# Math Help - Trig help

1. ## Trig help

two prolems I really need to get done fast;

1. sin2(y)= cos4(y) when 0 greater or equal y less than or equal 360 degrees

2. solve algerbraically

sin^2(x)/ 1- cos(x) = sec (x) +1/ sec (x)

2. ## Trigonometry

Hello alpina
Originally Posted by alpina
two prolems I really need to get done fast;

1. sin2(y)= cos4(y) when 0 greater or equal y less than or equal 360 degrees

2. solve algerbraically

sin^2(x)/ 1- cos(x) = sec (x) +1/ sec (x)

$\sin 2y = \cos 4y$

$\Rightarrow \sin 2y = \sin(90 - 4y)$

Now $\sin A = \sin B \Rightarrow A = 360n +B$
or $180(2n+1) -B$, where $n$ is any integer

So you now need to investigate $n = 0, \pm 1, \pm 2, \dots$, and put $A = 2y$ and $B = 90 - 4y$.

For instance: $n = 0$ gives $2y = 90 - 4y$ or $2y = 180 - (90 - 4y)$

$\Rightarrow 6y = 90$
or $-2y = 90$

$\Rightarrow y = 15$ (or $y = -45$, which is out-of-range).

Then try $n = +1$, then $-1$, etc until you have all the values of $y$ in the range.

2 I assume that you need to prove the identity:

$\frac{\sin^2 x}{1-\cos x} = \frac{\sec x + 1}{\sec x}$

So: $\frac{\sin^2 x}{1-\cos x} =\frac{1 - \cos^2x}{1 - \cos x}$

$= 1 + \cos x$

$= \frac{\sec x}{\sec x} + \frac{1}{\sec x}$

$= \frac{\sec x + 1}{\sec x}$

3. Originally Posted by alpina
two prolems I really need to get done fast;

1. sin2(y)= cos4(y) when 0 greater or equal y less than or equal 360 degrees
Do you mean sin(2y)= cos(4y) or sin^2(y)= cos^4(y)? Since you use "^" in problem 2 I assume it is the first. cos(4y)= cos(2(2y))= cos^2(2y)- sin^2(2y)= (1- sin^2(2y))- sin^2(2y)= 1- 2sin^2(2y) so you can write the question as sin(2y)= 1- 2sin^2(2y). Let u= sin(2y) and the equation becomes u= 1- 2u^2 or 2u^2+ u- 1= 0. Solve that equation and, once you know u, solve sin(2y)= u for y.

2. solve algerbraically

sin^2(x)/ 1- cos(x) = sec (x) +1/ sec (x)

I assume you mean sin^2(x)/(1- cos(x))= (sec(x)+ 1)/sec(x). Multiplying on both sides by 1- cos(x) and sec(x), sin^2(x)sec(x)= (sec(x)+ 1)(1- cos(x)).
Since sec(x)= 1/cos(x), that is sin^2(x)/cos(x)= (1/cos(x)+ 1)(1- cos(x)).
Multiplying out the right side gives 1/cos(x)- cos(x)+ 1- 1= 1/cos(x)- cos(x).
And on the left, sin^2(x)= 1- cos^2(x) so sin^2(x)/cos(x)= (1- cos^2(x)/cos(x)= 1/cos(x)- cos(x).

Were you asked to "solve" this equation or prove that this is an identity?

Hello alpina
$\sin 2y = \cos 4y$

$\Rightarrow \sin 2y = \sin(90 - 4y)$

Now $\sin A = \sin B \Rightarrow A = 360n +B$ or $180(2n+1) -B$, where $n$ is any integer

So you now need to investigate $n = 0, \pm 1, \pm 2, \dots$, and put $A = 2y$ and $B = 90 - 4y$.

For instance: $n = 0$ gives $2y = 90 - 4y$ or $2y = 180 - (90 - 4y)$

$\Rightarrow 6y = 90$ or $-2y = 90$

$\Rightarrow y = 15$ (or $y = -45$, which is out-of-range).

Then try $n = +1$, then $-1$, etc until you have all the values of $y$ in the range.

2 I assume that you need to prove the identity:

$\frac{\sin^2 x}{1-\cos x} = \frac{\sec x + 1}{\sec x}$

So: $\frac{\sin^2 x}{1-\cos x} =\frac{1 - \cos^2x}{1 - \cos x}$

$= 1 + \cos x$

$= \frac{\sec x}{\sec x} + \frac{1}{\sec x}$

$= \frac{\sec x + 1}{\sec x}$

Thanks for your responses, and is there any other way to write the 1st problem? because that's not the way I usually solve them, so I am just wondering.

5. ## Trigonometry

Hello alpina
Originally Posted by alpina
Thanks for your responses, and is there any other way to write the 1st problem? because that's not the way I usually solve them, so I am just wondering.
Yes, as HallsOfIvy suggested, you can write $\cos 4y$ as $1 - 2 \sin^2 2y$ and get a quadratic equation in $\sin 2y$.

$\sin 2y = 1 - 2\sin^2 2y$

$\Rightarrow 2\sin^2 2y + \sin 2y - 1 = 0$

$\Rightarrow (2\sin 2y -1)(\sin 2y +1) = 0$

$\Rightarrow \sin 2y = \frac{1}{2}$ or $-1$

and take it from there.