Math, asked by namratapanda11, 5 days ago

pls answer pls no unnecessary answers ( this is a model paper class 12 ) any maths genius pls​

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Answered by Itzheartcracer
4

12 (a)

Given :-

cos⁻¹(1/2) + 2 sin⁻¹(1/2)

Solution :-

\sf cos^{-1}\bigg(\dfrac{1}{2}\bigg)+2sin^{-1}\bigg(\dfrac{1}{2}\bigg)

First of all, let's take the cos value

\sf \cos^{-1}\bigg(\dfrac{1}{2}\bigg)

Assumption!!

Assume cos⁻¹ = x (To make easy)

\sf \cos\;x\bigg(\dfrac{1}{2}\bigg)

We know

  • π = 180

Rewriting as

\sf \cos\;x=\cos\bigg(\dfrac{\pi}{3}\bigg)

\sf \cos\;x=\cos\bigg(\dfrac{180}{3}\bigg)

\sf \cos\;x=\cos(60)^{\circ}

  • cos 60° = 1/2

\sf \cos\;x=\cos\bigg(\dfrac{1}{2}\bigg)

Hence

We observed cos(π/3) = cos(1/2)

\sf \cos\;x=\cos\bigg(\dfrac{\pi}{3}\bigg)

Canceling cos from LHS and RHS

\sf x=\dfrac{\pi}{3}

Range cos⁻¹ [0,π] & principal value of  cos⁻¹ = π/3

Now, taking sin value

\sf sin^{-1}\bigg(\dfrac{1}{2}\bigg)

Assumption!!

Assume sin⁻¹ = x (To make easy)

\sf \sin\;x=\dfrac{1}{2}

We know

  • π = 180

\sf \sin\;x=\sin\bigg(\dfrac{\pi}{6}\bigg)

\sf \sin\;x=\sin\bigg(\dfrac{180}{6}\bigg)

\sf \sin\;x=\sin(30)^{\circ}

  • sin 30° = 1/2

\sf \sin\;x=\sin\bigg(\dfrac{1}{2}\bigg)

We observed sin⁻¹(1/2) = sin(π/6)

\sf \sin\;x=\sin\bigg(\dfrac{\pi}{6}\bigg)

Canceling sin form LHS and RHS

\sf x=\dfrac{\pi}{6}

So,

\sf 2\;\sin^{-1}\bigg(\dfrac{1}{2}\bigg)=2(x)

\sf 2\;\sin^{-1}\bigg(\dfrac{1}{2}\bigg)=2\bigg(\dfrac{\pi}{6}\bigg)

\sf 2\;\sin^{-1}\bigg(\dfrac{1}{2}\bigg)=\dfrac{\pi}{3}

Finding value of cos⁻¹(1/2) + 2 sin⁻¹(1/2)

\sf \dfrac{\pi}{3}+\dfrac{\pi}{3}

\sf\dfrac{\pi+\pi}{3}

\sf\dfrac{2\pi}{3}

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