JEE MAIN Trigonometry Previous Year Questions (PYQs) – Page 1 of 8
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Given that the inverse trigonometric functions assume principal values only.
Let $x,y\in[-1,1]$ such that $\cos^{-1}x-\sin^{-1}y=\alpha$, with $-\dfrac{\pi}{2}\le\alpha\le\pi$.
Then, the minimum value of $x^{2}+y^{2}+2xy\sin\alpha$ is:
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Solution
Let the maximum value of $(\sin^{-1}x)^2 + (\cos^{-1}x)^2$ for $x \in \left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]$ be $\frac{m\pi^2}{n}$, where $\gcd(m,n)=1$. Then m+n is equal to ______.
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Solution
$(\sin^{-1}x)^2 + (\cos^{-1}x)^2$
$= (\sin^{-1}x + \cos^{-1}x)^2 - 2\sin^{-1}x\cos^{-1}x$
$= \frac{\pi^2}{4} - 2(\sin^{-1}x)\left(\frac{\pi}{2} - \sin^{-1}x\right)$
$= 2\left(\sin^{-1}x - \frac{\pi}{4}\right)^2 + \frac{\pi^2}{8}$
Maximum occurs at $\sin^{-1}x = -\frac{\pi}{3}$
$\Rightarrow 2\left(\frac{\pi}{3} + \frac{\pi}{4}\right)^2 + \frac{\pi^2}{8}$
$= 2\left(\frac{7\pi}{12}\right)^2 + \frac{\pi^2}{8}$
$= \frac{49\pi^2}{72} + \frac{9\pi^2}{72} = \frac{29\pi^2}{36}$
$\Rightarrow m = 29,; n = 36$
$\Rightarrow m+n = 65$
If $\sin \theta + \cos \theta = {1 \over 2}$, then 16(sin(2$\theta$) + cos(4$\theta$) + sin(6$\theta$)) is equal to :
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Solution
The value of $\cos \dfrac{\pi}{22}\cdot \cos \dfrac{\pi}{23}\cdot \ldots \cdot \cos \dfrac{\pi}{210}\cdot \sin \dfrac{\pi}{210}$ is –
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Solution
Let $f_{k}(x)=\dfrac{1}{k}(\sin^{k}x+\cos^{k}x)$ where $x\in\mathbb{R}$ and $k\ge 1$.
Then $f_{4}(x)-f_{6}(x)$ equals :
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Solution
If $\tan A$ and $\tan B$ are the roots of the quadratic equation $3x^{2}-10x-25=0$, then the value of
$3\sin^{2}(A+B)-10\sin(A+B)\cos(A+B)-25\cos^{2}(A+B)$ is :
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Solution
$ \text{The expression } \dfrac{\tan A}{1-\cot A,} + \dfrac{\cot A}{1-\tan A,} \text{ can be written as:} $
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Solution
Let $f(x) = 3{\sin ^4}x + 10{\sin ^3}x + 6{\sin ^2}x - 3$, $x \in \left[ { - {\pi \over 6},{\pi \over 2}} \right]$. Then, f is :
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Solution
If $\displaystyle \frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\alpha + \beta\sqrt{5}}{2}$, where $\alpha, \beta \in \mathbb{N}$, then $\alpha + \beta$ is equal to ______.
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Solution
Use identities
$\sin(A+B)\sin(A-B) = \sin^2 A - \sin^2 B$
$\cos(A+B)\cos(A-B) = \cos^2 A - \cos^2 B$
$\displaystyle \frac{\cos 60^\circ \cos 36^\circ}{\sin 30^\circ \sin 18^\circ}
= \frac{\frac{\sqrt{5}+1}{4}}{\frac{\sqrt{5}-1}{4}}
= \frac{\sqrt{5}+1}{\sqrt{5}-1}$
$= \frac{(\sqrt{5}+1)^2}{5-1}
= \frac{6+2\sqrt{5}}{4}
= \frac{3+\sqrt{5}}{2}$
$\Rightarrow \alpha = 3,; \beta = 1$
$\Rightarrow \alpha + \beta = 4$
Suppose $\theta\in[0,\tfrac{\pi}{4}]$ is a solution of $4\cos\theta-3\sin\theta=1$. Then $\cos\theta$ is:
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Solution
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