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1)
cos^4 = 3/8+1/2cos2x+1/8cos4x

2) Cos^6 theta +sin^6 theta = 1-3/4sin^22theta

Please prove both I will appreciate you

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Sagot :

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[tex]\qquad[/tex] [tex]\purple{\twoheadrightarrow\bf 2cos^2A = 1 + cos2A }[/tex]

[tex]\qquad[/tex] [tex]\purple{\twoheadrightarrow\bf a^3 +b^3 = (a+b)^3 -3ab(a+b)}[/tex]

[tex]\qquad[/tex] [tex]\purple{\twoheadrightarrow\bf 2 sinA cosA = sin2A}[/tex]

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1)

[tex]\qquad[/tex] [tex]\twoheadrightarrow\bf L.H.S[/tex]

[tex]\qquad[/tex] [tex]\pink{\twoheadrightarrow\bf cos^4 x}[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf \dfrac{1}{4} (2cos^2x)^2[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf \dfrac{1}{4}(1+cos2x)^2[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf \dfrac{1}{4}(1+2cos2x+cos^22x)[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf \dfrac{1}{4}+\dfrac{1}{2}cos2x+\dfrac{1}{8}\times 2cos^22x[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf \dfrac{1}{4}+\dfrac{1}{2}cos2x+\dfrac{1}{8}\times(1+cos4x)[/tex]

[tex]\qquad[/tex] [tex]\pink{\twoheadrightarrow\bf \dfrac{1}{4}+\dfrac{1}{2}cos2x+\dfrac{1}{8}\times cos4x} [/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\bf R.H.S[/tex]

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2)

[tex]\qquad[/tex] [tex]\twoheadrightarrow\bf L.H.S[/tex]

[tex]\qquad[/tex] [tex]\pink{\twoheadrightarrow\bf cos^6\theta +sin^6\theta }[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf (cos^2\theta)^3 +(sin^2\theta)^3[/tex]

[tex]\twoheadrightarrow\sf (cos^2\theta +sin^2\theta)^3 -3cos^2\theta sin^2\theta (cos^2\theta +sin^2\theta)[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf 1-3\times \dfrac{1}{4}\times 4(sin\theta cos\theta) ^2[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf 1-\dfrac{3}{4}(2sin\theta cos\theta) ^2[/tex]

[tex]\qquad[/tex] [tex]\pink{\twoheadrightarrow\sf 1-\dfrac{3}{4}sin^22\theta}[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\bf R.H.S[/tex]

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