Question

,ⓦⓗⓔⓝ sɪɴ ᴛʜᴇᴛᴀ +ᴄᴏs ᴛʜᴇᴛᴀ ,_<ᴛʜᴇᴛᴀ _<90• ᴛᴀᴋᴇs ᴛʜᴇ ɢʀᴇᴀᴛᴇsᴛ ᴠᴀʟᴜᴇ.



sᴏʟᴠᴇ ɪᴛ....ɪᴛs ᴜʀɢᴇɴᴛ....​

Answer 1

Follow the above attachment

Answer 2

☆ Given Question : -

Find the greatest value of sinθ + cosθ, 0° < θ < 90°.

$\huge \orange{AηsωeR} ✍$

$\begin{gathered}\begin{gathered}\bf{\underline{\underline{Given :-}}}\end{gathered}\end{gathered}$

• sinθ + cosθ, 0° < θ < 90°.

$\begin{gathered}\begin{gathered}\bf{\underline{\underline{To Find :-}}}\end{gathered}\end{gathered}$

• Greatest value of sinθ + cosθ.

☆ Concept Used :-

Let us consider a function f(x).

• Now, we have to find these points at which derivative of f(x) is zero. For this, we have to solve f'(x)=0. By solving this we will get some values of x for which derivative of f(x) is zero. These are the points of maxima or minima.

• To know that which point is maxima and which point is minima we have to double derivative the function.

• After this, we will put the solutions of f'(x)=0 in f''(x) and find the sign of f''(x)

• If sign of f''(x) is positive then it is a point of minima , if the sign is negative then it is a point of maxima.

• Further, if f''(x)=0 then we have to repeat above steps for higher order derivatives of f(x).

$\begin{gathered}\Large{\bold{\pink{\underline{CaLcUlAtIoN\::}}}} \end{gathered}$

$\bf \:f(θ) = sinθ + cosθ \: ⟼ \: (1)$

☆Differentiate w. r. t. θ, we get

$\bf \:  ⟼ f'(θ) = cosθ - sinθ \: ⟼(2)$

☆For maximum or minimum value,

$\bf \:f'(θ) = 0$

$\bf\implies \:cosθ - sinθ = 0$

$\bf \:  ⟼ sinθ = cosθ$

$\bf \:  ⟼ \dfrac{sinθ}{cosθ} = 1$

$\bf \:  ⟼ tanθ = 1$

☆As θ lies in Ist quadrant, we get

$\bf\implies \:θ = \dfrac{\pi}{4}$

☆Differentiate equation (2) w. r. t. θ, we get

$\bf\implies \:f''(θ) = - sinθ - cosθ$

$\begin{gathered}\bf\red{So,} \end{gathered}$

$\bf \:  ⟼ f''(\dfrac{\pi}{4} ) = - sin(\dfrac{\pi}{4} ) - cos(\dfrac{\pi}{4})$

$\bf \:  ⟼ f''(\dfrac{\pi}{4} ) = - \dfrac{1}{ \sqrt{2} } - \dfrac{1}{ \sqrt{2} }$

$\bf \:  ⟼ f''(\dfrac{\pi}{4} ) = - \dfrac{2}{ \sqrt{2} } < 0$

$\bf\implies \:f(θ) \: is \: maximum \: at \: θ = \dfrac{\pi}{4}$

And maximum value is

$\bf \:  ⟼ f(\dfrac{\pi}{4} ) = sin(\dfrac{\pi}{4} ) + \: cos(\dfrac{\pi}{4})$

$\bf \:  ⟼ f(\dfrac{\pi}{4} ) = \dfrac{1}{ \sqrt{2} } + \dfrac{1}{ \sqrt{2} }$

$\bf \:  ⟼ f(\dfrac{\pi}{4} ) = \dfrac{2}{ \sqrt{2} } = \sqrt{2}$

$\large{\boxed{\boxed{\sf{So, greatest \: value \: of \: sinθ + cosθ \: is \: \sqrt{2} }}}}$