Question

hii please answer. ..trigno

Answer 1

Answer:

Sin 2A = 2 SinA.CosA

Question: To Prove Cos²2A + 4Sin²A.Cos²A = 1

LHS:

⇒ Cos²2A + 4 Sin²A.Cos²A

4 Sin²A.Cos²A can also be written as : ( 2 SinA.CosA )²

⇒ Cos²2A + ( 2 SinA.CosA )²

According to identity we know that Sin2A = 2 SinA.CosA

⇒ Cos²2A + ( Sin 2A )²

⇒ Cos²2A + Sin²2A

Let us Take 2A to be some Ф

⇒ Cos²Ф + Sin²Ф

We know the identity that, Cos²A + Sin²A = 1

Hence on applying this we get that,

⇒ Cos²Ф + Sin²Ф = 1

⇒ 1 = 1

⇒ LHS = RHS

Hence Proved !!

Answer 2

Solution :→   →  ↓

↓   ←     ←     ←   ↓

Given:

$cos^22\theta + 4 cos^2\theta sin^2\theta$

[REMEMBER : $cos2\theta = 2cos^2\theta-1$

$\implies cos^22\theta = ( 2 cos^2\theta-1)^2$]

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$\implies cos^22\theta + 4 cos^2\theta sin^2\theta = (2 cos^2\theta - 1)^2+4cos^2\theta sin^2\theta$

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$\implies (2 cos^2\theta-sin^2\theta-cos^2\theta)^2+4cos^2\theta sin^2\theta$

[ We know that :

$sin^2\theta+cos^2\theta=1$

$\implies -1=-sin^2\theta-cos^2\theta$ ]

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$\implies (2 cos^2\theta-sin^2\theta-cos^2\theta)^2+4cos^2\theta sin^2\theta$

$\implies (cos^2\theta-sin^2\theta)^2+4sin^2\theta cos^2\theta$

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$\implies cos^4\theta+sin^4\theta - 2cos^2 \theta sin^2\theta+cos^2\theta sin^2\theta$

[ Using the formula:

$(a-b)^2=a^2+b^2-2ab$ ]

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$\implies cos^4\theta+sin^4\theta + 2cos^2\theta sin^2\theta$

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$\implies (cos^2\theta + sin^2\theta)^2$

[ Using the formula:

$(a+b)^2=a^2+b^2+2ab$ ]

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$\implies 1$

[ $sin^2\theta+cos^2\theta=1$ ]

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Hence LHS = RHS [ Proved ]

Hope it helps you:-)

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