I'm not asking for the answers flat out I just wondered if someone would be able to explain these questions as I'm not sure either what they;re asking or how I start to answer it!
1.) If Sinx = 3/5 and x is obtuse, calculate cosx and tanx.
I thought I had to put cos^{1}(3/5) and tan^{1}(3/5) And get the obtuse angles so unless ive done this wrong there are no obtuse angles for these? Any thoughts guys.
2.) Find a,b,c if
(1+ax+bx^{2})(1+3x)^{5}  1+16x+103x^{2}+cx^{3}
Haven't seen a question like this before.
3.) Solve 1+logq=log22log5
Probable easy for you guys but logs are my kryptonite!

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 19032013 13:33

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 19032013 13:40
for the last one you could express 1 as a logarithm then make a single logarithm on each side. Once you have done this then you can remove the logs and keep the other bits.

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 19032013 15:04
1) Assuming you're allowed to use a calculator here, do and you will find the primary solution, which you will note is acute. However, there is more than one solution to this. See if you can find an obtuse angle from this result. Drawing the sin curve and noting some symmetries might help. Once you have your answer, you can put in back into your calculator to find and you should get some nice fractions as your answers.
also note this is a 3 4 5 triangle, but you will need to think about it a bit using this method, since the angle is specified as obtuse.
2) i assume it's supposed to read
i may be missing a simpler way to do this, but try just expanding out the LHS and equating the coefficients.
3) are these logs with base 10? anyway, rearrange using the following rules of logarithms then antilog when you have just one log on one side and a number on the other.
the first rule you will need is:
and the second rule is
Last edited by Phredd; 19032013 at 18:01. 
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 4
 19032013 16:29
(Original post by Phredd)
1) Assuming you're allowed to use a calculator here, do and you will find the primary solution, which you will note is acute. However, there is more than one solution to this. See if you can find an obtuse angle from this result. Drawing the sin curve and noting some symmetries might help. Once you have your answer, you can put in back into your calculator to find and you should get some nice fractions as your answers.
also note this is a 3 4 5 triangle, but you will need to think about it a bit using this method, since the angle is specified as obtuse.
2) i assume it's supposed to read
i may be missing a simpler way to do this, but try just expanding out the LHS and equating the coefficients.
3) are these logs with base 10? anyway, rearrange using the following rules of logarithms then antilog when you have just one log on one side and a number on the other.
the first rule you will need is:
and the second rule is
[latex] \log a  \log b = \log \left( \dfrac{a}{b} \right) [\latex]
Q1  what is arcsin? 
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 5
 19032013 16:44
(Original post by breakeven)
I'm not asking for the answers flat out I just wondered if someone would be able to explain these questions as I'm not sure either what they;re asking or how I start to answer it!
1.) If Sinx = 3/5 and x is obtuse, calculate cosx and tanx.
I thought I had to put cos^{1}(3/5) and tan^{1}(3/5) And get the obtuse angles so unless ive done this wrong there are no obtuse angles for these? Any thoughts guys.
2.) Find a,b,c if
(1+ax+bx^{2})(1+3x)^{5}  1+16x+103x^{2}+cx^{3}
Haven't seen a question like this before.
3.) Solve 1+logq=log22log5
Probable easy for you guys but logs are my kryptonite!
I would highly recommend that you draw the graphs of sine and cosine, or a diagram of triangle in a unit circle if you have come across that representation. You can get the correct value of by symmetry.
Alternatively, you can avoid using inverse functions altogether. Using the identity , you can rearrange to find without using a calculator, noting that an obtuse angle gives a negative value for cosine (this is where the graph/diagram helps). Then you use the identity and you're done.
2) The term can be expanded quickly using the binomial theorem. You can speed up the process further by noting that the polynomial on the righthand side is a cubic, so you won't need to evaluate the or terms on the left. You then multiply the two remaining polynomials on the lefthand side and compare the coefficients to deduce a, b and c.
3) The answer will in fact depend on the base, but I will assume that you mean base 10. As Phredd said, you will first need . You will also need to note that , and finally that .
Don't hesitate to ask if this confuses you, although I hope it does not. 
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 19032013 16:46
(Original post by breakeven)
Q2  Its why I asked if there was a simpler way, seems like a pain in the butt to me,
Q1  what is arcsin? 
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 19032013 23:59
(Original post by Brister)
1) First of all, notice that you tried to take the inverse cosine/tangent of , but that fraction is equal to so you were in fact trying to take the inverse of a function of the angle (sorry if that sounds confusing). You should review the purpose of inverse functions. In any case, the inverse trigonometric functions are used to find angles. For this problem, the only inverse function that you could use is the inverse sine function to find your angle . Once you have the correct angle, you can then take the cosine/tangent of the angle using your calculator.
I would highly recommend that you draw the graphs of sine and cosine, or a diagram of triangle in a unit circle if you have come across that representation. You can get the correct value of by symmetry.
Alternatively, you can avoid using inverse functions altogether. Using the identity , you can rearrange to find without using a calculator, noting that an obtuse angle gives a negative value for cosine (this is where the graph/diagram helps). Then you use the identity and you're done.
2) The term can be expanded quickly using the binomial theorem. You can speed up the process further by noting that the polynomial on the righthand side is a cubic, so you won't need to evaluate the or terms on the left. You then multiply the two remaining polynomials on the lefthand side and compare the coefficients to deduce a, b and c.
3) The answer will in fact depend on the base, but I will assume that you mean base 10. As Phredd said, you will first need . You will also need to note that , and finally that .
Don't hesitate to ask if this confuses you, although I hope it does not.
For Q1 I Found Sin^{1} (3/5) = 36.9
And was I then supposed to do inverse cos and tan of 36.9?
Q2  I ended up with a=1, b=2, c=330
Took ages, and I think my some of my multiplications may be wrong but I'll check them tomorrow as its almost bedtime haha.
Q3  I got Log10q log2/log5^{2}
10q = 0.69
q=0.069
By the way, thanks so much for taking the time to help me, I appreciate it so much! <3 
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 20032013 06:34
For Q1, your method is fine but you were asked to find an obtuse angle. Use the symmetry of sine to find this from the angle you got.
Then you take the cos and tan of this angle. 
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 20032013 09:02
(Original post by Matureb)
For Q1, your method is fine but you were asked to find an obtuse angle. Use the symmetry of sine to find this from the angle you got.
Then you take the cos and tan of this angle.
Just done that, I got x=143.1
for Tan^{1} I got 89.6
for Cos^{1} I got error?
What have i done wrong?Last edited by breakeven; 20032013 at 09:16. 
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 10
 20032013 09:46
(Original post by breakeven)
Oh right, so it will be 18036.9 and then fine inverse cos and tan of that?
Just done that, I got x=143.1
for Tan^{1} I got 89.6
for Cos^{1} I got error?
What have i done wrong? 
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 20032013 10:06
(Original post by davros)
You found the angle by using the inverse sine. Now you have to find the cos and tan of the angle  you've been trying to to find the inverse of these which doesn't make sense! Just pretend someone gave you the angle in the first place, and work out cos x and tan x. 
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 20032013 12:43
Any insight into the log question?

