Question

How do you solve the equation ${{\sin }^{2}}\left( \theta \right)=\dfrac{1}{2}$ ?

Answer 1

Hint: To evaluate the given function ${{\sin }^{2}}\left( \theta \right)=\dfrac{1}{2}$ , let us take the square root of this function . Doing so, we will get $\sin \left( \theta \right)=\pm \dfrac{1}{\sqrt{2}}$ . This will get us two values, one positive and other one negative. To proceed further, we will also make use of the known fact that $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ . We know that the general value of $\sin \theta =\sin y$ is $\theta =n\pi +{{\left( -1 \right)}^{n}}y$ . By applying this value, we will get the required answer.

Complete step-by-step solution:
We have to evaluate ${{\sin }^{2}}\left( \theta \right)=\dfrac{1}{2}$ . Since the function is squared on the LHS, so let us take the square root of this function. Then, we will get
$\sin \left( \theta \right)=\pm \dfrac{1}{\sqrt{2}}$
We now have two values, one positive and other one negative. We can write it as
$\sin \left( \theta \right)=\dfrac{1}{\sqrt{2}},\sin \left( \theta \right)=-\dfrac{1}{\sqrt{2}}$
We know that $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ .Hence, we can write the above form as
$\sin \left( \theta \right)=\sin \dfrac{\pi }{4},\sin \left( \theta \right)=-\sin \dfrac{\pi }{4}$
Let us recall that $\sin \left( -\theta \right)=-\sin \theta $ . Therefore, the above expression becomes
 $\sin \left( \theta \right)=\sin \dfrac{\pi }{4},\sin \left( \theta \right)=\sin \left( -\dfrac{\pi }{4} \right)$
We know that the general value of $\sin \theta =\sin y$ is $\theta =n\pi +{{\left( -1 \right)}^{n}}y$ . Hence, we can write the value of $\theta $ as
$\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4},n\pi -{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$
Let us combine this to generalize. We can write the above form as
$\theta =n\pi \pm {{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$
Hence, we have got the value of ${{\sin }^{2}}\left( \theta \right)=\dfrac{1}{2}$ as $\theta =n\pi \pm {{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$ .

Note: Students have a chance of making mistakes when taking the square root. Never miss out the negative value. The students must be thorough with the identities, rules and general solutions of trigonometric functions. They may make mistakes when writing the general solution of $\sin \theta =\sin y$ as $\theta =n\pi \pm {{\left( -1 \right)}^{n}}y$ . Also, they may confuse the value of $\sin \left( -\theta \right)=-\sin \theta $ . From the value $\theta =n\pi +{{\left( -1 \right)}^{n}}y$ , we can see that when $\sin \theta =0$ , the general value will be $n\pi $ .