Several Pre-calc problems

[colin]

Can anyone help me with these pre-calc problems?

1) Find sin-1(sin(8pi/11))

2) How man solutions of the equation sin(x) tan(x) = sin(x) are in the interval [0, 2pi)?

3) For x > 0, evaluate tan[arcsin(6/x)]

4) At 2:00pm, a ship leaves port and travels N20E at a rate oof 20 mph. At 2:30pm, another ship leaves the same port and travels S65W at 30 mph. How far apart are the 2 ships at 4:30?

I am using c2 = a2 + b2 – 2ab * cos(gamma) which is = (50)2 + (60)2 - 2(50)(60) cos(125), which is approximately 37mi., which I don’t believe is the right answer.

5) Find the solutions of the eqn. 3cosx = 2sin2 x that are in the interval [0, 2pi).




Thank you SO much if anyone can help with any of these.

 

[stapel]

1) Look at the definition of the inverse sine function. Find the angle within the function's domain that gives the same sine value as 8pi/11. Then apply the definition of inverse sine.

2) Either sin(x) = 0, or it doesn't. If it does, that gives you one list of solution values. If it doesn't, then you can divide through by sin(x), and find the solutions to the resulting tangent equation.

3) Draw a right triangle. Note that "arcsin(6/x) = ß" means "sin(ß) = 6/x". Use this to label the triangle with x, 6, and angle ß. Use the Pythagorean Theorem to find an expression (in terms of 6 and x) for the third side of the triangle. Then take the tangent.

4) Please describe the triangles you've drawn, including how you've labeled the various distances, and explain precisely where in this process you are stuck.

5) Use the double-angle identity to convert "sin(2x)" into "2sin(x)cos(x)". Then solve in a manner similar to (2).

If you get stuck, please reply showing (or describing very clearly) what you have tried. Thank you.

Eliz.

 

[colin]

OK, I think I've solved 2 and 3. For 2 I got 4 solutions, and for 3 I got 6/(sqrt(x^2 - 36)).

1) Look at the definition of the inverse sine function. Find the angle within the function's domain that gives the same sine value as 8pi/11. Then apply the definition of inverse sine.

4) Please describe the triangles you've drawn, including how you've labeled the various distances, and explain precisely where in this process you are stuck.

5) Use the double-angle identity to convert "sin(2x)" into "2sin(x)cos(x)". Then solve in a manner similar to (2).

1) I know the definition of the inverse sine, but am having trouble finding an angle that has the same sine value as 8pi/11. Any tips?

4) I have a=50, b=40, and gamma (across from c)=125. However, when I plug that in, I don't get a given correct answer (the multiple choice responses are 87, 45, 67, and 98 miles). 98 miles gives the closest response, but is still off by about a hundred.

5) Question: do I convert the sin2x first, or the sin^2x into (1-cos2x)/2 first? If the first way, would it come out to be 2 sin^2(x) cos ^2(x)?


Thank you again so much for your help.

 

[stapel]

1) Just use what you know about the graph of sine. Draw the basic period graph. Mark 0, pi/2, pi, (3/2)pi, and 2pi. Mark the (approximate) location of (8/11)pi. Figure out the other, smaller angle value that must give the same sine value.

4) Where are you getting "50", "40", and "125°" from? You said earlier that you were getting an answer of "37". How is "98" the "closest" of the listed options to your solution value?

5) Your original equation didn't have a sin²(x). What are you converting, to get one? Or is there a typo in your original post?

Please reply with clarifications. Thank you.

Eliz.

 

[colin]

I am very sorry, the original post had the typo; the correct problem is:

3cos(x) = 2sin²(x).

Again, my apologies.

 

[pka]

If you had 3y=2−2y<SUP>2</SUP>, could you solve it?
Note that 2sin<SUP>2</SUP>(x)=2−2cos<SUP>2</SUP>(x). Now let y=cos(x).

 

[stapel]

...the original post had the typo; the correct problem is:
3cos(x) = 2sin²(x).

Typoes happen. No biggie.

Since it's a sin²(x) instead of a sin(2x), follow pka's hint: Use the Pythagorean Identity to replace sin²(x) with 1 - cos²(x), and solve the resulting quadratic for cosine solutions. Then solve the two resulting trig equations.

Eliz.