# 6 Trigonometry Questions needing help...

• Aug 12th 2007, 05:12 PM
forkball42
6 Trigonometry Questions needing help...
So I have like 6 problems right now left or so on my Calculus homework sheet (somewhere around like 150 problems total) and I'm having trouble showing these solutions.

Problem #1: Express sin(arccos x) in a form that contains no trigonometric functions or their inverses.

Problem #2: Use the trigonometric identity, sin(A+B) = (sin A cos B) + (cos A sin B), to verify that, sin2x = 2sin x cos x

Problem #3: Show that sin⁵ x cos⁴ x = sin⁵ x - 2sin⁷ x + sin⁹ x.

Problem #4: Show that sec³ x tan⁵ x = sec x tan x(sec⁶ x - 2sec⁴ x + sec² x).

Problem #5: Show that sin⁴ x = ³⁄₈ - ½cos 2x + ¹⁄₈cos 4x

Problem #6: Show that (1/1+ sin x) = sec² x - sec x tan x

And just to restate, that I'm sorry if it seems that I'm posting alot of questions on here but there is alot of problems and I do not have a book or any example problems to work off)

Thanks for any and all help.

Okay so I know that I'm a new member and all, and I'm sorry if it seems like
• Aug 12th 2007, 05:44 PM
Krizalid
1) You know that $\sin x=\sqrt{1-\cos^2x}$, now compute $\sin(\arccos x)$

2) $\sin2x=\sin(x+x)$, now apply the given formula.

It doesn't understand very well the exponents of some identities, try to use LaTeX
• Aug 12th 2007, 05:58 PM
Soroban
Hello, forkball42!

Welcome aboard . . . Here's some help . . .

Quote:

1) Express $\sin(\arccos x)$ in a form
that contains no trigonometric functions or their inverses.

Let $\theta \:=\:\arccos x\quad\Rightarrow\quad \cos\theta \:=\:x \:=\:\frac{x}{1} \:=\:\frac{adj}{hyp}$

So $\theta$ is an acute angle in a right triangle
. . with: . $adj \,= \,x,\:hyp \,= \,1$

Using Pythagorus, we find that: . $opp \,=\,\sqrt{1-x^2}$
. . Hence: . $\sin\theta \:=\:\frac{opp}{hyp} \:=\:\frac{\sqrt{1-x^2}}{1} \:=\:\sqrt{1-x^2}$

Therefore: . $\sin(\arccos x) \:=\:\sqrt{1-x^2}$

Quote:

2) Use the identity: . $\sin(A+B) \:= \:\sin A\cos B + \cos A\sin B$
to verify that: . $\sin2x \:= \:2\sin x \cos x$

We have: . $\sin2x \;=\;\sin(x + x) \;= \;\sin x\cos x + \sin x\cos x \;=\;2\sin x\cos x$

Quote:

3) Show that sin⁵ x cos⁴ x = sin⁵ x - 2sin⁷ x + sin⁹ x.
As you can see, we can't read the exponents.
. . But I'll take a guess . . .

Is it: . $\sin^2\!x\cos^4\!x \;=\;\sin^6\!x - 2\sin^4\!x + \sin^2\!x$ ?

We need the identity: . $\sin^2\!\theta + \cos^2\!\theta \:=\:1\quad\Rightarrow\quad\cos^2\!\theta \:=\:1 -\sin^2\!\theta$

We have: . $\sin^2\!x\cos^4\!x \;=\;\sin^2\!x\left(\cos^2\!x\right)^2 \;=\;\sin^2\!x\left(1 - \sin^2\!x\right)^2$

. . $= \;\sin^2\!x\left(1 - 2\sin^2\!x + \sin^4\!x\right) \;=\;\sin^2\!x - 2\sin^4\!x + \sin^6\!x$

• Aug 13th 2007, 12:24 PM
forkball42
Thank you for the help on problems 1 and 2. However I don't understand why my powers will not show up for your computers but I have managed to correctly solve 3 of the 4 other problems.

The one I still need is problem 5.

Here it is again and I'll make it to where I know you can read it.

Problem #5: Show that sin^4 x = (3/8) - (1/2)cos 2x + (1/8)cos 4x.

Sorry if this seems harder to read but it is the only way I can post the problem.

Thanks again for everything!
• Aug 13th 2007, 01:47 PM
Soroban
Hello, forkball42!

For #5, we will need these two double-angle identities:

. . $\sin^2\!\theta \:=\:\frac{1-\cos2\theta}{2}\qquad\qquad\cos^2\!\theta \:=\:\frac{1+\cos2\theta}{2}$

Quote:

5) Show that: . $\sin^4\!x \:=\:\frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$

We have: . $\sin^4\!x \:=\:\left(\sin^2\!x\right)^2 \;=\;\left(\frac{1-\cos2x}{2}\right)^2\;=\;\frac{1}{4}\left(1 - 2\cos2x + \cos^2\!2x\right)$

. . $= \;\frac{1}{4}\left(1 - 2\cos2x + \frac{1+\cos4x}{2}\right) \;=\;\frac{1}{4}\left(1 - 2\cos2x + \frac{1}{2} + \frac{1}{2}\cos4x\right)$

. . $= \;\frac{1}{4}\left(\frac{3}{2} - 2\cos2x + \frac{1}{2}\cos4x\right) \;=\;\frac{3}{8} - \frac{1}{2}\cos2x + \frac{1}{8}\cos4x$

• Aug 13th 2007, 02:18 PM
galactus
If you don't mind, here's another approach. More long connected than Soroban's, but just to show there's more than one way to skin a trig identity.

we will use $cos(2x)=cos^{2}(x)-sin^{2}(x)$ and

$cos(4x)=cos^{2}(2x)-sin^{2}(2x)$

$\frac{3}{8}-\frac{4cos(2x)}{8}+\frac{cos(4x)}{8}$

$\frac{3}{8}-\left(\frac{4cos^{2}(x)-4sin^{2}(x)}{8}\right)+\left(\frac{cos^{2}(2x)-sin^{2}(2x)}{8}\right)$

$\frac{3-4cos^{2}(x)+4sin^{2}(x)+cos^{2}(2x)-sin^{2}(2x)}{8}$

$\frac{3-4(1-sin^{2}(x))+4sin^{2}(x)+(1-sin^{2}(2x))-sin^{2}(2x)}{8}$

$\frac{3-4+4sin^{2}(x)+4sin^{2}(x)+1-sin^{2}(2x)-sin^{2}(2x)}{8}$

$\frac{8sin^{2}(x)-2sin^{2}(2x)}{8}$

But $2sin^{2}(2x)=8sin^{2}(x)cos^{2}(x)$

$\frac{8sin^{2}(x)-8sin^{2}(x)cos^{2}(x)}{8}$

$\frac{8sin^{2}(x)(1-cos^{2}(x))}{8}$

$sin^{4}(x)$
• Aug 13th 2007, 04:21 PM
Soroban
Hello, Galactus!

Good job . . . lovely work!