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Inverse Trigonometric Functions question

2019 · 10 Apr · Shift 2 · Q31
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Inverse Trigonometric Functions question

2019 · 10 Apr · Shift 2 · Q31

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
If cos⁡−1x−cos⁡−1y2=α{\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alphacos−1x−cos−12y​=α,where –1 ≤\le≤ x ≤\le≤ 1, – 2 ≤\le≤ y ≤\le≤ 2, x ≤y2\le {y \over 2}≤2y​, then for all x, y, 4x2 – 4xy cos α\alphaα + y2 is equal to :
  1. A
    4 sin2 α\alphaα
  2. B
    2 sin2 α\alphaα
  3. C
    4 sin2 α\alphaα - 2x2y2
  4. D
    4 cos2 α\alphaα + 2x2y2
View written solutionFree

Correct answer: A

  1. Set convenient angles

Let A=cos⁡−1x,B=cos⁡−1(y2).A=\cos^{-1}x,\qquad B=\cos^{-1}\left(\frac y2\right).A=cos−1x,B=cos−1(2y​). Then the given condition is A−B=α.A-B=\alpha.A−B=α. Also, x=cos⁡A,y2=cos⁡B⇒y=2cos⁡B.x=\cos A,\qquad \frac y2=\cos B \Rightarrow y=2\cos B.x=cosA,2y​=cosB⇒y=2cosB.

Since x≤y2x\le \frac y2x≤2y​, we have cos⁡A≤cos⁡B.\cos A\le \cos B.cosA≤cosB. Because cos⁡−1\cos^{-1}cos−1 takes values in [0,π][0,\pi][0,π] where cosine is decreasing, this implies A≥B,A\ge B,A≥B, so α=A−B≥0.\alpha=A-B\ge 0.α=A−B≥0. This is consistent.

  1. Rewrite the required expression

We need to find 4x2−4xycos⁡α+y2.4x^2-4xy\cos\alpha+y^2.4x2−4xycosα+y2. Substitute x=cos⁡Ax=\cos Ax=cosA and y=2cos⁡By=2\cos By=2cosB:

=4\cos^2A-4(\cos A)(2\cos B)\cos\alpha+(2\cos B)^2.$$ So, $$4x^2-4xy\cos\alpha+y^2 =4\cos^2A-8\cos A\cos B\cos\alpha+4\cos^2B.$$ Factor out $4$: $$=4\left(\cos^2A-2\cos A\cos B\cos\alpha+\cos^2B\right).$$ Now use $$\alpha=A-B \Rightarrow \cos\alpha=\cos(A-B)=\cos A\cos B+\sin A\sin B.$$ Hence, \begin{align*} &\cos^2A-2\cos A\cos B\cos\alpha+\cos^2B \\ &=\cos^2A-2\cos A\cos B(\cos A\cos B+\sin A\sin B)+\cos^2B. \end{align*} This is a bit messy, so a better route is to use the identity below. --- 3. **Use a standard identity** Observe that $$4x^2-4xy\cos\alpha+y^2=(2x)^2-2(2x)(y)\cos\alpha+y^2.$$ Since $2x=2\cos A$ and $y=2\cos B$, this becomes $$= (2\cos A)^2-2(2\cos A)(2\cos B)\cos(A-B)+(2\cos B)^2.$$ So, $$=4\left[\cos^2A+\cos^2B-2\cos A\cos B\cos(A-B)\right].$$ Now use $$\cos(A-B)=\cos A\cos B+\sin A\sin B.$$ Then \begin{align*} &\cos^2A+\cos^2B-2\cos A\cos B\cos(A-B) \\ &=\cos^2A+\cos^2B-2\cos A\cos B(\cos A\cos B+\sin A\sin B). \end{align*} But an easier simplification is obtained from $$\cos A=\cos(B+\alpha)=\cos B\cos\alpha-\sin B\sin\alpha.$$ Thus, $$x=\cos A=\cos B\cos\alpha-\sin B\sin\alpha,$$ so $$2x=y\cos\alpha-2\sin B\sin\alpha.$$ Hence, $$2x-y\cos\alpha=-2\sin B\sin\alpha.$$ Now compute the target expression: \begin{align*} 4x^2-4xy\cos\alpha+y^2 &=(2x)^2-2(2x)(y)\cos\alpha+y^2 \\ &=(2x-y\cos\alpha)^2+y^2(1-\cos^2\alpha). \end{align*} Therefore, $$= ( -2\sin B\sin\alpha )^2 + y^2\sin^2\alpha.$$ Since $y=2\cos B$, \begin{align*} 4x^2-4xy\cos\alpha+y^2 &=4\sin^2B\sin^2\alpha + 4\cos^2B\sin^2\alpha \\ &=4(\sin^2B+\cos^2B)\sin^2\alpha \\ &=4\sin^2\alpha. \end{align*} --- 4. **Match with options** Thus, $$4x^2-4xy\cos\alpha+y^2=4\sin^2\alpha.$$ So the correct option is: $$\boxed{\text{A: }4\sin^2\alpha}$$
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