How do I show that cos4x=18cos⁡(4x)+12cos⁡(2x)+38 I know how to prove that cos2x=12+12cos⁡(2x) by substituting cos⁡2x with 2cos2x−1 according to the double angle identity cos⁡(2x)=2cos2x−1 However, how do I do that for cos4x? Do I do it by writing cos4x as cos2(x)⋅cos2(x) and thus get it by squaring the LHS of cos2x=12+12cos⁡(2x) Im not sure how to proceed.

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How do I show that cos4x=18cos⁡(4x)+12cos⁡(2x)+38  I know how to prove that  cos2x=12+12cos⁡(2x)  by substituting cos⁡2x with 2cos2x−1 according to the double angle identity  cos⁡(2x)=2cos2x−1  However, how do I do that for cos4x?  Do I do it by writing cos4x as  cos2(x)⋅cos2(x)  and thus get it by squaring the LHS of  cos2x=12+12cos⁡(2x)  Im not sure how to proceed.

plusmarcacw · Accepted Answer

Hint: By squaring we get  cos4(x)=14(cos2(2x)+2cos⁡(2x)+1)  and then  cos⁡(4x)=2cos2(2x)−1

Mazzuranavf · Accepted Answer

Since   cos2x=12+12cos⁡(2x)  we have   cos2(2x)=12+12cos⁡(4x)   and thus, if we square first equation we get   cos4x=14+12cos⁡(2x)+14cos2(2x)=   =14+12cos⁡(2x)+14(12+12cos⁡(4x))   =38+12cos⁡(2x)+18cos⁡(4x)

How do I show that cos4x=18cos⁡(4x)+12cos⁡(2x)+38 I know how to prove that cos2x=12+12cos⁡(2x) by substituting...

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