Question

How do you solve $ \sin 2\theta -1=\cos 2\theta $ ?

Answer 1

Hint: In this question, we are given a trigonometric equation and we need to solve it to find the values of $ \theta $ for which this equation satisfies. We will first rearrange the equation such that sine and cosine functions are on the same side and constant on the other side. After that we will transform the equation such that we have single trigonometric function one side, to solve easily we will use following properties to solve this sum,
\[\begin{align}
  & \left( i \right)\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} \\
 & \left( ii \right)\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} \\
 & \left( iii \right)\sin \left( A-B \right)=\sin A\cos B-\sin B\cos A \\
 & \left( iv \right)If\text{ }\sin x=\sin y\text{ then }x=n\pi +{{\left( -1 \right)}^{n}}y,n\in z \\
\end{align}\]

Complete step by step answer:
Here we are given a trigonometric equation as $ \sin 2\theta -1=\cos 2\theta $ . We need to evaluate the value of $ \theta $ which satisfies this equation. For this let us find the general solution of $ \theta $ . Let us rearrange the terms such that $ \sin 2\theta \text{ and }\cos 2\theta $ are on one side and 1 is on the other side. We get $ \sin 2\theta -\cos 2\theta =1 $ .
Let us divide both side by $ \sqrt{2} $ we get $ \dfrac{\sin 2\theta -\cos 2\theta }{\sqrt{2}}=\dfrac{1}{\sqrt{2}} $ .
Separating the terms on the left side of the equation we get $ \sin 2\theta \dfrac{1}{\sqrt{2}}-\cos 2\theta \dfrac{1}{\sqrt{2}}=\dfrac{1}{\sqrt{2}} $ .
We know that from the trigonometric ratio table that $ \sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} $ and also $ \cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} $ . So let us use them in left side of the equation we get $ \sin 2\theta \cos \dfrac{\pi }{4}-\cos 2\theta \sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} $ .
We can see that, left side of the equation is of the form sinAcosB - sinBcosA which is equal to sin(A-B). So we get $ \sin \left( 2\theta -\dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}} $ .
As we know that, $ \sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} $ so replacing $ \dfrac{1}{\sqrt{2}} $ on the right side of the equation with $ \sin \dfrac{\pi }{4} $ we get $ \sin \left( 2\theta -\dfrac{\pi }{4} \right)=\sin \dfrac{\pi }{4} $ .
Now we know that if sinx = siny then $ x=n\pi +{{\left( -1 \right)}^{n}}y,n\in z $ .
So here $ x=2\theta -\dfrac{\pi }{4}\text{ and }y=\dfrac{\pi }{4} $ .
Hence we get $ 2\theta -\dfrac{\pi }{4}=n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4} $ .
Adding $ \dfrac{\pi }{4} $ on both sides we get $ 2\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}+\dfrac{\pi }{4} $ .
Dividing both sides by 2 we get $ \theta =\dfrac{n\pi }{2}+{{\left( -1 \right)}^{n}}\dfrac{\pi }{8}+\dfrac{\pi }{8} $ .
Since general solution of $ \theta $ is a little complicated so let us find value of $ \theta $ between the unit circle i.e. between 0 to $ 2\pi $ .
Taking n = 0 we get $ \theta =\dfrac{\pi }{4} $ .
And taking n = 1 we get $ \theta =\dfrac{\pi }{2}-\dfrac{\pi }{8}+\dfrac{\pi }{8}=\dfrac{\pi }{2} $ .
Hence $ \theta =\dfrac{\pi }{4}\text{ and }\dfrac{\pi }{2} $ .
These are the required values of $ \theta $.

Note:
Students should keep in mind all the trigonometric properties before solving this sum. They can also learn the formula that $ \sin x-\cos x=\sqrt{2}\sin \left( x-\dfrac{\pi }{4} \right) $ to evaluate easily. Do not forget about negative signs in the general solution.