simplifying trig expressions using identities

[crescentcitycid]

I seem to be having a bit of trouble simplifying a couple trig functions. I think I'm missing something. Here's what I've got:

1} (1+cos(x)tan(x)csc(x))/(csc(x))
The cosecants cancel out, so I'm left with 1+cos(x)tan(x). I change tan(x) to sin(x)/cos(x) and multiply by cos(x) and I end up with 1+sin(x). Can it go any further?

2} (tan[sup:23f2r1cu]3[/sup:23f2r1cu](x)-sec[sup:23f2r1cu]2[/sup:23f2r1cu](x)tan(x))/cot(-x)
Starting with the numerator, I get (sin[sup:23f2r1cu]3[/sup:23f2r1cu](x)/cos[sup:23f2r1cu]3[/sup:23f2r1cu](x))-(1/cos[sup:23f2r1cu]2[/sup:23f2r1cu](x))(sin(x)/cos)(x)), then (sin[sup:23f2r1cu]3[/sup:23f2r1cu](x)/cos[sup:23f2r1cu]3[/sup:23f2r1cu](x))-(sin(x)/cos[sup:23f2r1cu]3[/sup:23f2r1cu](x), then finally (sin[sup:23f2r1cu]3[/sup:23f2r1cu](x)-sin(x))/cos[sup:23f2r1cu]3[/sup:23f2r1cu](x) in the numerator. In the denominator, I change cot(-x) to -(cos(x)/sin(x) then multiply (sin[sup:23f2r1cu]3[/sup:23f2r1cu](x)-sin(x))/cos[sup:23f2r1cu]3[/sup:23f2r1cu](x) by -(sin(x)/cos(x)) and come up with sin[sup:23f2r1cu]6[/sup:23f2r1cu](x)/cos[sup:23f2r1cu]4[/sup:23f2r1cu](x). I don't think that's correct, I think I'm screwing up the exponents.

 

[Mrspi]

I seem to be having a bit of trouble simplifying a couple trig functions. I think I'm missing something. Here's what I've got:

1} (1+cos(x)tan(x)csc(x))/(csc(x))
The cosecants cancel out, so I'm left with 1+cos(x)tan(x). I change tan(x) to sin(x)/cos(x) and multiply by cos(x) and I end up with 1+sin(x). Can it go any further?
I'm not sure what you mean by "the cosecants cancel out".....
Here's what I would do with the NUMERATOR....

1 + cos x * (sin x / cos x)*(1 / sin x)
or, simplifying the part AFTER the "+" sign, we have
1 + 1, which is 2

The NUMERATOR, then, is 2. And you've got 2 / (csc(x)), which can be simplified to 2 sin x


2} (tan[sup:2wirpjmp]3[/sup:2wirpjmp](x)-sec[sup:2wirpjmp]2[/sup:2wirpjmp](x)tan(x))/cot(-x)
Starting with the numerator, I get (sin[sup:2wirpjmp]3[/sup:2wirpjmp](x)/cos[sup:2wirpjmp]3[/sup:2wirpjmp](x))-(1/cos[sup:2wirpjmp]2[/sup:2wirpjmp](x))(sin(x)/cos)(x)), then (sin[sup:2wirpjmp]3[/sup:2wirpjmp](x)/cos[sup:2wirpjmp]3[/sup:2wirpjmp](x))-(sin(x)/cos[sup:2wirpjmp]3[/sup:2wirpjmp](x), then finally (sin[sup:2wirpjmp]3[/sup:2wirpjmp](x)-sin(x))/cos[sup:2wirpjmp]3[/sup:2wirpjmp](x) in the numerator.

Starting with the numerator, here's what I would do. Remove a common factor of tan x from both terms in the numerator:

tan x(tan[sup:2wirpjmp]2[/sup:2wirpjmp] x - sec[sup:2wirpjmp]2[/sup:2wirpjmp] x)

Now, from the Pythagorean Identities, you know that
tan[sup:2wirpjmp]2[/sup:2wirpjmp] x + 1 = sec[sup:2wirpjmp]2[/sup:2wirpjmp] x
From that, you can get a "replacement" for tan[sup:2wirpjmp]2[/sup:2wirpjmp] x - sec[sup:2wirpjmp]2[/sup:2wirpjmp] x
And the numerator simplifies nicely. Can you take it from there?

In the denominator, I change cot(-x) to -(cos(x)/sin(x) then multiply (sin[sup:2wirpjmp]3[/sup:2wirpjmp](x)-sin(x))/cos[sup:2wirpjmp]3[/sup:2wirpjmp](x) by -(sin(x)/cos(x)) and come up with sin[sup:2wirpjmp]6[/sup:2wirpjmp](x)/cos[sup:2wirpjmp]4[/sup:2wirpjmp](x). I don't think that's correct, I think I'm screwing up the exponents.

