Question

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

Answer 1

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)

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

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

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