Question

Find the range of \[\sin 2\alpha +\cos 2\alpha \]

Answer 1

Hint: We solve this problem by using the range of sine function.
We have the standard result that the range of \[\sin \theta \] is \[\left[ -1,1 \right]\]
We convert the given expression into a single sine function so that we can find the range of the given expression.
We have the formula of composite angles of sine function as
\[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\]

Complete step by step answer:
We are given with the expression as \[\sin 2\alpha +\cos 2\alpha \]
Let us assume that the given expression as
\[\Rightarrow p=\sin 2\alpha +\cos 2\alpha \]
Now, let us multiply and divide the RHS of above function with \[\sqrt{2}\] then we get
\[\begin{align}
  & \Rightarrow p=\dfrac{\sqrt{2}}{\sqrt{2}}\left( \sin 2\alpha +\cos 2\alpha \right) \\
 & \Rightarrow p=\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\sin 2\alpha +\dfrac{1}{\sqrt{2}}\cos 2\alpha \right) \\
\end{align}\]
We know that the standard value of trigonometric ratios that is
\[\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\]
By using this standard value for above equation then we get
\[\Rightarrow p=\sqrt{2}\left( \sin 2\alpha \cos \dfrac{\pi }{4}+\cos 2\alpha \sin \dfrac{\pi }{4} \right)\]

We know that the formula of composite angles of sine function as
\[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\]
By using this formula in above equation we get
\[\Rightarrow p=\sqrt{2}\sin \left( 2\alpha +\dfrac{\pi }{4} \right)\]
Let us assume that the angle inside the sine function as
\[\Rightarrow 2\alpha +\dfrac{\pi }{4}=\theta \]
By using the above assumption we get
\[\Rightarrow p=\sqrt{2}\sin \theta .......equation(i)\]
We know that the standard result that the range of \[\sin \theta \] is \[\left[ -1,1 \right]\]
By converting the above statement into mathematical equation we get
\[\Rightarrow -1\le \sin \theta \le 1\]
By multiplying \[\sqrt{2}\] in each side of the above equation we get
\[\begin{align}
  & \Rightarrow -\sqrt{2}\le \sqrt{2}\sin \theta \le \sqrt{2} \\
 & \Rightarrow -\sqrt{2}\le p\le \sqrt{2} \\
\end{align}\]
Here, we can see that the range of \[p\] is given as \[\left[ -\sqrt{2},\sqrt{2} \right]\]
Therefore, we can conclude that the range of given expression is \[\left[ -\sqrt{2},\sqrt{2} \right]\]


Note:
We have a shortcut for this problem.
We are given with the expression as \[\sin 2\alpha +\cos 2\alpha \]
Let us assume that the given expression as
\[\Rightarrow p=\sin 2\alpha +\cos 2\alpha \]
We have a standard formula that the range of expression \[a\sin \theta +b\cos \theta +c\] is given as\[\left[ c-\sqrt{{{a}^{2}}+{{b}^{2}}},c+\sqrt{{{a}^{2}}+{{b}^{2}}} \right]\]
By using the above formula to given expression that is \[\sin 2\alpha +\cos 2\alpha \] we get
\[\begin{align}
  & \Rightarrow \left[ 0-\sqrt{{{1}^{2}}+{{1}^{2}}},0+\sqrt{{{1}^{2}}+{{1}^{2}}} \right] \\
 & \Rightarrow \left[ -\sqrt{2},\sqrt{2} \right] \\
\end{align}\]
Here, we can see that the range of \[p\] is given as \[\left[ -\sqrt{2},\sqrt{2} \right]\]
Therefore, we can conclude that the range of given expression is \[\left[ -\sqrt{2},\sqrt{2} \right]\]