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 20032013 13:06
Take the 1 to the other side of the equation, and change it to a log.
Gives you log q = lots of logs.
Then use the log laws. 
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 20032013 16:14
(Original post by breakeven)
Any insight into the log question? 
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 20032013 18:29
(Original post by Phredd)
how far have you got? you've been given plenty of good advice so far
Sorry, I don't have feel dumb compared to the people on this site.
I'm sorry I still havent got it after, as you say, all the good advice. I've handed in the work now anyway and I don't think it was right. I'll keep revising logs though as I struggle with them alot 
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 16
 20032013 18:57
(Original post by breakeven)
Sorry, I don't have feel dumb compared to the people on this site.
I'm sorry I still havent got it after, as you say, all the good advice. I've handed in the work now anyway and I don't think it was right. I'll keep revising logs though as I struggle with them alot
log(ab) = log a + log b
log(a/b) = log a  log b
log(a^n) = n(log a)
You didn't say what the base of the logarithms was in your original post, but if you are dealing with natural logarithms (usually written as ln), then log e = 1, where e is the base of natural logarithms. If you were talking about base 10 logarithms then you would have log 10 = 1.
Also note that log 1 = 0 in any base.
Using these rules in combination you can rearrange the equation you had to get the unknown variable in terms of things you know. 
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 20032013 19:07
(Original post by breakeven)
...
ok i will go through this for you.
assuming this is log base 10:
using :
and rearranging:
we now need to use the fact that, by the definition of logs, , so
finally, using
and again:
therefore:
If I have gone too fast, and there are some steps that you don't follow feel free to ask 
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 18
 21032013 11:59
(Original post by breakeven)
Sorry, I don't have feel dumb compared to the people on this site.
I'm sorry I still havent got it after, as you say, all the good advice. I've handed in the work now anyway and I don't think it was right. I'll keep revising logs though as I struggle with them alot
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