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Exercise 3.3 · Q57

Q.Prove the following: cos⁡2x+cos⁡2(x+120°)+cos⁡2(x−120°)=32\cos^2x+\cos^2(x+120°)+\cos^2(x-120°)=\dfrac{3}{2}

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Step 1: cos⁡2x+cos⁡2(x+120°)+cos⁡2(x−120°)=32+12[cos⁡2x+cos⁡(2x+240°)+cos⁡(2x−240°)]\cos^2x+\cos^2(x+120°)+\cos^2(x-120°)=\dfrac{3}{2}+\dfrac{1}{2}\big[\cos2x+\cos(2x+240°)+\cos(2x-240°)\big].

Step 2: cos⁡(2x+240°)+cos⁡(2x−240°)=2cos⁡2xcos⁡240°=2cos⁡2x(−12)=−cos⁡2x\cos(2x+240°)+\cos(2x-240°)=2\cos2x\cos240°=2\cos2x\left(-\dfrac{1}{2}\right)=-\cos2x. …

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