Prove the following trig identities: \(\displaystyle{2}{{\sin}^{{2}}\theta}-{1}={{\sin}^{{2}}\theta}-{{\cos}^{{2}}\theta}\) \(\displaystyle{{\cos}^{{2}}{t}}={{\sin}^{{2}}{t}}+{2}{{\cos}^{{2}}{t}}-{1}\)

London Douglas

London Douglas

Answered question

2022-03-30

Prove the following trig identities:
2sin2θ1=sin2θcos2θ
cos2t=sin2t+2cos2t1

Answer & Explanation

ostijum8dd

ostijum8dd

Beginner2022-03-31Added 7 answers

2sin2θ1=sin2θcos2θ
Let's make the right hand side equal to the left hand side (LHS-RHS proof).
sin2θcos2θ
=1cos2θcos2θ
=12cos2θ
=12(1sin2θ)
=12+2sin2θ
=1+2sin2θ
=2sin2θ1
2sin2θ1=sin2θcos2θ
Theodore Davila

Theodore Davila

Beginner2022-04-01Added 14 answers

cos2t=sin2t+2cos2t1
This time, let's ASSUME the identity is true.
cos2t=sin2t+2cost1
cos2t=sin2t1
cos2tsin2t=1
cos2t+sin2t=1
cos2t=sin2t+2cos2t1

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