The value of cos 2( (π /4)+θ )- sin 2( (π /4)-θ ) is

Question

The value of cos 2( (π /4)+θ )- sin 2( (π /4)-θ ) is (A)  0  (B)  cos 2θ  (C)  sin 2θ  (D)  cos θ

Tardigrade Admin · Accepted Answer

We know that, cos 2 (A)- sin 2(B) = cos (A+B) cos (A-B) ∴ cos 2 ( (π /4)+θ )- sin 2( (π /4)-θ ) = cos ( (π /4)+θ +(π /4)-θ ) cos ( (π /4)+θ -(π /4)+θ ) = cos ( (π /2) ) cos (2 θ )=0

Q. The value of $cos^{2} (\frac{π}{4} + θ) - sin^{2} (\frac{π}{4} - θ)$ is

$0$

$cos 2 θ$

$s in 2 θ$

$cos θ$

We know that, $cos^{2} (A) - sin^{2} (B)$
$= cos (A + B) cos (A - B)$
$∴$ $cos^{2} (\frac{π}{4} + θ) - sin^{2} (\frac{π}{4} - θ)$
$= cos (\frac{π}{4} + θ + \frac{π}{4} - θ) cos (\frac{π}{4} + θ - \frac{π}{4} + θ)$
$= cos (\frac{π}{2}) cos (2 θ) = 0$

Q. The value of cos2(4π​+θ)−sin2(4π​−θ) is

0

cos2θ

sin2θ

cosθ

We know that, cos2(A)−sin2(B) =cos(A+B)cos(A−B) ∴ cos2(4π​+θ)−sin2(4π​−θ) =cos(4π​+θ+4π​−θ)cos(4π​+θ−4π​+θ) =cos(2π​)cos(2θ)=0

Q. The value of $cos^{2} (\frac{π}{4} + θ) - sin^{2} (\frac{π}{4} - θ)$ is

$0$

$cos 2 θ$

$s in 2 θ$

$cos θ$

We know that, $cos^{2} (A) - sin^{2} (B)$
$= cos (A + B) cos (A - B)$
$∴$ $cos^{2} (\frac{π}{4} + θ) - sin^{2} (\frac{π}{4} - θ)$
$= cos (\frac{π}{4} + θ + \frac{π}{4} - θ) cos (\frac{π}{4} + θ - \frac{π}{4} + θ)$
$= cos (\frac{π}{2}) cos (2 θ) = 0$