# A few trig problems- help please (double angle and simplifying trig expressions)

• Apr 15th 2010, 07:39 PM
OnlyEternity
A few trig problems- help please (double angle and simplifying trig expressions)
Hey, I have been trying to figure out how to do these problems for quite some time, but I'm stuck. Help is very much appreciated.

Verify:

2- ((sin^2(x)/tan^2(x))= sin^2x-1

I really cannot figure this out.

Also, how do you solve this problem for theta? (this is one where I'm supposed to find the reference angle):

2sin2u+sin=0

and last thing, how do you find the cot(M+N)? Is it the same as the sum formula for cos but under 1?

Thank you so much!
• Apr 15th 2010, 07:46 PM
pickslides
Quote:

Originally Posted by OnlyEternity
Also, how do you solve this problem for theta? (this is one where I'm supposed to find the reference angle):

2sin2u+sin=0

here's a kick start...

$2\sin 2u+\sin u=0$

$2\times 2\sin u\cos u+\sin u=0$

$4\sin u\cos u+\sin u=0$

$4\sin u (\cos u+1)=0$
• Apr 15th 2010, 07:47 PM
harish21
Quote:

Originally Posted by OnlyEternity
Hey, I have been trying to figure out how to do these problems for quite some time, but I'm stuck. Help is very much appreciated.

Verify:

2- ((sin^2(x)/tan^2(x))= sin^2x-1

I really cannot figure this out.

Also, how do you solve this problem for theta? (this is one where I'm supposed to find the reference angle):

2sin2u+sin=0

and last thing, how do you find the cot(M+N)? Is it the same as the sum formula for cos but under 1?

Thank you so much!

$\frac{sin^{2}n}{tan^{2}n} = \frac{sin^{2}n}{\frac{sin^{2}n}{cos^{2}n}}= \frac{sin^{2}n}{1} \times\frac{cos^{2}n}{sin^{2}n}$
• Apr 15th 2010, 07:53 PM
harish21
Quote:

Originally Posted by OnlyEternity
Hey, I have been trying to figure out how to do these problems for quite some time, but I'm stuck. Help is very much appreciated.

Verify:

2- ((sin^2(x)/tan^2(x))= sin^2x-1

I really cannot figure this out.

Also, how do you solve this problem for theta? (this is one where I'm supposed to find the reference angle):

2sin2u+sin=0

and last thing, how do you find the cot(M+N)? Is it the same as the sum formula for cos but under 1?

Thank you so much!

$cot(M+N)= \frac{1}{tan(M+N)}$

or,

$cot(M+N) = \frac{cos(M+N)}{sin(M+N)}$
• Apr 15th 2010, 07:53 PM
OnlyEternity
Quote:

Originally Posted by pickslides
here's a kick start...

$2\sin 2u+\sin u=0$

$2\times 2\sin u\cos u+\sin u=0$

$4\sin u\cos u+\sin u=0$

$4\sin u (\cos u+1)=0$

Thanks! So after you get to this point, you solve for the reference angle and get:
Sin's ref <= 0 and cos's=180
so that means that u can = 0 degrees, 180 degrees or 360 degrees for sin, or 180 degrees for cos?

Hopefully that makes sense

Also:
Quote:

Originally Posted by harish21
$\frac{sin^{2}n}{tan^{2}n} = \frac{sin^{2}n}{\frac{sin^{2}n}{cos^{2}n}}= \frac{sin^{2}n}{1} \times\frac{cos^{2}n}{sin^{2}n}$

I got that far, but it's at this point where I don't know what to do to make it = to sin^2(x)-1

at that point I get 2-cos^2(x), but I need it to be sin^2(x)-1
• Apr 15th 2010, 07:59 PM
pickslides
Quote:

Originally Posted by OnlyEternity
Sin's ref <= 0 and cos's=180
so that means that u can = 0 degrees, 180 degrees or 360 degrees for sin, or 180 degrees for cos?

(Nod)
• Apr 15th 2010, 08:00 PM
OnlyEternity
Quote:

Originally Posted by pickslides
(Nod)

Great, thanks a lot! It makes more sense now.
• Apr 15th 2010, 08:12 PM
harish21
$2-cos^{2}x= 2-(1-sin^{2}x)=2-1+sin^{2}x = 1+sin^{2}x$

But your answer says the RHS= $sin^{2}x-1$, for which the equation would be false!