# Math Help - trig help needed

1. ## trig help needed

Hi
I have a few questions in can't get the right answers:

1)Simply the expression $sin^4x-cos^4x$

2)Simply the expression $tan^3x+tanx$
This is what i have done:
= $\frac{sinx^3}{cosx^3}+\frac{sinx}{cosx}$

= $\frac{sinx^3}{cos^3}+\frac{sinxcosx^2}{cosx^3}$

what should i do next?
3)if tanx=-4,x [ $\frac{-\pi}{2},0$] find: sec(x)

P.S

2. Originally Posted by Paymemoney
Hi
I have a few questions in can't get the right answers:

1)Simply the expression $sin^4x-cos^4x$ Note that this is the difference of squares.

2)Simply the expression $tan^3x+tanx$ Factor out a tanx, then note that (tan^2x+1)=sec^2x. What you have done is unneccesary.
This is what i have done:
= $\frac{sinx^3}{cosx^3}+\frac{sinx}{cosx}$

= $\frac{sinx^3}{cos^3}+\frac{sinxcosx^2}{cosx^3}$

what should i do next?
3)if tanx=-4,x [ $\frac{-\pi}{2},0$] find: sec(x)

P.S
.

3. Originally Posted by Paymemoney
Hi
I have a few questions in can't get the right answers:

1)Simply the expression $sin^4x-cos^4x$
You can begin by saying $\sin^4x-\cos^4x = \sin^2x\times \sin^2x - \cos^2x \times \cos^2x$

Now you can apply some identities for sin and cos all squared.

4. Originally Posted by VonNemo19
1)Simply the expression Note that this is the difference of squares.
How so?

5. Originally Posted by Paymemoney
Hi
I have a few questions in can't get the right answers:

1)Simply the expression $sin^4x-cos^4x$

2)Simply the expression $tan^3x+tanx$
This is what i have done:
= $\frac{sinx^3}{cosx^3}+\frac{sinx}{cosx}$

= $\frac{sinx^3}{cos^3}+\frac{sinxcosx^2}{cosx^3}$

what should i do next?
3)if tanx=-4,x [ $\frac{-\pi}{2},0$] find: sec(x)

P.S
1) $\sin^4{x} - \cos^4{x} = (\sin^2{x} + \cos^2{x})(\sin^2{x} - \cos^2{x})$

$= 1(\sin^2{x} - \cos^2{x})$

$= \sin^2{x} - \cos^2{x}$

$= \cos{2x}$.

2) $\tan^3{x} + \tan{x} = \tan{x}(\tan^2{x} + 1)$

$= \tan{x}\sec^2{x}$

3) $\tan{x} = -4$

$\tan^2{x} = 16$

$\tan^2{x} + 1 = 17$

$\sec^2{x} = 17$

$\sec{x} = \pm \sqrt{17}$.

Since $x$ is in the fourth quadrant, $\cos{x} > 0$ and so $\sec{x} > 0$.

Therefore $\sec{x} = \sqrt{17}$.

6. Originally Posted by Paymemoney
Hi
I have a few questions in can't get the right answers:

1)Simply the expression $sin^4x-cos^4x$

2)Simply the expression $tan^3x+tanx$
This is what i have done:
= $\frac{sinx^3}{cosx^3}+\frac{sinx}{cosx}$

= $\frac{sinx^3}{cos^3}+\frac{sinxcosx^2}{cosx^3}$

what should i do next?
3)if tanx=-4,x [ $\frac{-\pi}{2},0$] find: sec(x)

P.S
1) $sin^4x-cos^4x$

$(sin^2{x} + cos^2{x})(sin^2{x}-cos^2{x})
\Rightarrow
(1)(sin^2{x}-cos^2{x})
\Rightarrow
-cos{2x}$

7. Originally Posted by pickslides
How so?
Let $a=\sin^2{x}$ and $b=\cos^2{x}$

We then have $a^2-b^2$. Your way of doing things was equally valid. It looks as though prove it has combined our efforts.

8. Originally Posted by Prove It
3) $\tan{x} = -4$

$\tan^2{x} = 16$

$\tan^2{x} + 1 = 17$

$\sec^2{x} = 17$

$\sec{x} = \pm \sqrt{17}$.

Since $x$ is in the fourth quadrant, $\cos{x} > 0$ and so $\sec{x} > 0$.

Therefore $\sec{x} = \sqrt{17}$.
just to clarify what your doing is working backwards to find out sec(x)?

9. He is using the fact that

$
\tan^2{x} + 1 = \sec^2{x}
$

10. can someone explained to me how do you get $\frac{\sqrt17}{17}$ when you find cos(x) from tan(x)=-4.I understand that you get $cos(x)=\frac{1}{sec(x)}$then
$cos(x)=\frac{1}{\sqrt17}$ but where does the 17 come from?

11. Originally Posted by Paymemoney
can someone explained to me how do you get $\frac{\sqrt17}{17}$ when you find cos(x) from tan(x)=-4.I understand that you get $cos(x)=\frac{1}{sec(x)}$then
$cos(x)=\frac{1}{\sqrt17}$ but where does the 17 come from?
don't worry i undertsand how they got it, but why do you have to simply down to $\frac{\sqrt17}{17}$

12. 1)

.
Hey, shouldn't that be $-\cos(2x)$?

13. Originally Posted by bandedkrait
Hey, shouldn't that be $-\cos(2x)$?
It seems so

Table of Trigonometric Identities

14. Originally Posted by Paymemoney
don't worry i undertsand how they got it, but why do you have to simply down to $\frac{\sqrt17}{17}$
$\frac{1}{\sqrt{17}} = \frac{1}{\sqrt{17}}\times 1 = \frac{1}{\sqrt{17}}\times \frac{\sqrt{17}}{\sqrt{17}}= \frac{\sqrt17}{17}$