If alpha - beta-90-0 and alpha -2beta- then cos-2 alpha - sin-2 beta is10dfrac-1-2-2

Question

If  alpha - beta-90-0 and alpha -2beta- then  cos-2 alpha - sin-2 beta is10dfrac-1-2-2

Toppr Admin · Accepted Answer

-x3B1-x3B2-90-x2218-x21D2-2-x3B2-x3B2-90-x2218-x21D2-x3B2-30-x2218-x2234-x3B1-90-x2218-x2212-30-x2218-60-x2218-x2234-cos2-x3B1-sin2-x3B2-cos260-x2218-sin230-x2218-12-2-12-2-12

If $α + β = 90^{0}$ and $α = 2 β,$ then $c o s^{2} α + s i n^{2} β$ is
$1$
$0$
$\frac{1}{2}$
$2$

$α + β = 90^{\circ} \Rightarrow 2 β + β = 90^{\circ} \Rightarrow β = 30^{\circ} ∴ α = 90^{\circ} - 30^{\circ} = 60^{\circ} ∴ {cos}^{2} α + {sin}^{2} β = {cos}^{2} 60^{\circ} + {sin}^{2} 30^{\circ} = {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{2} = \frac{1}{2}$

If α+β=900 and α=2β, then cos2α+sin2β is10122

α+β=90∘⇒2β+β=90∘⇒β=30∘∴α=90∘−30∘=60∘∴cos2α+sin2β=cos260∘+sin230∘=(12)2+(12)2=12

If $α + β = 90^{0}$ and $α = 2 β,$ then $c o s^{2} α + s i n^{2} β$ is
$1$
$0$
$\frac{1}{2}$
$2$

$α + β = 90^{\circ} \Rightarrow 2 β + β = 90^{\circ} \Rightarrow β = 30^{\circ} ∴ α = 90^{\circ} - 30^{\circ} = 60^{\circ} ∴ {cos}^{2} α + {sin}^{2} β = {cos}^{2} 60^{\circ} + {sin}^{2} 30^{\circ} = {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{2} = \frac{1}{2}$