Considering only the principal values of the inverse trigonometric functions, the value of \frac32 \cos ^-1 \sqrt\frac22+\pi^2+\frac14 \sin ^-1 \frac2 \sqrt2 \pi2+\pi^2+\tan ^-1 \frac\sqrt2\pi is

Inverse Trigonometric Functions PYQ — JEE Advanced 2022 · Shift 1 · Q19 with Solution

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Correct answer: 2.35TO2.37

Let $E=\frac{3}{2}\cos^{-1}\sqrt{\frac{2}{2+\pi^2}}+\frac{1}{4}\sin^{-1}\frac{2\sqrt{2}\pi}{2+\pi^2}+\tan^{-1}\frac{\sqrt{2}}{\pi}.$ We use only principal values.

1. Introduce a convenient angle

Let $\theta=\tan^{-1}\frac{\sqrt{2}}{\pi}.$ Since $\pi>0$ and $\sqrt2>0$ , we have $\theta\in\left(0,\frac\pi2\right)$ .

Then $\tan\theta=\frac{\sqrt2}{\pi}.$ So, from a right triangle,

\qquad \cos\theta=\frac{\pi}{\sqrt{\pi^2+2}}.$$ Hence $$\sqrt{\frac{2}{2+\pi^2}}=\frac{\sqrt2}{\sqrt{\pi^2+2}}=\sin\theta.$$ Therefore, $$\cos^{-1}\sqrt{\frac{2}{2+\pi^2}}=\cos^{-1}(\sin\theta).$$ Because $\theta\in(0,\pi/2)$, we know $$\sin\theta=\cos\left(\frac\pi2-\theta\right),$$ with $\frac\pi2-\theta\in(0,\pi/2)$, which lies in the principal range of $\cos^{-1}$, namely $[0,\pi]$. So, $$\cos^{-1}(\sin\theta)=\frac\pi2-\theta.$$ Thus the first term becomes $$\frac32\cos^{-1}\sqrt{\frac{2}{2+\pi^2}}=\frac32\left(\frac\pi2-\theta\right)=\frac{3\pi}{4}-\frac{3\theta}{2}.$$ ## 2. Simplify the sine-inverse term Now, $$\frac{2\sqrt2\pi}{2+\pi^2}$$ looks like $2\sin\theta\cos\theta$ because $$2\sin\theta\cos\theta=2\cdot \frac{\sqrt2}{\sqrt{\pi^2+2}}\cdot \frac{\pi}{\sqrt{\pi^2+2}} =\frac{2\sqrt2\pi}{\pi^2+2}.$$ Therefore, $$\frac{2\sqrt2\pi}{2+\pi^2}=\sin2\theta.$$ So the second term is $$\frac14\sin^{-1}(\sin2\theta).$$ We must use principal values of $\sin^{-1}$, whose range is $\left[-\frac\pi2,\frac\pi2\right]$. Since $$\theta=\tan^{-1}\frac{\sqrt2}{\pi},$$ we have $0<\theta<\frac\pi4$ because $\frac{\sqrt2}{\pi}<1$. Hence $$0<2\theta<\frac\pi2,$$ so $2\theta$ lies in the principal range of $\sin^{-1}$. Therefore, $$\sin^{-1}(\sin2\theta)=2\theta.$$ Thus the second term becomes $$\frac14\cdot 2\theta=\frac\theta2.$$ ## 3. Simplify the whole expression The third term is simply $$\tan^{-1}\frac{\sqrt2}{\pi}=\theta.$$ So, $$E=\left(\frac{3\pi}{4}-\frac{3\theta}{2}\right)+\frac\theta2+\theta.$$ Combine the $\theta$ terms: $$-\frac{3\theta}{2}+\frac\theta2+\theta = -\theta+\theta=0.$$ Hence, $$E=\frac{3\pi}{4}.$$ ## 4. Numerical value Using $\pi\approx 3.14159265$, $$E=\frac{3\pi}{4}\approx \frac{3\times 3.14159265}{4}=2.35619\ldots$$ So the required value is $$\boxed{\frac{3\pi}{4}}$$ and numerically, $$\boxed{2.356\text{ (approximately)}}.$$ ## 5. Comparison with stored correct answer Stored correct answer: $2.35$ to $2.37$. Our value is $2.35619\ldots$, which lies in this interval. So the derived answer agrees with the stored correct answer.

