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Mathematics 8 Online
OpenStudy (anonymous):

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)

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

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 = ?

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

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 \[1+\frac{ 1 }{\cos^2x }(\sin^2x)\]

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) = ?

OpenStudy (anonymous):

hmm

OpenStudy (anonymous):

sin(x)tan(x) ??

OpenStudy (anonymous):

wait.. no its..

OpenStudy (anonymous):

x sec(2) sin^2(x)

OpenStudy (anonymous):

?

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):

?

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

OpenStudy (anonymous):

so all i put is 1 + tan2u = sec2u for the answer?

OpenStudy (jennychan12):

just write out the work

OpenStudy (anonymous):

ok thank you

OpenStudy (jennychan12):

no problem!

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