• Mar 6th 2006, 09:48 AM
Confuzzled?
1)Prove this identity
$\frac{1-cos\theta}{sin\theta}= \frac{sin\theta}{1+cos\theta}$

2)Solve this for $-\Pi\leq\theta\leq\Pi$

$sin2\theta=cos2\theta$

thanks if ya can :)
• Mar 6th 2006, 10:50 AM
earboth
Quote:

Originally Posted by Confuzzled?
1)Prove this identity
$\frac{1-cos\theta}{sin\theta}= \frac{sin\theta}{1+cos\theta}$

2)Solve this for $-\Pi\leq\theta\leq\Pi$

$sin2\theta=cos2\theta$

thanks if ya can :)

Hello,

to 1): Multiply both sides of the equation by $sin\theta$ and $1+cos\theta$. You'll get:

$1-cos^2\theta= sin^2\theta$

By the law (rule?) of Pythagoras both sides are equal.

to 2) sin(x) and cos(x) are equal for $x=\frac{\pi}{4}+k \cdot \pi$. So in your case you get:

$2\theta=\frac{\pi}{4}+k \cdot \pi$
$\theta=\frac{\pi}{8}+k \cdot \frac{\pi}{2}$

For the given domain: $\theta = \frac{5·\pi}{8} \vee \theta = - \frac{3·\pi}{8} \vee \theta = \frac{\pi}{8}$

Greetings

EB
• Mar 6th 2006, 02:38 PM
ThePerfectHacker
Quote:

Originally Posted by earboth
Hello,

to 1): Multiply both sides of the equation by $sin\theta$ and $1+cos\theta$. You'll get:

$1-cos^2\theta= sin^2\theta$

By the law (rule?) of Pythagoras both sides are equal.

to 2) sin(x) and cos(x) are equal for $x=\frac{\pi}{4}+k \cdot \pi$. So in your case you get:

$2\theta=\frac{\pi}{4}+k \cdot \pi$
$\theta=\frac{\pi}{8}+k \cdot \frac{\pi}{2}$

For the given domain: $\theta = \frac{5·\pi}{8} \vee \theta = - \frac{3·\pi}{8} \vee \theta = \frac{\pi}{8}$

Greetings

EB

Umm... You cannot multiply an identity and make it true and conclude that the hypothesis was true- That is called taking the converse of the statement.
• Mar 6th 2006, 10:43 PM
ticbol
Quote:

Originally Posted by Confuzzled?
1)Prove this identity
$\frac{1-cos\theta}{sin\theta}= \frac{sin\theta}{1+cos\theta}$

2)Solve this for $-\Pi\leq\theta\leq\Pi$

$sin2\theta=cos2\theta$

thanks if ya can :)

1) Let A = theta, so,

(1 -cosA)/sinA = sinA/(1 +cosA)

Easy is if we cross multiply,
(1 -cosA)(1 +cosA) = sinA*sinA
1 -cos^2(A) = sin^2(A)
sin^2(A) = sin^2(A)
and that is true.

The usual way, by developing one side of the equation to become exactly the other side,
LHS =
= (1 -cosA)/sinA
Multiply both numerator and denominator by (1 +cosA),
= [(1 -cosA)(1 +cosA)]/[sinA(1 +cosA)]
= [1 -cos^2(A)]/[sinA(1 +cosA)]
= [sin^2(A)]/[sinA(1 +cosA)]
= sinA/(1 +cosA)
= RHS

Therefore, proven.

----------------------------
2) Solve for angle A, if -pi < A < pi.

sin(2A) = cos(2A)
Divide both sides by cos(2A),
tan(2A) = 1
Positive tangent, so angle 2A is in the 1st or 3rd quadrants.
2A = arctan(1) ------***

In the interval/domain "-pi < A < pi", all measured counterclockwise,
----the 1st quadrant is from 0 to pi/2.
----the 3rd quadrant is from -pi to -pi/2.

2A = pi/4, in the 1st quadrant, so,
A = pi/8 -----------------------------***

2A = -pi +pi/4 = -3pi/4, in the 3rd quadrant, so,
A = -3pi/8 -----------------------------------------***

Check when A = pi/8,
sin(2*pi/8) =? cos(2*pi/8)
sin(pi/4) =? cos(pi/4)
1/sqrt(2) =? 1/sqrt(2)
Yes, so, OK.

Check when A = -3pi/8,
sin(2* -3pi/8) =? cos(2* -3pi/8)
sin(-3pi/4) =? cos(-3pi/4)
using a calculator,
-0.7071 =? -0.7071
Yes, so, OK.

• Mar 7th 2006, 10:04 AM
Confuzzled?
thank you guys for your help, it makes much more sense now. But i've stumbled across another question that I'm struggling with. Can anyone help me with this one:
Solve for $0<\theta<2\pi$
$tan3\theta(tan3\theta+1)=2$
• Mar 7th 2006, 11:00 AM
ticbol
Quote:

Originally Posted by Confuzzled?
thank you guys for your help, it makes much more sense now. But i've stumbled across another question that I'm struggling with. Can anyone help me with this one:
Solve for $0<\theta<2\pi$
$tan3\theta(tan3\theta+1)=2$

You should have just posted this one separately. Adding separate or new problems to an existing question is not a good practice.

But let me answer this new one here, for I am about to leave for work and so I cannot wait for a new/separate posting.

tan(3A)*[tan(3A) +1] = 2
tan^2(3A) +tan(3A) -2 = 0
That is like x^2 +x -2 = 0 if x=tan(3A), so, factoring,
[tan(3A) +2]*[tan(3A) -1] = 0

tan(3A) +2 = 0
tan(3A) = -2
Negative tangent, so angle 3A is in 2nd or 4th quadrants
3A = arctan(-2) -----***
Using a calculator,
In normal measuring, counterclockwise, that is (2pi -1.10715) = 5.176 rad
Hence,
A = 5.176/3 = 1.72535 radians ----------**

Counterclocwise, that is 2pi -4.2491 = 2.0341 rad
hence, A = 2.0341/3 = 0.678 rad --------**

Check when A = 1.72535 rad,
tan(3*1.72535)[tan(3*1.72535) +1] =? 2
(-2)[-2 +1] =? 2
-2[-1] =? 2
2 =? 2
Yes, so, OK.

Check when A = 0.678 rad,
tan(3*0.678)[tan(3*0.678) +1] =? 2
(-2)[-2 +1] =? 2
-2[-1] =? 2
2 =? 2
Yes, so, OK.

--------------
Ooppss, I forgot about the other factor.

tan(3A) -1 = 0
tan(3A) = 1
Positive tangent, so angle 3A is in 1st or 3rd quadrants
3A = arctan(1) ---***

3A = pi/4, in the 1st quadrant, so,
A = pi/4/3 = pi/12 rad ------**

3A = pi +pi/4 = 5pi/4, in the 3rd quadrant, so,
A = 5pi/4/3 = 5pi/12 rad -----**

Check when A = pi/12 rad,
tan(3*pi/12)[tan(3*pi/12) +1] =? 2
tan(pi/4)[tan(pi/4) +1] =? 2
(1)[1 +1] =? 2
1[2] =? 2
2 =? 2
Yes, so, OK.

Check when A = 5pi/12 rad,
tan(3*5pi/12)[tan(3*5pi/12) +1] =? 2
tan(5pi/4)[tan(5pi/4) +1] =? 2
(1)[1 +1] =? 2
1[2] =? 2
2 =? 2
Yes, so, OK.

Therfore, altogether,
A or theta = pi/12, 0.678, 5pi/12, or 1.72535, all in radians. ---answer.