Question

The value of cos^4 θ+sin^4 θ+2cos^2 θ sin^2 θ when θ=45∘ is

A) 1

B) 2

C) 1/√2

D) 2√2

Answer 1

➡HERE IS YOUR ANSWER⬇

Now,

♧♧Solution♧♧

Process 1 :::

${cos}^{4} \alpha + {sin}^{4} \alpha + 2 {sin}^{2} \alpha \: {cos}^{2} \alpha \\ \\ = {( {sin}^{2} \alpha + {cos}^{2} \alpha })^{2} \\ \\ = {1}^{2} \\ \\ = 1$

Process 2 :::

${cos}^{4} \alpha + {sin}^{4} \alpha + 2 {sin}^{2} \alpha \: {cos}^{2} \alpha \\ \\ = {(cos45)}^{4} + {(sin45)}^{4} + 2 {(sin45)}^{2} {(cos45)}^{2} \\ \\ = {( \frac{1}{ \sqrt{2} }) }^{4} + {( \frac{1}{ \sqrt{2} }) }^{4} + 2 {( \frac{1}{ \sqrt{2} }) }^{2} {( \frac{1}{ \sqrt{2} }) }^{2} \\ \\ (since \: \: sin45 = cos45 = \frac{1}{ \sqrt{2} }) \\ \\ = \frac{1}{4} + \frac{1}{4} + \frac{2}{4} \\ \\ = \frac{1 + 1 + 2}{4} \\ \\ = \frac{4}{4} \\ \\ = 1$

♧♧FORMULA USED♧♧

${sin}^{2} \alpha + {cos}^{2} \alpha = 1 \\ \\ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}$

⬆HOPE THIS HELPS YOU⬅

Answer 2

Hey brother!!!



here your solution with this attachment :-






hence Option number "A" is correct!!!





hope it helps:D



thanks