Question

The sum of two oppsité number is​

Answer 1

$\large\underline{\sf{Solution-}}$

Cosider,

$\rm \: 7 \cdot \sin \bigg(5 \cdot x - \dfrac{\pi}{2} \bigg ) + \cos \bigg( 3 \cdot x + \dfrac{\pi}{2} \bigg)$

We know,

$\boxed{\tt{ sin( - x) = - sinx \: }}$

So, using this result, the above expression can be rewritten as

$\rm \: = \: - 7 \sin \bigg(\dfrac{\pi}{2} - 5x \bigg ) + \cos \bigg(\dfrac{\pi}{2} + 3x \bigg)$

$\rm \: = \: - 7cos5x - sin3x$

So,

$\rm \: \dfrac{\partial}{\partial x} \bigg[7 \cdot \sin \bigg(5 \cdot x - \dfrac{\pi}{2} \bigg ) + \cos \bigg( 3 \cdot x + \dfrac{\pi}{2} \bigg) \bigg ]$

can be rewritten as

$\rm \: = \: \dfrac{\partial}{\partial x}\bigg( - 7cos5x - sin3x\bigg)$

$\rm \: = \: - 7\dfrac{\partial}{\partial x}cos5x - \dfrac{\partial}{\partial x}sin3x$

We know,

$\boxed{\tt{ \dfrac{d}{dx}sinx \: = \: cosx \: }}$

and

$\boxed{\tt{ \dfrac{d}{dx}cosx \: = \: - \: sinx \: }}$

So, using these results, we get

$\rm \: = \: - 7 \: ( - sin5x)\dfrac{d}{dx}5x - cos3x\dfrac{\partial}{\partial x}3x$

$\rm \: = \: 7sin5x \times 5 - cos3x \times 3$

$\rm \: = \: 35sin5x - 3cos3x$

Hence,

$\rm \: \dfrac{\partial}{\partial x} \bigg[7 \cdot \sin \bigg(5 \cdot x - \dfrac{\pi}{2} \bigg ) + \cos \bigg( 3 \cdot x + \dfrac{\pi}{2} \bigg) \bigg ]_{x = 2\pi}$

$\rm \: = \: 35sin10\pi \: - \: 3cos6\pi$

We know,

$\boxed{\tt{ sin \: n\pi = 0 \: \: \: where \: n \: is \: integer}}$

and

$\boxed{\tt{ cos \: n\pi = {( - 1)}^{n} \: \: \: where \: n \: is \: integer}}$

So, using these results, we get

$\rm \: = \: 0 - 3 {( - 1)}^{6}$

$\rm \: = \: - 3 \times 1$

$\rm \: = \: - 3$

Hence,

$\boxed{\tt{ \rm \: \dfrac{\partial}{\partial x} \bigg[7 \cdot \sin \bigg(5 \cdot x - \dfrac{\pi}{2} \bigg ) + \cos \bigg( 3 \cdot x + \dfrac{\pi}{2} \bigg) \bigg ]_{x = 2\pi} = - 3}}$

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ADDITIONAL INFORMATION

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

−

Cosider,

$\rm \: 7 \cdot \sin \bigg(5 \cdot x - \dfrac{\pi}{2} \bigg ) + \cos \bigg( 3 \cdot x + \dfrac{\pi}{2} \bigg)7$

⋅

We know,

$\begin{gathered}\boxed{\tt{ sin( - x) = - sinx \: }} \\ \end{gathered}$

So, using this result, the above expression can be rewritten as

$\rm \: = \: - 7 \sin \bigg(\dfrac{\pi}{2} - 5x \bigg ) + \cos \bigg(\dfrac{\pi}{2} + 3x \bigg)=−7sin$

$\rm \: = \: - 7cos5x - sin3x=−7cos5x−sin3x$

So,

$\rm \: \dfrac{\partial}{\partial x} \bigg[7 \cdot \sin \bigg(5 \cdot x - \dfrac{\pi}{2} \bigg ) + \cos \bigg( 3 \cdot x + \dfrac{\pi}{2} \bigg) \bigg ]$

can be rewritten as

$\rm \: = \: \dfrac{\partial}{\partial x}\bigg( - 7cos5x - sin3x\bigg)=$

(−7cos5x−sin3x)

$\rm \: = \: - 7\dfrac{\partial}{\partial x}cos5x - \dfrac{\partial}{\partial x}sin3x=−7$

We know,

$\begin{gathered}\boxed{\tt{ \dfrac{d}{dx}sinx \: = \: cosx \: }} \\ \end{gathered}$

$\begin{gathered}\boxed{\tt{ \dfrac{d}{dx}cosx \: = \: - \: sinx \: }} \\ \end{gathered}$

So, using these results, we get

$\rm \: = \: - 7 \: ( - sin5x)\dfrac{d}{dx}5x - cos3x\dfrac{\partial}{\partial x}3x=−7(−sin5x)$

$\rm \: = \: 7sin5x \times 5 - cos3x \times 3=7sin5x×5−cos3x×3$

$\rm \: = \: 35sin5x - 3cos3x=35sin5x−3cos3x$

Hence,

$\rm \: \dfrac{\partial}{\partial x} \bigg[7 \cdot \sin \bigg(5 \cdot x - \dfrac{\pi}{2} \bigg ) + \cos \bigg( 3 \cdot x + \dfrac{\pi}{2} \bigg) \bigg ]_{x = 2\pi}$

$\rm \: = \: 35sin10\pi \: - \: 3cos6\pi=35sin10π−3cos6π$

We know,

$\begin{gathered}\boxed{\tt{ sin \: n\pi = 0 \: \: \: where \: n \: is \: integer}} \\ \end{gathered}$

$\begin{gathered}\boxed{\tt{ cos \: n\pi = {( - 1)}^{n} \: \: \: where \: n \: is \: integer}} \\ \end{gathered}$

So, using these results, we get

$\rm \: = \: 0 - 3 {( - 1)}^{6}=0−3(−1)$

$\rm \: = \: - 3 \times 1=−3×1$

$\rm \: = \: - 3=−3$

Hence,

$\begin{gathered}\boxed{\tt{ \rm \: \dfrac{\partial}{\partial x} \bigg[7 \cdot \sin \bigg(5 \cdot x - \dfrac{\pi}{2} \bigg ) + \cos \bigg( 3 \cdot x + \dfrac{\pi}{2} \bigg) \bigg ]_{x = 2\pi} = - 3}} \\ \end{gathered}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

ADDITIONAL INFORMATION

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}\end{gathered}$