what is the arc cos 3pi/2 most similar to?
Given 3pi/2, thats impossible to solve without a calculator. the ARC COS of something is supposed to give you the angle. arc cos of something with Pi in it is seriously impossible.
Well, first of all, that's not a valid number as another poster said. I think you meant sqrt(3)/2 of which the arc cos of it is pi/6. Then arc cos sqrt(3)/2 is most similar to arcsin 1/2.
So let's say it's NOT sqrt(3)/2. Let's say it IS something with pi... let's say pi/5 ~ 0.628. Then what?
Well I would look for an answer with arcsin. The trick is that you have to think backwards a bit.
Think of cos(x) and sin(x) with the same value of x. Think about the triangle. You have legs A and B and hypotenuse C. You have angle a which is opposite side A. You have angle b opposite side B. You have angle c which is opposite side C and is 90 degrees. If you take cos(a) then you have for an answer adj/hyp = B/C. If you take sin(a) then you have for an answer opp/hyp = A/C. But since we're working with a unit circle, the value of C is 1. So we really have cos(a) = B and sin(a) = A. What do we know about the relationship between A and B? Well, A^2 + B^2 = C^2. And C^2 = 1. So we have A^2 + B^2 = 1.
What are A and B again? Remember cos(a) = B so arc cos(B) = a. But that means the value of B is pi/5.
So we need A such that (pi/5)^2 + A^2 = 1.
1 - pi^2/25 = A^2.
Using estimation you get roughly 1-10/25 = 1-2/5 = 3/5 = A^2. Since 3/5 = 0.60 and since I overestimated the value of pi^2, I can consider A^2 closer to 0.64 and so A ~ 0.8.
arc cos(pi/5) ~ 51.07 degrees.
arc sin(0.8) ~ 53.1 degrees.
That's pretty close... close enough to choose the correct answer. The actual answer is ~0.778.
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As far as word problems go, it's just a matter of translating english to math. About 90% or more of a word problem is garbage... things like names of people, where they are heading for distance problems, objects they are buying, and so on, do not matter for what you're trying to do. (I mean this in the sense that it does not matter what their name is. It doesn't matter what they are buying. You can use their name to pick out a variable. You look for their name to determine which variable you use later in the problem. But the name can be picked at random.) You're looking for relevant numbers... for words like 'is' or 'are' which scream out 'equal sign goes here'. When you see words like that you should stop and examine the parts of the problem to either side. For example, the stupid age problems...
'Bob is 4 years older than Dan now. In 6 years Bob will be three times Dan's age 2 years ago. How old is Bob now?'
Break it down...
'Bob is...' Enough said. We have an 'is'. That's an equal sign. What comes before it? Bob. Let Bob's current age = b. Then we have ' b = ' What comes after that? '4 years' That's a 4. 'b = 4'. What comes next? 'older than' That's a plus sign. 'b = 4 +'. What's next? 'Dan now'. Dan's current age which we'll put as a 'd'. 'b = 4 + d'.
Done. Period. That's the first equation. Two variables need two equations.
'In 6 years'. Stop. That means we're looking 6 more years for each person. In 6 years Bob will be b + 6 and Dan will be d + 6. Now we have to use those two values when we consider their future ages. 'Bob will be'. Stop again. We have 'Bob will' which means we want 'b + 6'. We then have 'be' which is an equal sign. 'b + 6 ='. Next. 'three times'. That's multiplication. 'b + 6 = 3*( )'. 'Dan's age two years ago.' Now we are talking about d - 2. So 'b + 6 = 3(d-2).
'How old is Bob now?' That's asking for the value of b.
Solve it...
b = 4 + d
b + 6 = 3(d-2)
Plug in...
4 + d + 6 = 3d - 6
10 + d = 3d - 6
16 = 2d
d = 8
b = 4 + d
b = 4 + 8
b = 12.
Bob is 12, Dan is 8. Bob is 4 years older than Dan. In 6 years Bob will be 18. Two years ago Dan was 6. So Bob's age in 6 years is 3 times as old as Dan's age 2 years ago.
Problem solved.
