Compound Angles or Double Angles?

**unstopabl3** · July 24th 2010, 07:42 AM

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) $2+sinycosy=2sin^2y$

2) $2tan2x+sec40=1$

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

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!

**Prove It** · July 24th 2010, 08:34 AM

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 $\pi$ .

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

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

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

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

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

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

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

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

$z^{\circ} = \left\{90^{\circ}, 170^{\circ} \right\} + 240^{\circ}n$ .

**unstopabl3** · July 24th 2010, 08:46 AM

Thanks, yes these are all in degrees :/
Can someone help me with the other 2 now

Thanks!

**unstopabl3** · July 24th 2010, 09:08 AM

Originally Posted by Prove It

$z^{\circ} = \left\{90^{\circ}, 170^{\circ} \right\} + 240^{\circ}$ .

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

Thanks anyways

**Archie Meade** · July 24th 2010, 09:18 AM

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) $2+sinycosy=2sin^2y$

Thanks!

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

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

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

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

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

$cosy(cosy+0.5siny)=0$

gives 2 solutions.

Finish by writing $cosy=-0.5siny\ \Rightarrow\ -2=tany$

**Archie Meade** · July 24th 2010, 12:06 PM

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

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

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

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

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

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

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

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

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

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

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.

**Mathelogician** · July 24th 2010, 12:18 PM

Originally Posted by Archie Meade

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

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

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

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

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

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

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

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

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.

**Prove It** · July 24th 2010, 09:32 PM

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 $240^{\circ}$ until you have all possible solutions in your given domain.

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

**unstopabl3** · July 25th 2010, 09:05 AM

Originally Posted by Prove It

I did not make a mistake.

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

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

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

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