Alexandra 25 Dec 2015, 21:03
Show (sinx+cosx)^2tanx=tanx+2sin^2x
`RHS = tanx+2sin^2x`
`=tan x + 2(1-cos^2x)`
`=tanx + 2 - 2cos^2 x`
But I can't make it look like the LHS.
X
Show (sinx+cosx)^2tanx=tanx+2sin^2x
Relevant page <a href="/analytic-trigonometry/1-trigonometric-identities.php">1. Trigonometric Identities</a> What I've done so far `RHS = tanx+2sin^2x` `=tan x + 2(1-cos^2x)` `=tanx + 2 - 2cos^2 x` But I can't make it look like the LHS.
Murray 27 Dec 2015, 00:13
It's correct so far, but I don't think it helped.
Expand out the bracket on the LHS and see if you recognize anything.
X
It's correct so far, but I don't think it helped. Expand out the bracket on the LHS and see if you recognize anything.
Alexandra 27 Dec 2015, 21:48
OK.
` (sinx+cosx)^2tanx ` `= (sin^2x + 2sinx cosx + cos^2 x)tanx`
`=(1 + 2sinxcosx)tanx`
Is that right? But what do I do now?
X
OK. ` (sinx+cosx)^2tanx ` `= (sin^2x + 2sinx cosx + cos^2 x)tanx` `=(1 + 2sinxcosx)tanx` Is that right? But what do I do now?
Murray 28 Dec 2015, 14:53
Just remember `tanx = (sinx)/(cosx)`
X
Just remember `tanx = (sinx)/(cosx)`
Alexandra 29 Dec 2015, 05:44
`(1+2sinxcosx)tanx` `=(1+2sinxcosx)(sinx)/(cosx)`
`= (sinx)/(cosx) + 2sin^2x`
`=tanx + 2sin^2 x`
`=RHS`
Thanks a lot.
X
`(1+2sinxcosx)tanx` `=(1+2sinxcosx)(sinx)/(cosx)` `= (sinx)/(cosx) + 2sin^2x` `=tanx + 2sin^2 x` `=RHS` Thanks a lot.
X
You're welcome!
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