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Use the double angle identity for ^2 x.

Mathematics

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Answer

7cos⁡2x−24sin⁡2x+77\cos 2x - 24\sin 2x + 7

Step-by-step solution

4 steps
  1. Use the double angle identity for cos⁡2x\cos^2 x.

    Recall the identity cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1.
    Rearranging this, we get cos⁡2x=1+cos⁡2x2\cos^2 x = \frac{1+\cos 2x}{2}.
    Substitute this into the first term of the expression:

    14cos⁡2x=14(1+cos⁡2x2)14\cos^2 x = 14\left(\frac{1+\cos 2x}{2}\right) =7(1+cos⁡2x)= 7(1+\cos 2x) =7+7cos⁡2x= 7 + 7\cos 2x
  2. Use the double angle identity for sin⁡xcos⁡x\sin x \cos x.

    Recall the identity sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x.
    Substitute this into the second term of the expression:

    −48sin⁡xcos⁡x=−24(2sin⁡xcos⁡x)-48\sin x \cos x = -24(2\sin x \cos x) =−24sin⁡2x= -24\sin 2x
  3. Combine the transformed terms.

    Add the results from Step 1 and Step 2:

    14cos⁡2x−48sin⁡xcos⁡x=(7+7cos⁡2x)−24sin⁡2x14\cos^2 x - 48\sin x \cos x = (7 + 7\cos 2x) - 24\sin 2x

    Rearrange the terms to match the form Acos⁡2x+Bsin⁡2x+CA\cos 2x + B\sin 2x + C:

    =7cos⁡2x−24sin⁡2x+7= 7\cos 2x - 24\sin 2x + 7
  4. Identify the constants AA, BB, and CC.

    Comparing 7cos⁡2x−24sin⁡2x+77\cos 2x - 24\sin 2x + 7 with Acos⁡2x+Bsin⁡2x+CA\cos 2x + B\sin 2x + C:
    A=7A = 7
    B=−24B = -24
    C=7C = 7

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