x3_drummerchick
  • x3_drummerchick
Verify the trig function: can someone explain to me how my professor got from step 1 to 2? will give medals!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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x3_drummerchick
  • x3_drummerchick
i attached a photo of the problem
1 Attachment
x3_drummerchick
  • x3_drummerchick
@freckles @Nnesha
Nnesha
  • Nnesha
so he is working on the right side

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x3_drummerchick
  • x3_drummerchick
yes
Nnesha
  • Nnesha
i'm sure you're familiar with the special identity \[\huge\rm \sin^2x+\cos^2x=1\] solve this identity for sin^2x
x3_drummerchick
  • x3_drummerchick
1-cos^2x=sin^2x
Nnesha
  • Nnesha
right so now we can replace sin^2x with 1-cos^2 \[\huge\rm \tan^2x \color{reD}{sin^2x} \] \[\huge\rm \tan^2x \color{reD}{(1-cos^2x)} \]
x3_drummerchick
  • x3_drummerchick
right
x3_drummerchick
  • x3_drummerchick
but how did they go from that ^ to tan^2-(sin^2/cos^2)?
Nnesha
  • Nnesha
that's what i'm trying to understand .... ,-,
Nnesha
  • Nnesha
can you take screenshot of next step
x3_drummerchick
  • x3_drummerchick
he said something about rewriting it as tan, but it does make sense
x3_drummerchick
  • x3_drummerchick
doesnt *
Nnesha
  • Nnesha
it doesn't make sense i guess he is trying to rewrite tan^2 in terms of sin and cos
x3_drummerchick
  • x3_drummerchick
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Nnesha
  • Nnesha
\[\huge\rm \tan^2x \color{reD}{(1-cos^2x)} \]\[\large\rm \frac{ \sin^2x }{ \cos^2x }(1-\cos^2x)\]
x3_drummerchick
  • x3_drummerchick
im still confused :(
Nnesha
  • Nnesha
hm let me think
Nnesha
  • Nnesha
ahh i see so he distributed by sin^2/cos^2 \[\rm \frac{ \sin^2 x}{ \cos^2x }-\frac{\sin^2}{\cos^2x}*\cos^2x\] and then he changed sin^2/cos^2 to tan^2
x3_drummerchick
  • x3_drummerchick
im sorry, could you rewrite it out from the start, im still confused
Nnesha
  • Nnesha
\[\huge\rm \tan^2x \color{reD}{(1-cos^2x)} \]\[\large\rm \color{blue}{ \frac{ \sin^2x }{ \cos^2x} }(1-\cos^2x)\] rewrite tan as sin/cos \[\rm \color{blue}{ \frac{ \sin^2 x}{ \cos^2x }}-\frac{\sin^2}{\cos^2x}*\cos^2x\] distribute parentheses by sin^2/cos^2 \[\rm \color{blue}{\tan^2x} -\frac{sin^2x}{cos^2x}*cos^2x\] rewrite sin^2/cos^2 as tan^2
x3_drummerchick
  • x3_drummerchick
im still lost on this, im sorry
1 Attachment
x3_drummerchick
  • x3_drummerchick
ohhh wait i think i get ittt
x3_drummerchick
  • x3_drummerchick
1 Attachment
x3_drummerchick
  • x3_drummerchick
he distributed tan from the get go?
Nnesha
  • Nnesha
okay so i applied distributive property \[\large\rm \color{ReD}{a}(b+c)=\color{ReD}{a}*b+\color{ReD}{a}*c=\color{red}{a}b+\color{ReD}{a}c\] multiply both terms in the parentheses by outside term that's how |dw:1446496572293:dw| \[\rm \color{blue}{ \frac{ \sin^2 x}{ \cos^2x }}-\frac{\sin^2}{\cos^2x}*\cos^2x\]
x3_drummerchick
  • x3_drummerchick
ohh okay, i dont know why that took so long to process xD thank you !!
Nnesha
  • Nnesha
lol np i was confused too actually he skipped one step :>
Nnesha
  • Nnesha
this one :D \[\rm \color{blue}{ \frac{ \sin^2 x}{ \cos^2x }}-\frac{\sin^2}{\cos^2x}*\cos^2x\]
x3_drummerchick
  • x3_drummerchick
yeah he did, i hate when they do that. we just started these today :/ hopefully it doesnt get too bad later lol
Nnesha
  • Nnesha
ye trig is fun my favorite topic!!!

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