write cos2x as sinx please help with this ​

Question

write cos2x as sinx
please help with this ​

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Thiên Hương 4 years 2021-09-05T04:16:02+00:00 1 Answers 6 views 0

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  1. mocmien2
    0
    2021-09-05T04:17:37+00:00
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    Answer:

    \cos(2\, x) = 1 - 2\, (\sin(x))^2.

    Step-by-step explanation:

    Angle sum identity for cosine: \cos(a + b) = \cos(a) \, \cos(b) - \sin(a) \, \sin(b).

    Pythagorean identity: (\cos(a))^{2} + (\sin(a))^{2} = 1 for all real a.

    Subtract (\cos(x))^{2} from both sides of the Pythagorean identity to obtain: (\sin(a))^{2} = 1 - (\cos(a))^{2}.

    Apply angle sum identity to rewrite \cos(2\, x).

    \begin{aligned}&\cos(2\, x)\\ &= \cos(x + x) \\ &= \cos(x) \, \cos(x) - \sin(x)\, \sin(x) \\ &= (\cos(x))^{2} + (\sin(x))^{2}\end{aligned}.

    (\sin(a))^{2} = 1 - (\cos(a))^{2} follows from the Pythagorean identity. Hence, it would be possible to replace the (\cos(x))^{2} in the previous expression with (1 - (\sin(x))^{2}).

    \begin{aligned}&(\cos(x))^{2} - (\sin(x))^{2}\\ &= \left[1 - (\sin(x))^{2}\right] - (\sin(x))^{2} \\ &= 1 - 2\, (\sin(x))^{2} \end{aligned}.

    Conclusion:

    \begin{aligned}&\cos(2\, x) \\ &= (\cos(x))^{2} + (\sin(x))^{2} \\ &=1 - 2\, (\sin(x))^{2}\end{aligned}

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