# 2 Homework Problems

• Nov 23rd 2008, 09:33 PM
ChargersFan84
2 Homework Problems
I'd like to thank Chris for the help on my last post. I just have two more problems I could use assistance on and I'll be all set.

Find the values of all trigonometric functions of theta if tan (theta) = the square root of 7/3 and sec(theta) = 4/3 (Just the 7 is the square root)

Let x=3sin(theta), -pi/2<x<pi/2. Simplify the expression: x/square foot of 9 + x squared. (whole bottom is under square root.) I have put 3sin(theta) in for the x and the x squared, is this the right step? I just cannot get the hang of this! (Headbang) Thanks in advance for the help!
• Nov 23rd 2008, 09:54 PM
Chris L T521
Quote:

Originally Posted by ChargersFan84
I'd like to thank Chris for the help on my last post. I just have two more problems I could use assistance on and I'll be all set.

Find the values of all trigonometric functions of theta if tan (theta) = the square root of 7/3 and sec(theta) = 4/3 (Just the 7 is the square root)

Constructing a triangle, we see that $\text{opp}=\sqrt{7}$, $\text{adj}=3$ and $\text{hyp}=4$

You can now easily find the values of the 4 remaining trigonometric functions.

Quote:

Let x=3sin(theta), -pi/2<x<pi/2. Simplify the expression: x/square foot of 9 + x squared. (whole bottom is under square root.) I have put 3sin(theta) in for the x and the x squared, is this the right step? I just cannot get the hang of this! (Headbang) Thanks in advance for the help!
Substituting this into the expression, we get $\frac{3\sin\vartheta}{\sqrt{9-9\sin^2\vartheta}}=\frac{3\sin\vartheta}{3\sqrt{1-\sin^2\vartheta}}=\frac{\sin\vartheta}{\sqrt{\cos^ 2\vartheta}}=\dots$

Can you take it from here?

--Chris
• Nov 24th 2008, 08:06 AM
ChargersFan84
Thank you so much. I really appreciate it.
• Nov 24th 2008, 08:19 AM
David24
Quote:

Originally Posted by ChargersFan84
I'd like to thank Chris for the help on my last post. I just have two more problems I could use assistance on and I'll be all set.

Find the values of all trigonometric functions of theta if tan (theta) = the square root of 7/3 and sec(theta) = 4/3 (Just the 7 is the square root)

Let x=3sin(theta), -pi/2<x<pi/2. Simplify the expression: x/square foot of 9 + x squared. (whole bottom is under square root.) I have put 3sin(theta) in for the x and the x squared, is this the right step? I just cannot get the hang of this! (Headbang) Thanks in advance for the help!

hey mate, we are given that,

tan(A) = sqrt(7)/3 and sec(A) = 4/3

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

Therefore

sqrt(7)/3 = sin(A) * (4/3)
thus sin(A) = sqrt(7)/4
which you then can solve using your calculator (there may be a Identity Triangle to gain a direct value of A - I'll have a play around and if I find anything I'll let you know,

Hope this helps,

David