trig identities: eliminating θ

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June 26th 2010, 11:29 PM
shimmerblade

trig identities: eliminating θ

Hello. Thanks for any help in advance.

just having trouble with the method of eliminating θ from the following pairs of equations:
a. x = 2 cosθ, y = 2 sin θ

b. x = b tan, y = a sec θ
June 27th 2010, 12:22 AM
Prove It

Quote:

Originally Posted by shimmerblade

Hello. Thanks for any help in advance.

just having trouble with the method of eliminating θ from the following pairs of equations:
a. x = 2 cosθ, y = 2 sin θ

b. x = b tan, y = a sec θ

The key is to use the many forms of the Pythagorean Identity - in this case $\sin^2{\theta} + \cos^2{\theta} = 1$ and $\tan^2{\theta} + 1 = \sec^2{\theta}$ .

(a) $x = 2\cos{\theta}, y = 2\sin{\theta}$

$x^2 = 4\cos^2{\theta}, y^2 = 4\sin^2{\theta}$

$x^2 + y^2 = 4\cos^2{\theta} + 4\sin^2{\theta}$

$x^2 + y^2 = 4(\cos^2{\theta} + \sin^2{\theta})$

$x^2 + y^2 = 4\cdot 1$

$x^2 + y^2 = 4$ .

Can you follow a similar process for (b)?
June 27th 2010, 04:20 PM
shimmerblade

for b. x = b tan θ, y = a sec θ would it be something like

(b tan θ)^2 + 1 = (a sec θ)^2

b^2 tan^2 θ + b^2 a^2 = a^2 sec^2 θ

b^2 (tan^2 θ + a^2)^2 = a^2 (sec^2 θ)

b^2 x^2 + a^2 b^2 = a^2 y^2

not too sure....=\
June 27th 2010, 04:22 PM
Prove It

Quote:

Originally Posted by shimmerblade

for b. x = b tan θ, y = a sec θ would it be something like

(b tan θ)^2 + 1 = (a sec θ)^2

b^2 tan^2 θ + b^2 a^2 = a^2 sec^2 θ

b^2 (tan^2 θ + a^2)^2 = a^2 (sec^2 θ)

b^2 x^2 + a^2 b^2 = a^2 y^2

not too sure....=\

You're on the right track. Remember that $\tan^2{\theta} + 1 = \sec^2{\theta}$ , so $\sec^2{\theta} - \tan^2{\theta} = 1$ .

If $x = b\tan{\theta}, y = a\sec{\theta}$

then $x^2 = b^2\tan^2{\theta}, y^2 = a^2\sec^2{\theta}$

$a^2x^2 = a^2b^2\tan^2{\theta}, b^2y^2 = a^2b^2\sec^2{\theta}$ .

$a^2b^2\sec^2{\theta} - a^2b^2\tan^2{\theta} = b^2y^2 - a^2x^2$

$a^2b^2(\sec^2{\theta} - \tan^2{\theta}) = b^2y^2 - a^2x^2$

$a^2b^2\cdot 1 = b^2y^2 - a^2x^2$

$a^2b^2 = b^2y^2 - a^2x^2$

$1 = \frac{b^2y^2 - a^2x^2}{a^2b^2}$

$\frac{b^2y^2}{a^2b^2} - \frac{a^2x^2}{a^2b^2} = 1$

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ .
June 28th 2010, 03:44 PM
shimmerblade

ok thanks for your help. (Happy)