(sin2θ +cos2θ)² = 4sinθcosθ - 8sin³θcosθ +1 and sin4x = 8cos³xsinx - 4cosxsinx
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7 विचारों·
06 मार्च 2024
A couple of trickier trig identities that have been requested by students.
Hope you find these helpful!
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