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 :)

Printable View

- Mar 6th 2006, 09:48 AMConfuzzled?Please help with trigonometric functions!
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 :) - Mar 6th 2006, 10:50 AMearbothQuote:

Originally Posted by**Confuzzled?**

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 - Mar 6th 2006, 02:38 PMThePerfectHackerQuote:

Originally Posted by**earboth**

- Mar 6th 2006, 10:43 PMticbolQuote:

Originally Posted by**Confuzzled?**

(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. - Mar 7th 2006, 10:04 AMConfuzzled?
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 $\displaystyle 0<\theta<2\pi$

$\displaystyle tan3\theta(tan3\theta+1)=2$ - Mar 7th 2006, 11:00 AMticbolQuote:

Originally Posted by**Confuzzled?**

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.