How do you prove the identity:

\[ \cos^2x - \sin^2x = 1 - 2\sin^2x \]

s in ( 2 α ) = 2 ⋅ s in α ⋅ cos α cos ( 2 α ) = 1 − 2 s i n 2 α − − − − − − − − − − − − − − − − − − − − cos 2 x − s in 2 x = 1 − 2 s in 2 x cos 2 x + s in 2 x = 1 1 − 2 s i n 2 x + 2 s in x ⋅ cos x = 1 − 2 s in x ( s in x − cos x ) = 0 − 2 s in x = 0 or s in x − cos x = 0 1 ) − 2 s in x = 0 ⇔ s in x = 0 ⇔ x 1 ​ = kπ an d k ∈ I 2 ) s in x − cos x = 0 ⇔ s in x = cos x / : cos x ∧ cos x  = 0 t an x = 1 ⇔ x 2 ​ = 4 π ​ + kπ an d k ∈ I

Answered by kate200468 | 2024-06-10

We proved the identity cos 2 x − sin 2 x = 1 − 2 sin 2 x by using the Pythagorean identity. We substituted cos 2 x with 1 − sin 2 x and simplified it to match the right-hand side of the equation. This shows the relationship between the squares of sine and cosine.
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Answered by kate200468 | 2024-11-29