# Pythagorean Identities

• Nov 25th 2006, 09:07 AM
kcsteven
Pythagorean Identities
I am still having trouble with these, maybe you could explain what I am doing wrong, here are a few I have been working on.
find tan angle
tan angle in QII, sec angle=-4/sqrt(5)
tan 2 angle in QI, cos 2 angle=3/5
tan 2 angle in QI, csc 2 angle=4/3
tan angle/2 in QII, Sec angle/2=-4
sin angle/2=1/3

here is what I did
sec angle=-4/sqrt(5), sec=1/cos
tan^2 angle+1=(-4/sqrt(5))^2
tan^2 angle +1= 16/5
since sec is neg is QII, tan=-sqrt(11/5)

cos2 angle=3/5, cos is pos in QI
cos^2 angle=1-sin^2 angle
(3/5)^2=1-sin^2 angle
sin angle= 4/5
tan=sin angle/con angle= 4/5/3/5= 4/3
tan= 2(4/3)= 8/3

tan2 angle in QI, csc2 angle= 4/3, csc is pos in QI
csc angle = 1/sin angle
cos^2 angle=1-1/4/3
cos^2= 9/4, cos= 3/4
csc2 angle, cos angle=9/4
tan^2 angle +1=(1/9/4)^2
tan^2 angle +1= 16/81
I can see this is going badly.

tan angle/2 in QII, sec angle/2=-4, sec in QII is neg
sec= 1/cos angle
tan^2 angle+1=(-8)^2
tan^2 angle=63
tan=sqrt(63)
tan angle/2 in QII= -6sqrt(7)

sin angle/2= 1/3
+-sqrt(1-cos a/2)= +-sqrt(1-1/3/2)= +-sqrt(1/3)

Could you please show me what I am doing wrong. I have been working on this for a few hours and I think a thirty minute break to clear my head might do some good before I start doing some more problems. I would appreciate all and any suggestions you can give because I reall want to understand these identities.
Thank You very much!!!!
Keith Stevens
• Nov 25th 2006, 11:41 AM
Soroban
Hello, Keither!

Instead of wrestling with various identities,
. . why not use the method I've used . . . "build the triangle."

Quote:

Find $\displaystyle \tan\theta$

$\displaystyle (1)\;\theta \text{ in Q2},\;\sec\theta = -\frac{4}{\sqrt{5}}$

Since $\displaystyle \sec\theta = \frac{hyp}{adj}$, we have: .$\displaystyle \sec\theta \:=\:\frac{4}{-\sqrt{5}} \:=\:\frac{hyp}{adj}$

From $\displaystyle opp^2 + adj^2\:=\:hyp^2$, we have: .$\displaystyle opp^2 +(-\sqrt{5})^2\:=\:4^2\quad\Rightarrow\quad opp = \pm\sqrt{11}$
In $\displaystyle Q2$, opposite is positive: $\displaystyle opp = \sqrt{11}$

We know all three sides of the triangle: .$\displaystyle opp = \sqrt{11},\;adj = -\sqrt{5},\;hyp = 4$

We can write any function now: .$\displaystyle \tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{\sqrt{11}}{-\sqrt{5}} \:=\:\boxed{-\frac{\sqrt{55}}{5}}$

Quote:

$\displaystyle (2)\;2\theta\text{ in Q1},\;\cos2\theta = \frac{3}{5}$

This time using identities is faster . . .

We have: .$\displaystyle \sin^2\!\theta \:=\:\frac{1 - \cos2\theta}{2}\quad\Rightarrow\quad \sin\theta \:=\:\pm\sqrt{\frac{1-\cos2\theta}{2}}$

. . Then: .$\displaystyle \sin\theta\:=\:\pm\sqrt{\frac{1 - \frac{3}{5}}{2}} \:=\:\pm\sqrt{\frac{\frac{2}{5}}{2}} \:=\:\pm\sqrt{\frac{1}{5}}\:=\:+\frac{1}{\sqrt{5}}$ . . . (positive in Q1)

We have: .$\displaystyle \cos^2\!\theta \:=\:\frac{1 + \cos2\theta}{2}\quad\Rightarrow\quad\cos\theta \:=\:\pm\sqrt{\frac{1 + \cos2\theta}{2}}$

