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(sin2θ +cos2θ)² = 4sinθcosθ - 8sin³θcosθ +1 and sin4x = 8cos³xsinx - 4cosxsinx

5 Views· 06 March 2024
Alagai Augusten
Alagai Augusten
6 Subscribers
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A couple of trickier trig identities that have been requested by students.
Hope you find these helpful!

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