trigenometry help

Posted on 2007-08-30 01:51 By jennifernavarre (0)

Categories:

I am a nursing student and do not understand math very well at all. i need help with 5 math problems . I would greatly appreciate any help out there. thanks in advance.(they are multiple choice problems.)

1) Convert this relationship to a form which has no trigonometric
relationships in the answer, just numbers.

[2*(tan of 15 degrees)]/[1 - tan^2 of 15 degrees]

a) sqrt3

b) 3

c) (sqrt3)/3

d) 3/(sqrt3)

2) Choose the answer which completes the identity below.( i believe this is D, am i correct?)

(The square of the tangent of x) + 1 = ?

or alternatively written in symbols this is [tan^2(x)] + 1 = ?

a) Sin^2(x) + Cos^2(x)

b) Cot(x)

c) Cos(x)

d) Sec^2(x)

3.) Simplify the expression

[cot(A)]*[sin(A)]

a) Cos(A)

b) Sin(A)

c) Tan(A)

d) Cot(A)

4.) What is the tangent of 105 degrees? (i believe this is D also??)

a) 2 + sqrt(3)

b) 2 - sqrt(3)

c) - 2 + sqrt(3)

d) - 2 - sqrt(3)

5.)

Solve for x

arccos x = arcsin (2/7)

Note that since I cannot write "sin to the -1 power" in Educator (or likewise cos to the -1 power), I am using the alternate notation arcsin and arccos, which mean exactly the same thing.

(4sqrt5)/7 = (4/7)sqrt5

(2sqrt5)/7 = (2/7)sqrt5

(sqrt5)/7 = (1/7)sqrt5

(3sqrt5)/7 = (3/7)sqrt5

Nurses need all the help we can give them

Comment on 2007-08-30 02:40 from Mariner ( Maple Rating 4

562)

Normally, I don't give help with homework, but these questions seem a bit out of line with the kind of math that I would expect a nurse to need. So, try these answers:

d
d
a
d
(3/7)sqrt5

Good luck with your chosen career.

J. Tarr

A little more explanation

Comment on 2007-08-30 11:55 from jmurphy (

11)

Hi,

I thought I would try and give a little more insight on how to arrive at the correct answer. Hopefully this will help you with future assignments and tests :)

Q1. To calculate the correct answer for this question you should use the Unit Circle - very useful in trig calculations. From this we can determine that:

tan(15)=(1/4)/(sqrt(3)/4)

After some simplification on the right hand side we arrive at:
tan(15)=1/sqrt(3)

We can then substitute
1/sqrt(3) for tan(15) into

(2*tan(15))/(1-(tan(15))^2)

And arrive at the final answer:
3/sqrt(3)

Q2. To solve this problem we need to use the trig identities. Most people memorize these however they can be derived - this way you don't have to remember as much :)

We will derivation method so you can see where the answer comes from.
Some things we need to know:
tan(x)=sin(x)/cos(x)
cos(x)^2+sin(x)^2=1
sec(x)=1/cos(x)

Ok so we know cos(x)^2+sin(x)^2=1 but our question is in terms of tan(x)^2

Fist we figure out how we can rewrite cos(x)^2+sin(x)^2=1 in terms of
tan(x)^2

We know that tan(x)=sin(x)/cos(x) and tan(x)^2=sin(x)^2/cos(x)^2 so a good start

would be to divide the equation

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

cos(x)^2

To get 1+sin(x)^2/cos(x)^2=1/cos(x)^2

Now we can simplify

1+tan(x)^2=sec(x)^2

From this we can conclude our answer of sec(x)^2

Q3. Again, we use trig identities to solve the problem.

cot(A)*sin(A)

=(1/tan(A))*sin(A)

=(1/(sin(A)/cos(A)))*sin(A)

=(cos(A)/sin(A))*sin(A)

=cos(A)

I will leave it at this. It seems like you were able to get the last 2 on your own. I know this is a bit hard on the eyes but hopefully you will find it useful.

Jenna