Prove that cos2-4-sin2-4- is - independent-

Now, cos^2(π4+θ) sin^2(π4 θ) #160; =cos{π4+θ+π4 θ}.cos{π4+θ π4 θ} [ Since cos^2A sin^2B=cos(A+B).cos(A B) ] =cosπ2.cos 2θ =0 . [ Since ...

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Trigonometric Ratios of Compound Angles

Prove that ...

Question

Prove that ${cos}^{2} (\frac{π}{4} + θ) - {sin}^{2} (\frac{π}{4} - θ)$ is $θ$ independent.

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The value of $c o s^{2} (\frac{π}{6} + θ) - s i n^{2} (\frac{π}{6} - θ)$ is

c o s^{2} (\frac{π}{2} + θ) - s i n^{2} (\frac{π}{2} - θ) =

Prove that:

$c o s^{2} (\frac{π}{4} - θ) - s i n^{2} (\frac{π}{4} - θ) = s i n 2 θ$

{cos}^{2} (\frac{π}{6} + θ) - {sin}^{2} (\frac{π}{6} - θ)

θ

c o s θ = \frac{8}{17},

c o s (\frac{π}{6} + θ) + c o s (\frac{π}{4} - θ) + c o s (\frac{2 π}{3} - θ)

= (\frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}}) \frac{23}{17}

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