Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation
1 + sec^2(x) sin^2(x) = sec^2(x)
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OpenStudy (anonymous):
please help!! so lost!!
OpenStudy (jennychan12):
rewrite everything in terms of sine and cosine
OpenStudy (anonymous):
huh?
OpenStudy (anonymous):
im just st
OpenStudy (jennychan12):
rewrite sec^2(x) in terms of cosine
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OpenStudy (anonymous):
starting to learn all this so im still so comfused
OpenStudy (jennychan12):
u do know the six trig functions right?
OpenStudy (anonymous):
kinda.. not memorized
OpenStudy (jennychan12):
-_-
you kinda HAVE to memorize those for this section
OpenStudy (jennychan12):
okay so what does sec^2x = ?
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OpenStudy (anonymous):
1/cos^2x
OpenStudy (jennychan12):
okay now plug that into sec^2(x)
for these types of questions, simplify the "funny looking" or the longer side first
OpenStudy (jennychan12):
so plug that into the left side
OpenStudy (anonymous):
now im comfused.. i have to make the right side of the equation match the left
OpenStudy (jennychan12):
yeah. forget about the right side for now. pretend it's not there for now.
just plug in 1/cos^2(x) into sec^2(x) on the left side
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OpenStudy (jennychan12):
like this
\[1+\frac{ 1 }{\cos^2x }(\sin^2x)\]
OpenStudy (anonymous):
ohhokay
OpenStudy (jennychan12):
now just simplify everything
and use one of the pythagorean identies in your last step
OpenStudy (anonymous):
im not really sure how to do all that
OpenStudy (jennychan12):
simplify
\[1+\frac{ 1 }{\cos^2x }(\sin^2x)\]
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OpenStudy (anonymous):
i feel stupid lol i dont no really how to
OpenStudy (jennychan12):
it's ok.
\[1+\frac{ \sin^2x }{\cos^2x }\]
just multiply the sin^2(x) in
OpenStudy (anonymous):
ohokay
OpenStudy (anonymous):
where do i go from there?
OpenStudy (jennychan12):
sin^2(x)/cos^2(x) = ?
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OpenStudy (anonymous):
hmm
OpenStudy (anonymous):
sin(x)tan(x) ??
OpenStudy (anonymous):
wait.. no its..
OpenStudy (anonymous):
x sec(2) sin^2(x)
OpenStudy (anonymous):
?
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OpenStudy (jennychan12):
just simplify this for now
\[1+\frac{ \sin^2x }{\cos^2x }\]
don't worry about the 1
just simplify
\[\frac{ \sin^2x }{\cos^2x }\]
OpenStudy (anonymous):
im not sure really how to
OpenStudy (jennychan12):
ohh
you do know that sin^2(x)/cos^(2)x = tan^2(x) right?
OpenStudy (anonymous):
oh yes
OpenStudy (anonymous):
?
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OpenStudy (jennychan12):
yeah this section is all about algebraic manipulation.
then you have this right?
\[1+\tan^2x \]
use one of the pythagorean identities.
OpenStudy (anonymous):
1 + tan2u = sec2u
OpenStudy (jennychan12):
yep.
you have the answer!
OpenStudy (anonymous):
wait.. thats the answer?
OpenStudy (jennychan12):
yeah. you proved that the left side is equal to right side.
lol it's called trig proofs for a reason
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