. . Then: .$\displaystyle \cos\theta\:=\:\pm\sqrt{\frac{1 + \frac{3}{5}}{2}} \:=\:\pm\sqrt{\frac{\frac{8}{5}}{2}}\:=\:\pm\sqrt{ \frac{4}{5}} \:=\:+\frac{2}{\sqrt{5}}$ . . . (positive in Q1)

Therefore: .$\displaystyle \tan\theta \:=\:\frac{\sin\theta}{\cos\theta} \:=\:\frac{\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} \:=\:\boxed{\frac{1}{2}}$

• Nov 25th 2006, 11:59 AM
topsquark
Quote:

Originally Posted by kcsteven
I am still having trouble with these, maybe you could explain what I am doing wrong, here are a few I have been working on.
find tan angle
tan angle in QII, sec angle=-4/sqrt(5)
tan 2 angle in QI, cos 2 angle=3/5
tan 2 angle in QI, csc 2 angle=4/3
tan angle/2 in QII, Sec angle/2=-4
sin angle/2=1/3

here is what I did
sec angle=-4/sqrt(5), sec=1/cos
tan^2 angle+1=(-4/sqrt(5))^2
tan^2 angle +1= 16/5
since sec is neg is QII, tan=-sqrt(11/5)

This one is fine.

From here on either your notation is inaccurate, or you were trying to answer the wrong questions.

Quote:

Originally Posted by kcsteven
cos2 angle=3/5, cos is pos in QI
cos^2 angle=1-sin^2 angle
(3/5)^2=1-sin^2 angle
sin angle= 4/5
tan=sin angle/con angle= 4/5/3/5= 4/3
tan= 2(4/3)= 8/3

You are given $\displaystyle cos(2x) = \frac{3}{5}$

So
$\displaystyle sin^2(2x) = 1 - cos^2(2x) = 1 - \left ( \frac{3}{5} \right )^2 = \frac{16}{25}$

or
$\displaystyle sin(2x) = \frac{4}{5}$ (Positive since x is in the first quadrant.)

Thus
$\displaystyle tan(2x) = \frac{sin(2x)}{cos(2x)} = \frac{ \frac{4}{5} }{ \frac{3}{5} } = \frac{4}{3}$

Quote:

Originally Posted by kcsteven
tan2 angle in QI, csc2 angle= 4/3, csc is pos in QI
csc angle = 1/sin angle
cos^2 angle=1-1/4/3
cos^2= 9/4, cos= 3/4
csc2 angle, cos angle=9/4
tan^2 angle +1=(1/9/4)^2
tan^2 angle +1= 16/81
I can see this is going badly.

$\displaystyle csc(2x) = \frac{4}{3}$

So
$\displaystyle sin(2x) = \frac{1}{csc(2x)} = \frac{3}{4}$

Then, just like the last problem:
$\displaystyle cos(2x) = \frac{ \sqrt{7} }{4}$

So
$\displaystyle tan(2x) = \frac{3}{\sqrt{7}}$

Quote:

Originally Posted by kcsteven
tan angle/2 in QII, sec angle/2=-4, sec in QII is neg
sec= 1/cos angle
tan^2 angle+1=(-8)^2
tan^2 angle=63
tan=sqrt(63)
tan angle/2 in QII= -6sqrt(7)

$\displaystyle sec(x/2) = - 4$

So
$\displaystyle cos(x/2) = \frac{1}{sec(x/2)} = -\frac{1}{4}$

Thus
$\displaystyle sin(x/2) = \frac{ \sqrt{15} }{4}$ (This is positive since x is in the second quadrant.)

So
$\displaystyle tan(x/2) = - \frac{ \sqrt{15} }{1} = -\sqrt{15}$

Quote:

Originally Posted by kcsteven
sin angle/2= 1/3
+-sqrt(1-cos a/2)= +-sqrt(1-1/3/2)= +-sqrt(1/3)

Do we want $\displaystyle tan(x/2)$ again? It can't be done until we know what quadrant the angle is in.

-Dan