Simplifying Trig Expression: (sec(t) - cos(t))/sec(t) = (f(t))^2

[3.141592654]

Homework Statement

Simplify and write the trigonometric expression in terms of sine and cosine:

(sec (t) - cos (t))/sec (t) = (f(t))^2

Homework Equations

sec (t)=1/cos (t)

The Attempt at a Solution

(sec (t) - cos (t))/sec (t)

= ((1/cos (t))-cos(t)) / (1/cos (t))

= ((1-cos^2(t))/cos(t)) / (1/cos (t))

From here, can I take the entire numerator, ((1-cos^2(t))/cos(t)), and divide it by one? This way I can do division of two rational numbers to get:

= cos(t)*((1-cos^2(t))/cos(t))

=1-cos^2(t) = sin^2(t) = (f(t))^2

so f(t)=sin (t)

I am quite sure this is the right answer, but I am wondering if the method I used is correct math. Thanks for your help everyone.

 

[konthelion]

That looks correct and your method is spot on.

 

[dlgoff]

= ((1/cos (t))-cos(t)) / (1/cos (t)) goes directly to 1-cos^2(t) by inverting the denominator and multiplying.

Save a step or two.

 

[3.141592654]

Thanks guys. I have another question as well, this one is as follows:

1. Homework Statement

sin(x)tan(x) = A) tan (x)
B) cos (x)
C) (1-cos^2(x)) / cos(x)

2. Homework Equations

tan (x) = sin(x)/cos(x)

3. The Attempt at a Solution

sin(x)tan(x)

=sin(x) * (sin(x)/cos(x))

=sin^2(x) / cos(x)

This is as far as I could simplify, and I can't see how it equals either a, b, or c. Did I make a mistake or is there a step I'm not seeing? Thanks again for your help.

 

[rock.freak667]

Know any useful trig identities with sin^2(x) and 1 in it?

 

[3.141592654]

haha yea as soon as I posted this I remembered good 'ol Pythagoras

 

[3.141492654]

alright, I have one more question.

1. Homework Statement

By using known trig identities, sin(2x)/(1+cos(2x)) can be written as:

A) tan(2x)
B) tan(x)
C) csc(2x)
D) sec(x)
E) all of the above
F) none of the above

2. Homework Equations

cos x = sin x/cos x

3. The Attempt at a Solution

sin (2x)/ (1+cos(2x))

=sin(2x) + (sin (2x) / cos(2x))

=sin (2x)+tan (2x)

This is all I have gotten, and don't feel like getting tan is probably going to help. I also proved that sin x=1+cos x but I haven't gotten anywhere with that either.

Thanks for any help!

 

[rock.freak667]

Try your double angle formulas for sin2x and cos2x

 

[3.141492654]

Alright, so:

2. Homework Equations

sin (2x) = 2sin(x)cos(x)

cos (2x) = cos^2(x)-sin^2(x)

3. The Attempt at a Solution

Sin (2x) / (1+cos(2x))

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

= 2 / (1+cos(x)-sin(x))

= 2sec(x)-csc(x)

= sec(2x)-csc(x)

This seems right, but it doesn't fit with any of the given answers? Did I trip up somewhere along the way?

 

[rock.freak667]

Alright, so:

2. Homework Equations

sin (2x) = 2sin(x)cos(x)

cos (2x) = cos^2(x)-sin^2(x)

3. The Attempt at a Solution

Sin (2x) / (1+cos(2x))

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

= 2 / (1+cos(x)-sin(x))

= 2sec(x)-csc(x)

= sec(2x)-csc(x)

This seems right, but it doesn't fit with any of the given answers? Did I trip up somewhere along the way?

the parts in red are incorrect.

From this line


$$\frac{2sinxcosx}{1+cos^2x-sin^2x}$$


use $sin^2x+cos^2x=1$ and you'll get through.