Help urgently (1 Viewer)

5uckerberg · Jul 1, 2022

Farhanthestudent005 said:
View attachment 35884

Which Dr Du class are you in?

a) Okay with this question we are required to find the auxiliary angle which is in the form $R\sin\left(x+\alpha\right)$ for $\sqrt{2}\sin{x}+\sqrt{2}\cos{x}$

Well, $R\sin\left(x+\alpha\right) = R\sin{x}\cos{\alpha}+R\cos{x}\sin{\alpha}$ where $R > 0 \& 0\leq{\alpha}\leq{\frac{\pi}{2}}$

Now, put $\sqrt{2}\sin{x}+\sqrt{2}\cos{x}$ and $R\sin{x}\cos{\alpha}+R\cos{x}\sin{\alpha}$ side by side what do you see.

$R\sin{\alpha}=\sqrt{2}$
$R\cos{\alpha}=\sqrt{2}$
$\tan{\alpha}=1\rightarrow{\tan^{-1}1=\frac{\pi}{4}$ noting the condition $0\leq{\alpha}\leq{\frac{\pi}{2}}$
$R^{2}\left(\sin^{2}\alpha+\cos^{2}\alpha\right)=4$
$R^{2}=4\rightarrow{R=2}$ noting the condition $R > 0$ . Putting it together we will have $\sqrt{2}\sin{x}+\sqrt{2}\cos{x}=2\sin\left(x+\frac{\pi}{4}\right)$ .

b) Now we have to sketch $2\sin\left(x+\frac{\pi}{4}\right)$ in the range of $0\leq{x}\leq{2\pi}$ .
Hint: Compare this to the original function $\sin{x}$ and note the differences. Then sketch this character.

That is only the first question. I imagine the later question will be harder than this one I presume. Anyways @Farhanthestudent005 just send them here. We are waiting for them.

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Help urgently (1 Viewer)

Farhanthestudent005

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5uckerberg

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