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more trig identities i dont get
- Thread starter dcgirl
- Start date
G'day, Dcgirl.
You can use \(\displaystyle \L sin^2{x} \, + \, \cos^2{x} \, = \, 1\) to rewrite \(\displaystyle \L \cos^4{x}\) (if you begin with the LHS) as you see appropriately. Expand the resulting quadratic and use the identity once more to show LHS = RHS.
Begin with the more complicated-looking side, the LHS.
You should recognise the identity \(\displaystyle \L \tan^2{x} \, + \, 1 \, = \, ?\).
Write \(\displaystyle \L \tan{x}\) (the denominator) and \(\displaystyle \L \sec^2{x}\) (in the numerator) in terms of \(\displaystyle \L \sin{x}\) and \(\displaystyle \L \cos{x}\) to simplify. You should see the RHS popping out from that simplification.
sin^2x+cos^4x=cos^2x+sin^4x
You can use \(\displaystyle \L sin^2{x} \, + \, \cos^2{x} \, = \, 1\) to rewrite \(\displaystyle \L \cos^4{x}\) (if you begin with the LHS) as you see appropriately. Expand the resulting quadratic and use the identity once more to show LHS = RHS.
(sinx+cosx)(tan^2x+1)/tanx=1/cosx+1/sinx
Begin with the more complicated-looking side, the LHS.
You should recognise the identity \(\displaystyle \L \tan^2{x} \, + \, 1 \, = \, ?\).
Write \(\displaystyle \L \tan{x}\) (the denominator) and \(\displaystyle \L \sec^2{x}\) (in the numerator) in terms of \(\displaystyle \L \sin{x}\) and \(\displaystyle \L \cos{x}\) to simplify. You should see the RHS popping out from that simplification.