Boilanubjaini8f

Answered

2022-06-28

We must to calculate

$\underset{x\to \pi /2}{lim}\frac{\mathrm{sin}x-1}{2x-\pi}.$

$\underset{x\to \pi /2}{lim}\frac{\mathrm{sin}x-1}{2x-\pi}.$

Answer & Explanation

Zayden Wiley

Expert

2022-06-29Added 21 answers

We must to calculate

$\underset{x\to \pi /2}{lim}\frac{\mathrm{sin}x-1}{2x-\pi}$

If we use a change of variables $\text{}h=x-\frac{\pi}{2},\text{}$ then this becomes:

$\underset{h\to \text{}0}{lim}\frac{\mathrm{sin}(\frac{\pi}{2}+h)-\mathrm{sin}\left(\frac{\pi}{2}\right)}{2h}$

$\stackrel{(\ast )}{=}\frac{1}{2}\text{}{f}^{\prime}\left(\frac{\pi}{2}\right),$

where $\text{}f(x)=\mathrm{sin}(x).\text{}$ All we have done at (*) is use the definition of $\text{}{f}^{\prime}(x).$

But $\text{}{f}^{\prime}\left(\frac{\pi}{2}\right)=0,\text{}$ and therefore the answer is $\text{}\frac{1}{2}\times 0=0.$

Also, from the comments, you went wrong here:

$\underset{t\to 0}{lim}\frac{\mathrm{sin}(t+\frac{\pi}{2})-\mathrm{sin}\frac{\pi}{2}}{2t}=\underset{t\to 0}{lim}\frac{2\mathrm{sin}(t+\pi )\mathrm{cos}t}{2t}$

This is incorrect. The correct formula is:

$\text{}\mathrm{sin}\alpha -\mathrm{sin}\beta =2\mathrm{cos}\frac{\alpha +\beta}{2}\mathrm{sin}\frac{\alpha -\beta}{2},\text{}$ which would get you:

$\text{}\mathrm{sin}(t+\frac{\pi}{2})-\mathrm{sin}\left(\frac{\pi}{2}\right)=2\mathrm{cos}\left(\frac{1}{2}(t+\pi )\right)\mathrm{sin}\left(\frac{1}{2}t\right)$

$\underset{x\to \pi /2}{lim}\frac{\mathrm{sin}x-1}{2x-\pi}$

If we use a change of variables $\text{}h=x-\frac{\pi}{2},\text{}$ then this becomes:

$\underset{h\to \text{}0}{lim}\frac{\mathrm{sin}(\frac{\pi}{2}+h)-\mathrm{sin}\left(\frac{\pi}{2}\right)}{2h}$

$\stackrel{(\ast )}{=}\frac{1}{2}\text{}{f}^{\prime}\left(\frac{\pi}{2}\right),$

where $\text{}f(x)=\mathrm{sin}(x).\text{}$ All we have done at (*) is use the definition of $\text{}{f}^{\prime}(x).$

But $\text{}{f}^{\prime}\left(\frac{\pi}{2}\right)=0,\text{}$ and therefore the answer is $\text{}\frac{1}{2}\times 0=0.$

Also, from the comments, you went wrong here:

$\underset{t\to 0}{lim}\frac{\mathrm{sin}(t+\frac{\pi}{2})-\mathrm{sin}\frac{\pi}{2}}{2t}=\underset{t\to 0}{lim}\frac{2\mathrm{sin}(t+\pi )\mathrm{cos}t}{2t}$

This is incorrect. The correct formula is:

$\text{}\mathrm{sin}\alpha -\mathrm{sin}\beta =2\mathrm{cos}\frac{\alpha +\beta}{2}\mathrm{sin}\frac{\alpha -\beta}{2},\text{}$ which would get you:

$\text{}\mathrm{sin}(t+\frac{\pi}{2})-\mathrm{sin}\left(\frac{\pi}{2}\right)=2\mathrm{cos}\left(\frac{1}{2}(t+\pi )\right)\mathrm{sin}\left(\frac{1}{2}t\right)$

Yesenia Sherman

Expert

2022-06-30Added 5 answers

Another way to rewrite it is

$\frac{\mathrm{sin}(t+\pi /2)-1}{2t}=\frac{\mathrm{sin}t\mathrm{cos}(\pi /2)+\mathrm{sin}(\pi /2)\mathrm{cos}t-1}{2t}=\frac{\mathrm{cos}t-1}{2t}.$

$\frac{\mathrm{sin}(t+\pi /2)-1}{2t}=\frac{\mathrm{sin}t\mathrm{cos}(\pi /2)+\mathrm{sin}(\pi /2)\mathrm{cos}t-1}{2t}=\frac{\mathrm{cos}t-1}{2t}.$

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