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1. Introduce a convenient angle

Let

\theta=\tan^{-1}\frac{\sqrt{2}}{\pi}.

Since

\pi>0

and

\sqrt2>0

, we have

\theta\in\left(0,\frac\pi2\right)

Then

\tan\theta=\frac{\sqrt2}{\pi}.

So, from a right triangle,

\qquad \cos\theta=\frac{\pi}{\sqrt{\pi^2+2}}.$$ Hence $$\sqrt{\frac{2}{2+\pi^2}}=\frac{\sqrt2}{\sqrt{\pi^2+2}}=\sin\theta.$$ Therefore, $$\cos^{-1}\sqrt{\frac{2}{2+\pi^2}}=\cos^{-1}(\sin\theta).$$ Because $\theta\in(0,\pi/2)$, we know $$\sin\theta=\cos\left(\frac\pi2-\theta\right),$$ with $\frac\pi2-\theta\in(0,\pi/2)$, which lies in the principal range of $\cos^{-1}$, namely $[0,\pi]$. So, $$\cos^{-1}(\sin\theta)=\frac\pi2-\theta.$$ Thus the first term becomes $$\frac32\cos^{-1}\sqrt{\frac{2}{2+\pi^2}}=\frac32\left(\frac\pi2-\theta\right)=\frac{3\pi}{4}-\frac{3\theta}{2}.$$ ## 2. Simplify the sine-inverse term Now, $$\frac{2\sqrt2\pi}{2+\pi^2}$$ looks like $2\sin\theta\cos\theta$ because $$2\sin\theta\cos\theta=2\cdot \frac{\sqrt2}{\sqrt{\pi^2+2}}\cdot \frac{\pi}{\sqrt{\pi^2+2}} =\frac{2\sqrt2\pi}{\pi^2+2}.$$ Therefore, $$\frac{2\sqrt2\pi}{2+\pi^2}=\sin2\theta.$$ So the second term is $$\frac14\sin^{-1}(\sin2\theta).$$ We must use principal values of $\sin^{-1}$, whose range is $\left[-\frac\pi2,\frac\pi2\right]$. Since $$\theta=\tan^{-1}\frac{\sqrt2}{\pi},$$ we have $0<\theta<\frac\pi4$ because $\frac{\sqrt2}{\pi}<1$. Hence $$0<2\theta<\frac\pi2,$$ so $2\theta$ lies in the principal range of $\sin^{-1}$. Therefore, $$\sin^{-1}(\sin2\theta)=2\theta.$$ Thus the second term becomes $$\frac14\cdot 2\theta=\frac\theta2.$$ ## 3. Simplify the whole expression The third term is simply $$\tan^{-1}\frac{\sqrt2}{\pi}=\theta.$$ So, $$E=\left(\frac{3\pi}{4}-\frac{3\theta}{2}\right)+\frac\theta2+\theta.$$ Combine the $\theta$ terms: $$-\frac{3\theta}{2}+\frac\theta2+\theta = -\theta+\theta=0.$$ Hence, $$E=\frac{3\pi}{4}.$$ ## 4. Numerical value Using $\pi\approx 3.14159265$, $$E=\frac{3\pi}{4}\approx \frac{3\times 3.14159265}{4}=2.35619\ldots$$ So the required value is $$\boxed{\frac{3\pi}{4}}$$ and numerically, $$\boxed{2.356\text{ (approximately)}}.$$ ## 5. Comparison with stored correct answer Stored correct answer: $2.35$ to $2.37$. Our value is $2.35619\ldots$, which lies in this interval. So the derived answer agrees with the stored correct answer.

Inverse Trigonometric Functions question

Inverse Trigonometric Functions question

1. Introduce a convenient angle

More from Inverse Trigonometric Functions

Inverse Trigonometric Functions question

Inverse Trigonometric Functions question

1. Introduce a convenient angle

More from Inverse Trigonometric Functions