If $s i n (s i n x + c o s x) = c o s (c o s x - s i n x)$ , then the largest possible value of $s i n x$ is

\frac{1}{\sqrt{2}}

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1

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\frac{\sqrt{16 - π^{2}}}{4}

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\frac{π}{4}

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Solution

The correct option is A $\frac{1}{\sqrt{2}}$
Use foot of $sin θ = cos (90 - θ)$
$sin (sin x + ϕ) = cos (90 - (sin x + cos))$
$cos (90 - (sin x + cos x)) = cos (cos x - sin x)$
$90 - sin x - cos x = cos x - sin x$
$2 cos x = 90$
$cos x = 90$
$x = \frac{π}{4}$
$sin x = sin \frac{π}{4} = \frac{1}{\sqrt{2}}$

Suggest Corrections

Similar questions

${lim}_{x \to \frac{π}{4}} \frac{\sqrt{2} - c o s x - s i n x}{(4 x - π)^{2}}$

L t x \to \frac{π}{4} \frac{\sqrt{2} - cos x - sin x}{(4 x - π)^{2}} =

cos x + sin x = \frac{1}{2}

x \in (0, π)

tan x

lim x \to π \frac{1 - sin \frac{x}{2}}{cos \frac{x}{2} (cos \frac{x}{4} - sin \frac{x}{4})}

\frac{sin x}{1 + cos x} + \frac{sin x}{1 - cos x} = 4

0^{0} \leq x \leq 90^{0}

x

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