Question

Find the maximum value of sin θ + cos θ in the range $\left[ 0,\dfrac{\pi }{2} \right]$.

Answer 1

Hint: Try to convert the two terms into one term This will help to further solve the problem.Multiply and divide same number to the expression so as to merge it in 1 term. You can also do this question using differentiation of the function. Differentiate the given expression with respect to θ and equate the resulting expression to 0. Then find the value of θ in the defined range. If θ is found then check for the double derivative of the given expression. If it is negative at that θ then θ gives the maximum value.If θ is not found in the defined range, then check for the value on the boundary of interval. Whichever value comes greater is maximum value.

Complete step-by-step answer:

The given expression is
sin θ + cos θ
Multiplying and divide $\sqrt{2}$ to this expression we get
$\sqrt{2}\left( \dfrac{\sin \theta }{\sqrt{2}}+\dfrac{\cos \theta }{\sqrt{2}} \right)\,\,\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(i)}$
We know that
Sin 45° = $\dfrac{1}{\sqrt{2}}$
Cos 45° = $\dfrac{1}{\sqrt{2}}$
So replacing first $\dfrac{1}{\sqrt{2}}$ with Cos 45° and second $\dfrac{1}{\sqrt{2}}$with sin 45° in equation (i) we get
$\sqrt{2}\left( \sin \theta \cos {{45}^{{}^\circ }}+\cos \theta \sin {{45}^{{}^\circ }} \right)\,\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(ii)}$
We know that
Sin(A + B) = sinAcosB + cosAsinB
On applying this formula in equation (ii) we get the equation converted as
$\sqrt{2}\left( \sin \left( \theta +{{45}^{{}^\circ }} \right) \right)\,\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(iii)}$
We know that
-1 ≤ sin θ ≤ 1 for every θ$\in \mathbb{R}$
So,
$\begin{aligned} & -1\le \sin \left( \theta +{{45}^{{}^\circ }} \right)\le 1 \\\ & -\sqrt{2}\le \sqrt{2}\sin \left( \theta +{{45}^{{}^\circ }} \right)\le \sqrt{2} \\\ \end{aligned}$
So we get that maximum value of the expression is $\sqrt{2}$which occurs when

& \sin \left( \theta +{{45}^{{}^\circ }} \right)=1 \\\ & \Rightarrow \theta +{{45}^{{}^\circ }}={{90}^{{}^\circ }} \\\ & \Rightarrow \theta ={{45}^{{}^\circ }} \\\ \end{aligned}$$ $$\theta ={{45}^{{}^\circ }}$$ lies in the range $\left[ 0,\dfrac{\pi }{2} \right]$. So this is our maximum point and maximum value is $\sqrt{2}$. Note: There is one very major mistake that most of the students do. Some students may think that -1 ≤ sin θ ≤ 1 for every θ$\in \mathbb{R}$ And -1 ≤ cos θ ≤ 1 for every θ$\in \mathbb{R}$ So adding both the equation we get -2 ≤ sin θ + cos θ ≤ 2 for every θ$\in \mathbb{R}$ . . . (iv) So 2 is the maximum value. But this is wrong. Addition of inequality is true but it does not strictly satisfy the resulting expression. That means equation (iv) is correct but the expression does not value from -2 to 2.