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,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
1) Let A = theta, so,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.
You should have just posted this one separately. Adding separate or new problems to an existing question is not a good practice.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.