If y-sin n x-cos n xx- n real- then -2- y -2

We have y = sin nx + cos nx =√(2) [ 1/√((2)) sin nx+1/√((2))cos nx ] =√(2) [ cos π/4 sin nx + sin π/4 cos nx ] =√(2) sin ( nx + π/4 ) Since 1 ≤ sin (nx + π ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Inequality

If y=sin n x+...

Question

If $y = s i n n x + c o s n x$ (x, n real), then $- \sqrt{2} \leq y \leq \sqrt{2}$

True

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

False

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

We have
$y = s i n n x + c o s n x$
$= \sqrt{2} [\frac{1}{\sqrt{(2)}} s i n n x + \frac{1}{\sqrt{(2)}} c o s n x]$
$= \sqrt{2} [c o s \frac{π}{4} s i n n x + s i n \frac{π}{4} c o s n x]$
$= \sqrt{2} s i n (n x + \frac{π}{4})$
Since $- 1 \leq s i n (n x + \frac{π}{4}) \leq 1$
we have $- \sqrt{2} \leq \sqrt{2} s i n (n x + \frac{π}{4}) \leq \sqrt{2}$
that is $- \sqrt{2} \leq y \leq \sqrt{2}$

Suggest Corrections

Similar questions

Q. If

y = s i n n x + c o s n x

(x, n real), then

- \sqrt{2} \leq y \leq \sqrt{2}

Q. If

y = s i n n x + c o s n x

(x, n real), then

- \sqrt{2} \leq y \leq \sqrt{2}

Q. If

A_{n} = \int_{0}^{π / 2} \frac{sin (2 n - 1) x}{sin x} d x; B_{n} = \int_{0}^{π / 2} {(\frac{sin n x}{sin x})}^{2} d x;

for

n \in N

, then

Q. If

n

is a positive integer, and

I_{n} = \int \frac{sin (n x)}{cos x} d x

, then

I_{n} + I_{n - 2} =

Q. if

x = \frac{π}{2},

satisfies the equation

sin \frac{n x}{2} - cos \frac{n x}{2} = 1 - sin n x

& the inequality

∣ ∣ ∣ \frac{n x}{2} - \frac{π}{2} ∣ ∣ ∣ \leq \frac{3 π}{4},

then

Join BYJU'S Learning Program

If y=sin nx+cos nx (x, n real), then −√2≤y≤√2

We have y=sin nx+cos nx =√2[1√(2)sin nx+1√(2)cos nx] =√2[cos π4sin nx+sin π4cos nx] =√2sin(nx+π4) Since −1≤sin(nx+π4)≤1 we have −√2≤√2sin(nx+π4)≤√2 that is −√2≤y≤√2

If $y = s i n n x + c o s n x$ (x, n real), then $- \sqrt{2} \leq y \leq \sqrt{2}$