =3\pi +{{\left( -1 \right)}^{2}}\dfrac{\pi }{4}\]

\[=3\pi +\dfrac{\pi }{4}\]

We get, \[\theta =\dfrac{13\pi }{4}\]

So, we get the value of \[\theta \] as \[\dfrac{\pi }{4},\dfrac{3\pi }{4},\dfrac{13\pi }{4}.....\text{so on}\]

But in the question, we are given that \[\theta \] must be acute that means \[0<\theta <\dfrac{\pi }{2}\]. So, only one value of \[\theta \] i.e. \[\theta =\dfrac{\pi }{4}\] is acceptable. Also, for \[\theta =\dfrac{\pi }{4}\], we get \[\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\].

So, we get \[\theta =\dfrac{\pi }{4}={{45}^{o}}\].

Note: Some students often make this mistake of getting the result \[\theta =\alpha \] initially by just looking at \[\sin \theta =\sin \alpha \]. But they must note that \[\theta =\alpha \] is not the only result but it is one of the results. The general value of \[\theta =n\pi +{{\left( -1 \right)}^{n}{\alpha }}\] where \[\alpha \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\] and \[n\in I\] and there would be infinitely many values of \[\theta \] for giving \[\alpha \] not just \[\theta =\alpha \]. Also, students must check the value of \[\theta \] by substituting it in the given equation that is \[\sin \theta =\cos \theta \] and then only answer.

Complete step-by-step answer:

Here, we have to find the values of \[\theta \] if \[\sin \theta =\cos \theta \] given that \[\theta \] is acute. First of all, let us consider the equation given in the question, \[\sin \theta =\cos \theta \]

By squaring both sides of the above equation, we get

\[{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \]

Now, we know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] or \[{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \]. By substituting the value of \[{{\cos }^{2}}\theta \] in terms of \[{{\sin }^{2}}\theta \] in the above equation, we get,

\[{{\sin }^{2}}\theta =1-{{\sin }^{2}}\theta \]

By adding \[{{\sin }^{2}}\theta \] on both sides of the above equation, we get,

\[{{\sin }^{2}}\theta +{{\sin }^{2}}\theta =1\]

Or, \[2{{\sin }^{2}}\theta =1\]

By dividing 2 on both sides of the above equation, we get,

\[\Rightarrow \dfrac{2{{\sin }^{2}}\theta }{2}=\dfrac{1}{2}\]

By canceling the like terms in LHS of the above equation, we get,

\[\Rightarrow {{\sin }^{2}}\theta =\dfrac{1}{2}\]

By taking square root on both sides of the above equation, we get,

\[\sqrt{{{\sin }^{2}}\theta }=\sqrt{\dfrac{1}{2}}\]

We know that \[\sqrt{{{a}^{2}}}=\pm a\] and \[\sqrt{1}=1\]. By applying these in the above equation, we get,

\[\sin \theta =\pm \dfrac{1}{\sqrt{2}}\]

Since we are given that \[\theta \] is acute, so \[\sin \theta >0\].

Therefore, we take \[\sin \theta =\dfrac{1}{\sqrt{2}}\].

We know that \[\sin {{45}^{o}}=\dfrac{1}{\sqrt{2}}\]. So by substituting the value of \[\dfrac{1}{\sqrt{2}}\] in terms of sine in the above equation, we get,

\[\Rightarrow \sin \theta =\sin {{45}^{o}}=\sin \left( 45\times \dfrac{\pi }{180} \right)=\sin \dfrac{\pi }{4}\]

We know that if \[\sin \theta =\sin \alpha \], then

\[\theta =n\pi +{{\left( -1 \right)}^{n}{\alpha }},\alpha \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\text{ and }n\in I\]

By applying this in the above equation, we get,

\[\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}\]

By substituting n = 0, we get

\[\theta =0.\left( \pi \right)+{{\left( -1 \right)}^{0}}\dfrac{\pi }{4}\]

We get, \[\theta =\dfrac{\pi }{4}\]

By substituting n = 1, we get

\[\theta =\pi +{{\left( -1 \right)}^{1}}\dfrac{\pi }{4}\]

\[=\pi -\dfrac{\pi }{4}\]

We get, \[\theta =\dfrac{3\pi }{4}\]

By substituting n = 2, we get,

\[\theta =3\pi +{{\left( -1 \right)}^{2}}\dfrac{\pi }{4}\]

\[=3\pi +\dfrac{\pi }{4}\]

We get, \[\theta =\dfrac{13\pi }{4}\]

So, we get the value of \[\theta \] as \[\dfrac{\pi }{4},\dfrac{3\pi }{4},\dfrac{13\pi }{4}.....\text{so on}\]

But in the question, we are given that \[\theta \] must be acute that means \[0<\theta <\dfrac{\pi }{2}\]. So, only one value of \[\theta \] i.e. \[\theta =\dfrac{\pi }{4}\] is acceptable. Also, for \[\theta =\dfrac{\pi }{4}\], we get \[\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\].

So, we get \[\theta =\dfrac{\pi }{4}={{45}^{o}}\].

Note: Some students often make this mistake of getting the result \[\theta =\alpha \] initially by just looking at \[\sin \theta =\sin \alpha \]. But they must note that \[\theta =\alpha \] is not the only result but it is one of the results. The general value of \[\theta =n\pi +{{\left( -1 \right)}^{n}{\alpha }}\] where \[\alpha \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\] and \[n\in I\] and there would be infinitely many values of \[\theta \] for giving \[\alpha \] not just \[\theta =\alpha \]. Also, students must check the value of \[\theta \] by substituting it in the given equation that is \[\sin \theta =\cos \theta \] and then only answer.

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