How do you write #r=2sec(theta+pi/4)# in rectangular form?

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Answer 1 · 2017-01-25T18:54:44

Multiply both sides by the #cos(theta+pi/4)#
Use the identity #cos(a+b)=cos(a)cos(b)-sin(a)sin(b)#
Substitute x for #rcos(theta)# y for #rsin(theta)#

Given: #r = 2sec(theta+pi/4)#

Multiply both sides by #cos(theta+pi/4)#:

#rcos(theta+pi/4) = 2#

Use the identity #cos(a+b)=cos(a)cos(b)-sin(a)sin(b)# to substitute #cos(theta)cos(pi/4) - sin(theta)sin(pi/4)# for #cos(theta+pi/4)#:

#r(cos(pi/4)cos(theta) - sin(pi/4)sin(theta)) = 2#

The sine and cosine are both #sqrt(2)/2# at #pi/4#

#r(sqrt(2)/2cos(theta) - sqrt(2)/2sin(theta)) = 2#

Use the distributive property:

#sqrt(2)/2rcos(theta) - sqrt(2)/2rsin(theta) = 2#

Multiply both sides by #sqrt(2)#

#rcos(theta) - rsin(theta) = 2sqrt2#

Substitute x for #rcos(theta)# and y for #rsin(theta)#

#x - y = 2sqrt2#