# Compound Angles or Double Angles?

• Jul 24th 2010, 07:42 AM
unstopabl3
Compound Angles or Double Angles?
Hi, I am stuck these types of questions and I just can't make any sense of them.
Do I apply compound angles or double angles or the simple trigno functions?

Kindly show me a few steps and I will take it from there.

Q) Find all the angles between 0 and 360 which satisfy the following equations:

1) $\displaystyle 2+sinycosy=2sin^2y$

2) $\displaystyle 2tan2x+sec40=1$

3) $\displaystyle 1+2sin(3z/2+75)=0$

Another question is that when they say sin40 or sin10+40, the 40,10 and 40 are all angles right, even if they don't write the degree sign on them?

Thanks!
• Jul 24th 2010, 08:34 AM
Prove It
They should be written with the degree sign, unless the angles are measured in radians, but in that case you will nearly always be given multiples of $\displaystyle \pi$.

3) Assuming the angles are measured in degrees...

$\displaystyle 1 + 2\sin{\left(\frac{3z}{2}+ 75 \right)^{\circ}} = 0$

$\displaystyle 2\sin{\left(\frac{3z}{2} + 75\right)^{\circ}}= -1$

$\displaystyle \sin{\left(\frac{3z}{2} + 75\right)^{\circ}} = -\frac{1}{2}$

$\displaystyle \left(\frac{3z}{2}\right)^{\circ} + 75^{\circ} = \left\{180^{\circ} + 30^{\circ}, 360^{\circ} - 30^{\circ}\right\} + 360^{\circ} n$ where $\displaystyle n \in \mathbf{Z}$

$\displaystyle \left(\frac{3z}{2}\right)^{\circ} + 75^{\circ} = \left\{ 210^{\circ}, 330^{\circ} \right\} + 360^{\circ}n$

$\displaystyle \left(\frac{3z}{2}\right)^{\circ} = \left\{ 135^{\circ}, 255^{\circ}\right\} + 360^{\circ}n$

$\displaystyle (3z)^{\circ} = \left\{270^{\circ}, 510^{\circ} \right\} + 720^{\circ}n$

$\displaystyle z^{\circ} = \left\{90^{\circ}, 170^{\circ} \right\} + 240^{\circ}n$.
• Jul 24th 2010, 08:46 AM
unstopabl3
Thanks, yes these are all in degrees :/
Can someone help me with the other 2 now ;)

Thanks!
• Jul 24th 2010, 09:08 AM
unstopabl3
Quote:

Originally Posted by Prove It
$\displaystyle z^{\circ} = \left\{90^{\circ}, 170^{\circ} \right\} + 240^{\circ}$.

Well I think you made a mistake somewhere because I am getting the following angles: 90, 170 and 330

Thanks anyways ;)
• Jul 24th 2010, 09:18 AM
Quote:

Originally Posted by unstopabl3
Hi, I am stuck these types of questions and I just can't make any sense of them.
Do I apply compound angles or double angles or the simple trigno functions?

Kindly show me a few steps and I will take it from there.

Q) Find all the angles between 0 and 360 which satisfy the following equations:

1) $\displaystyle 2+sinycosy=2sin^2y$

Thanks!

$\displaystyle 1=sin^2y+cos^2y\ \Rightarrow\ 2=2\left(sin^2y+cos^2y\right)$

$\displaystyle 2\left(sin^2y+cos^2y\right)+sinycosy=2sin^2y$

$\displaystyle 2\left(cos^2y+sin^2y+0.5sinycosy)=2sin^2y$

$\displaystyle sin^2y=cos^2y+sin^2y+0.5sinycosy$

$\displaystyle cos^2y+cosy(0.5siny)=0$

$\displaystyle cosy(cosy+0.5siny)=0$

gives 2 solutions.

Finish by writing $\displaystyle cosy=-0.5siny\ \Rightarrow\ -2=tany$
• Jul 24th 2010, 12:06 PM
Quote:

Originally Posted by unstopabl3
Well I think you made a mistake somewhere because I am getting the following angles: 90, 170 and 330

Thanks anyways ;)

$\displaystyle \frac{3z}{2}+75^o=210^o,\ 330^o,\ 570^o,\ 690^o....$

$\displaystyle \frac{3z}{2}=135^o,\ 255^o,\ 495^o,\ 615^o....$

$\displaystyle 3z=270^o,\ 510^o,\ 990^o,\ 1230^o....$

$\displaystyle z=90^o,\ 170^o,\ 330^o.......$

(2) $\displaystyle 2tan2x+sec40^o=1$

$\displaystyle 2tan2x+\frac{1}{cos40^o}=1$

$\displaystyle 2tan2x=1-\frac{1}{cos40^o}$

$\displaystyle tan2x=\frac{cos40^o-1}{2cos40^0}$

That can be evaluated and you can finish with the identity $\displaystyle tan2x=\frac{2tanx}{1-tan^2x}$

to get a quadratic equation in tanx......

if you want some practice,
though that identity is used to discover tan2x from tanx.

Instead just take the inverse tan of both sides to get 2x between 0 and 720 degrees.
Since tan(angle) is the slope of a line, remember there is a second solution 180 degrees away
from the solution returned by a calculator.
• Jul 24th 2010, 12:18 PM
Mathelogician
Quote:

$\displaystyle \frac{3z}{2}+75^o=210^o,\ 330^o,\ 570^o,\ 690^o....$

$\displaystyle \frac{3z}{2}=135^o,\ 255^o,\ 495^o,\ 615^o....$

$\displaystyle 3z=270^o,\ 510^o,\ 990^o,\ 1230^o....$
$\displaystyle z=90^o,\ 170^o,\ 330^o.......$

(2) $\displaystyle 2tan2x+sec40^o=1$

$\displaystyle 2tan2x+\frac{1}{cos40^o}=1$

$\displaystyle 2tan2x=1-\frac{1}{cos40^o}$

$\displaystyle tan2x=\frac{cos40^o-1}{2cos40^0}$

That can be evaluated and you can finish with the identity $\displaystyle tan2x=\frac{2tanx}{1-tan^2x}$

to get a quadratic equation in tanx.

Using that identity makes it complicated!
Instead, it can be solved using the general solution for tanx. That's let tan2x=a => 2x= k.pi+Arctg(a) => x= ./5(k.pi+Arctg(a)) for evrey k in Z.
• Jul 24th 2010, 09:32 PM
Prove It
Quote:

Originally Posted by unstopabl3
Well I think you made a mistake somewhere because I am getting the following angles: 90, 170 and 330

Thanks anyways ;)

I did not make a mistake.

You keep adding multiples of $\displaystyle 240^{\circ}$ until you have all possible solutions in your given domain.

What is $\displaystyle 90^{\circ} + 240^{\circ}$?
• Jul 25th 2010, 09:05 AM
unstopabl3
Quote:

Originally Posted by Prove It
I did not make a mistake.

You keep adding multiples of $\displaystyle 240^{\circ}$ until you have all possible solutions in your given domain.

What is $\displaystyle 90^{\circ} + 240^{\circ}$?

Well 240 didn't have a "n" written infront of it in your first attempt. Maybe that's why I was confused.
Thanks for fixing that up :)