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

1. 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!

2. 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...

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

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

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

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

3. 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!
$\displaystyle \frac{sin^{2}n}{tan^{2}n} = \frac{sin^{2}n}{\frac{sin^{2}n}{cos^{2}n}}= \frac{sin^{2}n}{1} \times\frac{cos^{2}n}{sin^{2}n}$

4. 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!
$\displaystyle cot(M+N)= \frac{1}{tan(M+N)}$

or,

$\displaystyle cot(M+N) = \frac{cos(M+N)}{sin(M+N)}$

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

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

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

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

$\displaystyle 4\sin u (\cos u+1)=0$
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: Originally Posted by harish21 $\displaystyle \frac{sin^{2}n}{tan^{2}n} = \frac{sin^{2}n}{\frac{sin^{2}n}{cos^{2}n}}= \frac{sin^{2}n}{1} \times\frac{cos^{2}n}{sin^{2}n}$
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

6. 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? 7. Originally Posted by pickslides  Great, thanks a lot! It makes more sense now.

8. $\displaystyle 2-cos^{2}x= 2-(1-sin^{2}x)=2-1+sin^{2}x = 1+sin^{2}x$

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

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