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laila [671]
3 years ago
7

Verify the trigonometric identities

Mathematics
1 answer:
snow_lady [41]3 years ago
8 0
1)

here, we do the left-hand-side

\bf [sin(x)+cos(x)]^2+[sin(x)-cos(x)]^2=2
\\\\\\\
[sin^2(x)+2sin(x)cos(x)+cos^2(x)]\\\\+~ [sin^2(x)-2sin(x)cos(x)+cos^2(x)]
\\\\\\
2sin^2(x)+2cos^2(x)\implies 2[sin^2(x)+cos^2(x)]\implies 2[1]\implies 2

2)

here we also do the left-hand-side

\bf \cfrac{2-cos^2(x)}{sin(x)}=csc(x)+sin(x)
\\\\\\
\cfrac{2-[1-sin^2(x)]}{sin(x)}\implies \cfrac{2-1+sin^2(x)}{sin(x)}\implies \cfrac{1+sin^2(x)}{sin(x)}
\\\\\\
\cfrac{1}{sin(x)}+\cfrac{sin^2(x)}{sin(x)}\implies csc(x)+sin(x)

3)

here, we do the right-hand-side

\bf \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}=\cfrac{csc(x)-tan(x)}{sec(x)+cot(x)}
\\\\\\
\cfrac{csc(x)-tan(x)}{sec(x)+cot(x)}\implies \cfrac{\frac{1}{sin(x)}-\frac{sin(x)}{cos(x)}}{\frac{1}{cos(x)}-\frac{cos(x)}{sin(x)}}\implies \cfrac{\frac{cos(x)-sin^2(x)}{sin(x)cos(x)}}{\frac{sin(x)+cos^2(x)}{sin(x)cos(x)}}
\\\\\\
\cfrac{cos(x)-sin^2(x)}{\underline{sin(x)cos(x)}}\cdot \cfrac{\underline{sin(x)cos(x)}}{sin(x)+cos^2(x)}\implies \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}
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If f(x) = -x-2, what is f¹(x)?
astra-53 [7]

Answer:

  1. f¯¹(x) =9x+18, for the pictorial(image) question
  2. what is the send question is it asking derivative or inverse
  3. If it is inverse f¯¹(x)=-x-2 as it is
  4. Or if it is derivative f'(x)=-1

Step-by-step explanation:

<h2>For the image question f(x)=1/9x-2,f¯¹(x)=?</h2>
  • f(x)=1/9x-2............given
  • y=1/9x-2................swapping f(x) by y to Easily write
  • x=1/9y-2................interchanging x and y
  • 1/9y-2=x................changeling side of equation
  • 9(1/9y-2)=(x)9.......multiplying both sides by 9 to override the fraction on the right side
  • y-18=9x
  • y=9x-18.................Return to where it were
  • <em><u>f¯¹(x)=9x+18</u></em>..........swap back f¯¹(x) in the y

<h2>For the question f(x)=-x-2,f¯¹(x)=?</h2>
  • following the ☝️ arrangement
  • y=-x-2
  • x=-y-2
  • -y-2=x
  • -y=x+2
  • y=-x-2
  • f¯¹(x)=-x-2
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Step-by-step explanation:

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The authors of a paper describe an experiment to evaluate the effect of using a cell phone on reaction time. Subjects were asked
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Answer:

a) The 99% confidence interval would be given by (24.409;24.979)  

b) n=464

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Part a

The confidence interval for the mean is given by the following formula:  

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

df=n-1=47-1=46  

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,46)".And we see that t_{\alpha/2}=2.01  

Now we have everything in order to replace into formula (1):  

525-2.01\frac{75}{\sqrt{47}}=503.01  

525+2.01\frac{75}{\sqrt{47}}=546.99  

So on this case the 95% confidence interval would be given by (503.01;546.99)

Part b

The margin of error is given by this formula:  

ME=t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

And on this case we have that ME =7 msec, we are interested in order to find the value of n, if we solve n from equation (1) we got:  

n=(\frac{t_{\alpha/2} s}{ME})^2 (2)  

The critical value for 95% of confidence interval is provided, t_{\alpha/2}=2.01 from part a, replacing into formula (2) we got:  

n=(\frac{2.01(75)}{7})^2 =463.79 \approx 464  

So the answer for this case would be n=464 rounded up to the nearest integer  

8 0
3 years ago
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