If - - 30- prove that cos 2- - cos2 - - sin2 -

L. H. S. = cos2θ = cos 2∙ nbsp;30^∘ = cos nbsp;60^∘ = 12 And R. H. S. = cos2θ sin2θ = cos230^∘ nbsp; sin230^∘ = ( √(3)2 )^2 ( 12 )^2 =34 14 =12 Ther ...

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Standard IX

Mathematics

Trigonometric Ratios of Complementary Angles

If θ = 30∘, p...

Question

If θ = $30^{\circ}$ , prove that cos 2θ = cos² θ - sin² θ

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Solution

L. H. S. = cos2θ
= cos 2∙ $30^{\circ}$
= cos $60^{\circ}$
= $\frac{1}{2}$
And R. H. S. = cos²θ - sin²θ
= cos² $30^{\circ}$ - sin² $30^{\circ}$
$= {(\frac{\sqrt{3}}{2})}^{2} - {(\frac{1}{2})}^{2} = \frac{3}{4} - \frac{1}{4} = \frac{1}{2}$
Therefore, L.H.S = R.H.S

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If θ = 30∘, prove that cos 2θ = cos2 θ - sin2 θ

L. H. S. = cos2θ = cos 2∙ 30∘ = cos 60∘ = 12 And R. H. S. = cos2θ - sin2θ = cos230∘ - sin230∘ =(√32)2−(12)2=34−14=12 Therefore, L.H.S = R.H.S

If θ = $30^{\circ}$ , prove that cos 2θ = cos² θ - sin² θ

L. H. S. = cos2θ
= cos 2∙ $30^{\circ}$
= cos $60^{\circ}$
= $\frac{1}{2}$
And R. H. S. = cos²θ - sin²θ
= cos² $30^{\circ}$ - sin² $30^{\circ}$
$= {(\frac{\sqrt{3}}{2})}^{2} - {(\frac{1}{2})}^{2} = \frac{3}{4} - \frac{1}{4} = \frac{1}{2}$
Therefore, L.H.S = R.H.S