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

How do you solve the equation sin⁡2(θ)=12{{\sin }^{2}}\left( \theta \right)=\dfrac{1}{2} ?

Explanation

Solution

To evaluate the given function sin⁡2(θ)=12{{\sin }^{2}}\left( \theta \right)=\dfrac{1}{2} , let us take the square root of this function . Doing so, we will get sin⁡(θ)=±12\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⁡π4=12\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} . We know that the general value of sin⁡θ=sin⁡y\sin \theta =\sin y is θ=nπ+(−1)ny\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(θ)=12{{\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⁡(θ)=±12\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⁡(θ)=12,sin⁡(θ)=−12\sin \left( \theta \right)=\dfrac{1}{\sqrt{2}},\sin \left( \theta \right)=-\dfrac{1}{\sqrt{2}}
We know that sin⁡π4=12\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} .Hence, we can write the above form as
sin⁡(θ)=sin⁡π4,sin⁡(θ)=−sin⁡π4\sin \left( \theta \right)=\sin \dfrac{\pi }{4},\sin \left( \theta \right)=-\sin \dfrac{\pi }{4}
Let us recall that sin⁡(−θ)=−sin⁡θ\sin \left( -\theta \right)=-\sin \theta . Therefore, the above expression becomes
sin⁡(θ)=sin⁡π4,sin⁡(θ)=sin⁡(−π4)\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⁡θ=sin⁡y\sin \theta =\sin y is θ=nπ+(−1)ny\theta =n\pi +{{\left( -1 \right)}^{n}}y . Hence, we can write the value of θ\theta as
θ=nπ+(−1)nπ4,nπ−(−1)nπ4\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
θ=nπ±(−1)nπ4\theta =n\pi \pm {{\left( -1 \right)}^{n}}\dfrac{\pi }{4}
Hence, we have got the value of sin⁡2(θ)=12{{\sin }^{2}}\left( \theta \right)=\dfrac{1}{2} as θ=nπ±(−1)nπ4\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⁡θ=sin⁡y\sin \theta =\sin y as θ=nπ±(−1)ny\theta =n\pi \pm {{\left( -1 \right)}^{n}}y . Also, they may confuse the value of sin⁡(−θ)=−sin⁡θ\sin \left( -\theta \right)=-\sin \theta . From the value θ=nπ+(−1)ny\theta =n\pi +{{\left( -1 \right)}^{n}}y , we can see that when sin⁡θ=0\sin \theta =0 , the general value will be nπn\pi .