The Purplemath Forums

by **Martingale** on Mon Aug 23, 2010 12:45 am

burnbird16 wrote:But that's what I had originally, and I can't solve for y. Please, explain.

you can't solve...

Martingale wrote:
we know $\sin^2(t)+\cos^2(t)=1$

so replace with $x=\sin(t),<br />y=\cos(t)$

and solve for $y$

$x^2+y^2=1$ for $y$

subtract the $x^2$ then take the square root of both sides.

**Sponsor**

by **burnbird16** on Mon Aug 23, 2010 12:46 am

Oh, wait, I think I get it! do $x^2+y^2=1$ , then solve that for y! Awesome. Now, I only have one more problem...but it's in six parts. Basically, they are using simple trigonometric identities to create new ones. But I don't know how to get there. For example, I am to produce $sin^2x+cos^2x=1$ using $cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$

by **Martingale** on Mon Aug 23, 2010 12:55 am

burnbird16 wrote:Oh, wait, I think I get it! do $x^2+y^2=1$ , then solve that for y! Awesome. Now, I only have one more problem...but it's in six parts. Basically, they are using simple trigonometric identities to create new ones. But I don't know how to get there. For example, I am to produce $sin^2x+cos^2x=1$ using $cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$

let $y=-x$

and use the fact that $\cos(-x)=\cos(x)$

and $\sin(-x)=-\sin(x)$

by **burnbird16** on Mon Aug 23, 2010 12:59 am

Ah, okay, makes sense. Well, I'm gonna get to work on this, and I'll be back shortly. Again, thank you for all of your help, I couldn't have done this without your and Stapel's guidance.

by **burnbird16** on Mon Aug 23, 2010 1:42 am

Ugh, yet more issues arise.

How do I produce $cos(2x)=2cos^2(x)-1$ using the same $cos(x+y)$ identity as before? Also, how would I produce $|cos(x/2)|=sqrt((1+cos(x))/2)$ ?

by **Martingale** on Mon Aug 23, 2010 2:38 am

burnbird16 wrote:Ugh, yet more issues arise.

How do I produce $cos(2x)=2cos^2(x)-1$ using the same $cos(x+y)$ identity as before?

let $y=x$

so you get ...

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

now add and subtract a $\cos^2(x)$ into the right hand side of the equation

$\cos(2x)=\cos^2(x)+\cos^2(x)-\cos^2(x)-\sin^2(x)$

then group the terms the right way and use the Identity $\sin^2(x)+\cos^2(x)=1$ to get the identity you want

by **Martingale** on Mon Aug 23, 2010 2:41 am

burnbird16 wrote:Ugh, yet more issues arise.

How do I produce $cos(2x)=2cos^2(x)-1$ using the same $cos(x+y)$ identity as before? Also, how would I produce $|cos(x/2)|=sqrt((1+cos(x))/2)$ ?

to get $|\cos(x/2)|=\sqrt{\frac{1+\cos(x)}{2}}$

take

$\cos(2x)=2\cos^2(x)-1$

let $x=\frac{x}{2}$

then solve for the cosine that is on the right hand side.

by **burnbird16** on Mon Aug 23, 2010 2:46 am

Th-th-th-th-th-th-th-th-that's all, folks!

Thanks for all of your help, again, that finishes out the packet!

by **burnbird16** on Mon Aug 23, 2010 2:58 am

One tiny question though. How does the absolute value come into play? I'm to that point, but I don't understand the purpose of the absolute value there, or how to get it there.

by **Martingale** on Mon Aug 23, 2010 3:04 am

burnbird16 wrote:One tiny question though. How does the absolute value come into play? I'm to that point, but I don't understand the purpose of the absolute value there, or how to get it there.

for example .... suppose we have the function $f(x)=x^2$

taking the square root of both sides gives a new function...

then $g(x)=\sqrt{f(x)}=\sqrt{x^2}$

look what happens when we put in a few values

$g(0)=\sqrt{0^2}=\sqrt{0}=0$
$g(1)=\sqrt{1^2}=\sqrt{1}=1$
$g(-1)=\sqrt{(-1)^2}=\sqrt{1}=1$

let $x>0$
$g(\pm x)=\sqrt{(\pm x)^2}=\sqrt{x^2}=x$

so no matter what number I put in I get the positive version back...well...that is the absolute value function.

ie $g(x)=\sqrt{x^2}=|x|$

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