# Calculate the probability that a head will show

A kind of the trigonometry question...

Let θ be an angle in standard position such that cotθ=-4/3 and sinθ<0.
Determine the exact value of sec θ

Determine the number of solutions for (asinx-b)(acosx-a)(bsinx+a)=0
where 0≦x＜2$$\pi$$, if 0＜a＜b.

A fair coin is tossed 60 times. Using the normal approximation to the binomial distribution, calculate the probability that a head will show between 32 and 36 times.

Please give me as many explanations as possible~ thanks

A kind of the trigonometry question...

Let θ be an angle in standard position such that cotθ=-4/3 and sinθ<0.
Determine the exact value of sec θ

Determine the number of solutions for (asinx-b)(acosx-a)(bsinx+a)=0
where 0≦x＜2$$\pi$$, if 0＜a＜b.

A fair coin is tossed 60 times. Using the normal approximation to the binomial distribution, calculate the probability that a head will show between 32 and 36 times.

Please give me as many explanations as possible~ thanks

Well, where are sine and cot negative at the same time? Quadrant IV. So, draw a triangle whose adjacent side is 4, and whose opposite side is (negative) three. By doing this, you are in essence drawing cot(-4/3). Solve for the hypotenuse of this triangle, and find out what secant is.

For problem two:
sinx=b/a, cosx=1, sinx=-a/b. Since a<b, sinx=b/a does not exist, because |b/a|>1, which is not in the range of the arcsin function. Cosx=1 at one place (x=0), and sinx=b/a in two places: where sinx is positive. I'm not sure on this, but I think you can say that there are only two solutions, because the coefficient of the x is only 1. So, on the interval [0,2$$\Pi$$], there are three solutions.

For problem three:
I'm not sure on these...haven't done probability in a very long time. Here's my guess though. Type this on your calculator: binomcdf(60,0.5,36)-binomcdf(60,0.5,32). Another guess would be normalcdf(32,36) (also on your calculator).

Last edited:
Defennder
Homework Helper

and the thread title A kind of the trigonometry question...

Let θ be an angle in standard position such that cotθ=-4/3 and sinθ<0.
Determine the exact value of sec θ

Determine the number of solutions for (asinx-b)(acosx-a)(bsinx+a)=0
where 0≦x＜2$$\pi$$, if 0＜a＜b.

A fair coin is tossed 60 times. Using the normal approximation to the binomial distribution, calculate the probability that a head will show between 32 and 36 times.

Please give me as many explanations as possible~ thanks

It's ok ==~~ not a creative name...

and the thread title how's the title interesting @@

You mean the form?

Well, where are sine and cot negative at the same time? Quadrant IV. So, draw a triangle whose adjacent side is 4, and whose opposite side is (negative) three. By doing this, you are in essence drawing cot(-4/3). Solve for the hypotenuse of this triangle, and find out what secant is.

For problem two:
sinx=b/a, cosx=1, sinx=-a/b. Since a<b, sinx=b/a does not exist, because |b/a|>1, which is not in the range of the arcsin function. Cosx=1 at one place (x=0), and sinx=b/a in two places: where sinx is positive. I'm not sure on this, but I think you can say that there are only two solutions, because the coefficient of the x is only 1. So, on the interval [0,2$$\Pi$$], there are three solutions.

For problem three:
I'm not sure on these...haven't done probability in a very long time. Here's my guess though. Type this on your calculator: binomcdf(60,0.5,36)-binomcdf(60,0.5,32). Another guess would be normalcdf(32,36) (also on your calculator).

I got the second one ~~

however... I still don't know how to do the first one and the third one

Well, where are sine and cot negative at the same time? Quadrant IV. So, draw a triangle whose adjacent side is 4, and whose opposite side is (negative) three. By doing this, you are in essence drawing cot(-4/3). Solve for the hypotenuse of this triangle, and find out what secant is.

For problem two:
sinx=b/a, cosx=1, sinx=-a/b. Since a<b, sinx=b/a does not exist, because |b/a|>1, which is not in the range of the arcsin function. Cosx=1 at one place (x=0), and sinx=b/a in two places: where sinx is positive. I'm not sure on this, but I think you can say that there are only two solutions, because the coefficient of the x is only 1. So, on the interval [0,2$$\Pi$$], there are three solutions.

For problem three:
I'm not sure on these...haven't done probability in a very long time. Here's my guess though. Type this on your calculator: binomcdf(60,0.5,36)-binomcdf(60,0.5,32). Another guess would be normalcdf(32,36) (also on your calculator).

so many rules =o=

1) draw x-y axis.

plot cot (t) = -4/3 (it can be at two positions)
cot(t) = x/y
so you know what x and y would give that

once you have two vectors(or lines) find where the second condition is met,
you would left with one vector (or line)

using that line find your sec t

3. no clue (can't remember how to do those questions)

remove [solved] if possible from the thread ...

------------------------

It's font. I know when you use chinese font and write in english it looks really cool to me!

1) Draw X-y Axis.

Plot Cot (t) = -4/3 (it Can Be At Two Positions)
Cot(t) = X/y
So You Know What X And Y Would Give That

Once You Have Two Vectors(or Lines) Find Where The Second Condition Is Met,
You Would Left With One Vector (or Line)

Using That Line Find Your Sec T

3. No Clue (can't Remember How To Do Those Questions)

Remove [solved] If Possible From The Thread ...

------------------------

It's Font. I Know When You Use Chinese Font And Write In English It Looks Really Cool To Me!

Ｏｈ～～　Ｉ　ｄｉｄ　ｇｅｔ　ｔｈｅ　ｆｉｒｓｔ　ｏｎｅ。。。

Ｔｈａｎｋｓ　ｆｏｒ　ｔｈｅ　ｉｎｓｐｉｒａｔｉｏｎ

ｔｈｅ　ｔｈｉｒｄ　ｏｎｅ　ｉｓ　。。。ｓｔｉｌｌ　ｃｏｎｆｕｓｅｄ。。。

ｈａｈａ。。。　Ｉ　ｇｕｅｓｓ　ｉｔ’ｓ　ｃｕｔｅｒ。。　ｓｏ　ｔｈａｔ’ｓ　ｗｈｙ　ｉ　ｕｓｅ　ｉｔ　ａｌｌ　ｔｈｅ　ｔｉｍｅ