Value of x sinx+cosx=y+1y, x∈0, , thenA) x= 4C) x=5 4 - Maths - Introduction to Trigonometry

Question

Value of x

s i n x + c o s x = y + 1 y ,   x ∈ 0 , π ,   t h
e n A )   x = π 4 C )   x = 5 π 4

Manmeet Singh · Accepted Answer

To evaluate:

$\sin x + \cos x = \sqrt{y + \frac{1}{y}}$ ... (1)

We know that sum of a number and its reciprocal is always greater than 2.

∴ $y + \frac{1}{y} \geq 2$

So now,

$\sin x + \cos x \geq \sqrt{2} \Rightarrow \frac{\sin x}{\sqrt{2}} + \frac{\cos x}{\sqrt{2}} \geq 1 \Rightarrow \sin (x + \frac{π}{4}) \geq \sin \frac{π}{2} (U \sin g \sin A \cos B + \cos A \sin B = \sin (A + B), w h e r e A = x a n d B = \frac{π}{4}) \Rightarrow x + \frac{π}{4} \geq \frac{π}{2} \Rightarrow x \geq \frac{π}{4}$

Since x belongs to (0, $π$ )

∴ x = $\frac{π}{4}$

Now, putting this value of x in (1), we get

$\sin \frac{π}{4} + \cos \frac{π}{4} = \sqrt{y + \frac{1}{y}} \Rightarrow \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{y + \frac{1}{y}} \Rightarrow 2 = \frac{y^{2} + 1}{y} \Rightarrow y^{2} - 2 y + 1 = 0 \Rightarrow (y - 1)^{2} = 0 \Rightarrow y = 1$

∴ x = $\frac{π}{4}$ and y = 1

The correct option is 1.

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Value of x s i n x + c o s x = y + 1 y , x ∈ 0 , π , t h e n A ) x = π 4 C ) x = 5 π 4

Value of x

$s i n x + c o s x = \sqrt{y + \frac{1}{y}}, x \in [0, π], t h e n A) x = \frac{π}{4} C) x = \frac{5 π}{4}$