1)Prove this identity
$\displaystyle \frac{1-cos\theta}{sin\theta}= \frac{sin\theta}{1+cos\theta}$

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

$\displaystyle sin2\theta=cos2\theta$

thanks if ya can

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

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

$\displaystyle sin2\theta=cos2\theta$

thanks if ya can
Hello,

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

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

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

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

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

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

Greetings

EB

3. Originally Posted by earboth
Hello,

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

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

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

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

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

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

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.

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

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

$\displaystyle sin2\theta=cos2\theta$

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.

Therefore, A or theta = -3pi/4 or pi/4, in radians. ---------answer.

5. 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 $\displaystyle 0<\theta<2\pi$
$\displaystyle tan3\theta(tan3\theta+1)=2$

6. 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 $\displaystyle 0<\theta<2\pi$
$\displaystyle tan3\theta(tan3\theta+1)=2$
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,
3A = -1.10715 radians ---in the 4th quadrant
In normal measuring, counterclockwise, that is (2pi -1.10715) = 5.176 rad
Hence,
A = 5.176/3 = 1.72535 radians ----------**

3A = -1.10715 -pi = -4.2491 rad ----in the 2nd quadrant
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.

Therefore, A or theta = 0.678 or 1.72535, in radians. -----answer.

--------------
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.