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Vinit

The document discusses Pythagorean triplets for angles that are multiples or fractions of a given angle α, specifically 2α, 3α, and α/2. It provides calculations for tangent and cosine values based on given sine and cosine values, demonstrating how to derive new triplets for these angles. Examples illustrate the process of calculating the sides of new triangles and the corresponding trigonometric ratios.

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12 views9 pages

Vinit

The document discusses Pythagorean triplets for angles that are multiples or fractions of a given angle α, specifically 2α, 3α, and α/2. It provides calculations for tangent and cosine values based on given sine and cosine values, demonstrating how to derive new triplets for these angles. Examples illustrate the process of calculating the sides of new triangles and the corresponding trigonometric ratios.

Uploaded by

apridge.182132.iwcm
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PPTX, PDF, TXT or read online on Scribd
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Now we will discuss about the Pythagorean

triplets when the angle(α) is twice (i.e.2α),


or when it is 3α or is α/2.

Now, First of
For
alltwice the angle(2α):
Angle Triplet
α (p,b,h)
2α (2pb,b²-p²,h²)
g. (i) If Sinα = 3/5
Then, Tan2α = ?

lution: We have,
Sinα = 3/5
, p=3 and h=5
the Pythagorean triplet is (3,4,5)
w, the triplet for 2α:
(2×3×4,4²-3²,5²) = (24,7,25)
, p’=24, b’=7 and h’=25 (sides of new triangle)
en,
Tan2α = p’/b’ = 24/7
If Cosα = 40/41
Then, Cos2α = ?

ution: We have,
Cosα = 40/41
b=40 and h=41
the Pythagorean triplet is (9,40,41)
w, the triplet for 2α:
(2×9×40,40²-9²,41²) = (720,1519,1681)
p’=720, b’=1519 and h’=1681 (sides of new trian
en,
Cos2α = b’/h’ = 1519/1681
Now,
For thrice the angle(3α):

Angle Triplet
α (p,b,h)
3α (3ph²-4p³,4b³-3bh²,h³)
) If Tanα = 7/24
Then, Cos3α = ?

on: We have,
Tanα = 7/24
p=7 and b=24
Pythagorean triplet is (7,24,25)
he triplet for 3α:
3×7×(25)²-4×7³,4×(24)³-3×24×(25)²,(25)³) = (11753,1029
=11753, b’=10296 and h’=15625 (sides of new tria

Cos3α = b’/h’ = 10296/15625


f Sinα = 3/5
Then, Tan3α = ?

ion: We have,
Sinα = 3/5
p=3 and h=5
he Pythagorean triplet is (3,4,5)
the triplet for 3α:
(3×3×5²-4×3³,4×4³-3×4×5²,5³) = (117,-44,125)
’=117, b’=-44 and h’=125 (sides of new triangle)
,
Tan3α = p’/b’ = -117/44
Now,
For half the angle(α/2):

Angle Triplet
α (p,b,h)
α/2 (p,b+h,√(b+h)²+p²)
g. (i) If Sinα = 12/13
Then, Tanα/2 = ?

lution: We have,
Sinα = 12/13
, p=12 and h=13
the Pythagorean triplet is (12,5,13)
w, the triplet for α/2:
(12,5+13,√(5+13)²+12²) = (12,18,6√13)
, p’=12, b’=18 and h’=6√13 (sides of new triangle)
en,
Tanα/2 = p’/b’ = 12/18 = 2/3
) If Cosα = 4/5
Then, Cosα/2 = ?

lution: We have,
Cosα = 4/5
, b=4 and h=5
the Pythagorean triplet is (3,4,5)
ow, the triplet for α/2:
(3,4+5,√(4+5)²+3²) = (3,9,3√10)
, p’=3, b’=9 and h’=3√10 (sides of new triangle)
en,
Cosα/2 = b’/h’ = 9/3√10 = 3/√10

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