 

[crescentcitycid]

I seem to be having a bit of trouble simplifying a couple trig functions. I think I'm missing something. Here's what I've got:

1} (1+cos(x)tan(x)csc(x))/(csc(x))
The cosecants cancel out, so I'm left with 1+cos(x)tan(x). I change tan(x) to sin(x)/cos(x) and multiply by cos(x) and I end up with 1+sin(x). Can it go any further?
I'm not sure what you mean by "the cosecants cancel out".....
Here's what I would do with the NUMERATOR....

1 + cos x * (sin x / cos x)*(1 / sin x)
or, simplifying the part AFTER the "+" sign, we have
1 + 1, which is 2

The NUMERATOR, then, is 2. And you've got 2 / (csc(x)), which can be simplified to 2 sin x

I figured the csc(x) in the numerator and denominator cancelled each other out.

2} (tan[sup:1h3il4q2]3[/sup:1h3il4q2](x)-sec[sup:1h3il4q2]2[/sup:1h3il4q2](x)tan(x))/cot(-x)
Starting with the numerator, I get (sin[sup:1h3il4q2]3[/sup:1h3il4q2](x)/cos[sup:1h3il4q2]3[/sup:1h3il4q2](x))-(1/cos[sup:1h3il4q2]2[/sup:1h3il4q2](x))(sin(x)/cos)(x)), then (sin[sup:1h3il4q2]3[/sup:1h3il4q2](x)/cos[sup:1h3il4q2]3[/sup:1h3il4q2](x))-(sin(x)/cos[sup:1h3il4q2]3[/sup:1h3il4q2](x), then finally (sin[sup:1h3il4q2]3[/sup:1h3il4q2](x)-sin(x))/cos[sup:1h3il4q2]3[/sup:1h3il4q2](x) in the numerator.

Starting with the numerator, here's what I would do. Remove a common factor of tan x from both terms in the numerator:

tan x(tan[sup:1h3il4q2]2[/sup:1h3il4q2] x - sec[sup:1h3il4q2]2[/sup:1h3il4q2] x)

Now, from the Pythagorean Identities, you know that
tan[sup:1h3il4q2]2[/sup:1h3il4q2] x + 1 = sec[sup:1h3il4q2]2[/sup:1h3il4q2] x
From that, you can get a "replacement" for tan[sup:1h3il4q2]2[/sup:1h3il4q2] x - sec[sup:1h3il4q2]2[/sup:1h3il4q2] x
And the numerator simplifies nicely. Can you take it from there?

I think so, I end up with -tan[sup:1h3il4q2]3[/sup:1h3il4q2](x).

In the denominator, I change cot(-x) to -(cos(x)/sin(x) then multiply (sin[sup:1h3il4q2]3[/sup:1h3il4q2](x)-sin(x))/cos[sup:1h3il4q2]3[/sup:1h3il4q2](x) by -(sin(x)/cos(x)) and come up with sin[sup:1h3il4q2]6[/sup:1h3il4q2](x)/cos[sup:1h3il4q2]4[/sup:1h3il4q2](x). I don't think that's correct, I think I'm screwing up the exponents.

 

[Mrspi]

I seem to be having a bit of trouble simplifying a couple trig functions. I think I'm missing something. Here's what I've got:

1} (1+cos(x)tan(x)csc(x))/(csc(x))
The cosecants cancel out, so I'm left with 1+cos(x)tan(x). I change tan(x) to sin(x)/cos(x) and multiply by cos(x) and I end up with 1+sin(x). Can it go any further?
I'm not sure what you mean by "the cosecants cancel out".....
Here's what I would do with the NUMERATOR....

1 + cos x * (sin x / cos x)*(1 / sin x)
or, simplifying the part AFTER the "+" sign, we have
1 + 1, which is 2

The NUMERATOR, then, is 2. And you've got 2 / (csc(x)), which can be simplified to 2 sin x

I figured the csc(x) in the numerator and denominator cancelled each other out.

There's only one problem with your "figuring"....(and it is a BIG problem!)......csc x is NOT a factor of the numerator.

 

[soroban]

Hello, crescentcitycid!

\(\displaystyle [1]\;\;\frac{1+\cos x\!\cdot\!\tan x\!\cdot\!\csc x}{\csc x}\)

\(\displaystyle \displaystyle \frac{1 + \cos x\!\cdot\!\tan x\!\cdot\!\csc x}{\csc x} \;=\; \frac{1 + \cos x\cdot\dfrac{\sin x}{\cos x}\cdot\frac{1}{\sin x}}{\csc x} \;=\;\frac{1+1}{\frac{1}{\sin x}} \;=\;2\sin x\)

\(\displaystyle [2]\;\;\frac{\tan^3 - \sec^2\!x\!\cdot\!\tan x}{\cot(-x)}\)

\(\displaystyle \displaystyle{\frac{\tan^3x - \sec^2\!x\!\cdot\!\tan x}{\cot(-x)} \;=\;\frac{\tan x\overbrace{(\tan^2\!x - \sec^2\!x)}^{\text{This is -1}}}{-\cot x} \;=\;\frac{-\tan^2\!x}{-\frac{1}{\tan x}} \;=\;\tan^3\!x\)

 

[crescentcitycid]

Thanks Mrspi and soroban, I greatly appreciate the help.