Struggling on these C2 Questions

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+bx²)(1+3x)⁵ - 1+16x+103x²+cx³

Haven't seen a question like this before.

3.) Solve 1+logq=log2-2log5

Probable easy for you guys but logs are my kryptonite!

Reply 1

the bear

20

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.

Reply 2

username667907

4

1) Assuming you're allowed to use a calculator here, do

\arcsin\left(\dfrac{3}{5}\right)

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

cos x

and

tan x

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

(1+ax+bx^2)(1+3x)^5 = 1+16x+103x^2+cx^3

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 anti-log when you have just one log on one side and a number on the other.

the first rule you will need is:

n \log a = \log \left( a^n \right)

and the second rule is

\log a - \log b = \log \left( \dfrac{a}{b} \right)

(edited 11 years ago)

Reply 3

breakeven

OP

13

[QUOTE="Phredd;41870909"]1) Assuming you're allowed to use a calculator here, do

\arcsin\left(\dfrac{3}{5}\right)

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

cos x

and

tan x

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

(1+ax+bx^2)(1+3x)^5 = 1+16x+103x^2+cx^3

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 anti-log when you have just one log on one side and a number on the other.

the first rule you will need is:

n \log a = \log \left( a^n \right)

and the second rule is

Unparseable latex formula:

\log a - \log b = \log \left( \dfrac{a}{b} \right) [\latex]

Q2 - Its why I asked if there was a simpler way, seems like a pain in the butt to me,
Q1 - what is arcsin?

Reply 4

Brister

7

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+bx²)(1+3x)⁵ - 1+16x+103x²+cx³

Haven't seen a question like this before.

3.) Solve 1+logq=log2-2log5

Probable easy for you guys but logs are my kryptonite!

1) First of all, notice that you tried to take the inverse cosine/tangent of

\frac{3}{5}

, but that fraction is equal to

sinx

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

x

. 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

x

by symmetry.

Alternatively, you can avoid using inverse functions altogether. Using the identity

sin^2(x) + cos^2(x) = 1

, you can rearrange to find

cosx

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

tanx = \frac{sinx}{cosx}

and you're done.

2) The term

(1+3x)^5

can be expanded quickly using the binomial theorem. You can speed up the process further by noting that the polynomial on the right-hand side is a cubic, so you won't need to evaluate the

x^4

or

x^5

terms on the left. You then multiply the two remaining polynomials on the left-hand 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

n \log a = \log \left( a^n \right)

. You will also need to note that

1 = log 10

, and finally that

log a + log b = log (a \times b)

.

Don't hesitate to ask if this confuses you, although I hope it does not.

Reply 5

Brister

7

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?

arcsin

is just a different way of writing

sin^{-1}

and it tends to be less confusing.

Reply 6

breakeven

OP

13

Original post by Brister

1) First of all, notice that you tried to take the inverse cosine/tangent of

\frac{3}{5}

, but that fraction is equal to

sinx

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

x

. 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

x

by symmetry.

Alternatively, you can avoid using inverse functions altogether. Using the identity

sin^2(x) + cos^2(x) = 1

, you can rearrange to find

cosx

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

tanx = \frac{sinx}{cosx}

and you're done.

2) The term

(1+3x)^5

can be expanded quickly using the binomial theorem. You can speed up the process further by noting that the polynomial on the right-hand side is a cubic, so you won't need to evaluate the

x^4

or

x^5

terms on the left. You then multiply the two remaining polynomials on the left-hand 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

n \log a = \log \left( a^n \right)

. You will also need to note that

1 = log 10

, and finally that

log a + log b = log (a \times b)

.

Don't hesitate to ask if this confuses you, although I hope it does not.

Probably wrong but I had a go anyway.

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/log5²

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

Reply 7

Matureb

19

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.

Reply 8

breakeven

OP

13

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.

Oh right, so it will be 180-36.9 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?

(edited 11 years ago)

Reply 9

davros

16

Original post by breakeven

Oh right, so it will be 180-36.9 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?

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.

Reply 10

breakeven

OP

13

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.

Oh dear, now it makes sense. Thanks! <3

Reply 11

breakeven

OP

13

Any insight into the log question?

Reply 12

Matureb

19

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.

Reply 13

username667907

4

Original post by breakeven

Any insight into the log question?

how far have you got? you've been given plenty of good advice so far :smile:

Reply 14

breakeven

OP

13

Original post by Phredd

how far have you got? you've been given plenty of good advice so far :smile:

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

Reply 15

davros

16

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

You probably just need some practice with them to get more confidence. There aren't that many log rules to remember - the basic ones being:

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.

Reply 16

username667907

4

Original post by breakeven

...

you are not dumb: the people helping you have had lots of practice and experience, you just need to keep at it :smile:

ok i will go through this for you.
assuming this is log base 10:

1+\log q=\log 2 - 2\log 5

using

n\log a = \log a^n

:

1+\log q=\log 2- \log 25

and rearranging:

\log q = \log2 - \log25 - 1

we now need to use the fact that, by the definition of logs,

\log_{10} 10 = 1

, so

\log q = \log2 - \log25 - \log 10

finally, using

\log a - \log b = \log \left( \dfrac{a}{b} \right)

\log q = \log \left( \dfrac{2}{25} \right) - \log 10

and again:

\log q = \log \left( \dfrac{2}{250} \right)

therefore:

q = \dfrac{2}{250} = \dfrac{1}{125}

If I have gone too fast, and there are some steps that you don't follow feel free to ask :smile:

Reply 17

Brister

7

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

Don't be afraid to ask your teacher for clarification. It is much easier to help in person than through a computer, and teachers generally appreciate students who make an effort to understand. As Phredd says, this stuff really comes with practice. I think you will find that once you have spent enough time working with these functions, they are not so difficult after all. You will learn to see them as more than just symbols, which is especially important for trigonometric functions. My advice for those is to always do a quick sketch; it is usually much easier to make sense of a diagram than a symbol :smile:

Struggling on these C2 Questions

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