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Aleks04 [339]
2 years ago
15

Hello, precalc, need help on finding csc

Mathematics
1 answer:
Ainat [17]2 years ago
3 0

Recall the double angle identity for cosine:

\cos(2x) = \cos^2(x) - \sin^2(x) = 1 - 2 \sin^2(x)

It follows that

\sin^2(x) = \dfrac{1 - \cos(2x)}2 \implies \sin(x) = \pm \sqrt{\dfrac{1-\cos(2x)}2} \implies \csc(x) = \pm \sqrt{\dfrac2{1-\cos(2x)}}

Since 0° < 22° < 90°, we know that sin(22°) must be positive, so csc(22°) is also positive. Let x = 22°; then the closest answer would be C,

\csc(22^\circ) = \sqrt{\dfrac2{1-\cos(44^\circ)}} = \sqrt{\dfrac2{1-\frac5{13}}} = \dfrac{\sqrt{13}}2

but the problem is that none of these claims are true; cot(32°) ≠ 4/3, cos(44°) ≠ 5/13, and csc(22°) ≠ √13/2...

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Step-by-step explanation:

1) Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

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Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.35 of the area on the left and 0.65 of the area on the right it's z=-0.385. On this case P(Z<-0.385)=0.35 and P(Z>-0.385)=0.65

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