Question

The equation of the tangent to the curve $y = \sqrt{2} \sin\left(2x + \frac{\pi}{4}\right)$ at $x = \frac{\pi}{4}$ is

• A. $2x + y - \frac{\pi}{2} - 1 = 0$
• B. $2x - y - \frac{\pi}{2} + 1 = 0$
• C. $x + y - \frac{\pi}{2} - 1 = 0$
• D. $2x + y - \frac{\pi}{2} + 1 = 0$

Hint:

Always look at the signs of your options to save time! Since the slope is negative ($m = -2$), moving it to the left side makes the coefficient of $x$ positive ($+2x$). Since $y$ also has a positive coefficient, the line must start with $2x + y$, instantly narrowing your choices down to options (A) and (D)!

Answer 1

The Correct Option is A

Step 1: Find the point of contact.
At $x = \dfrac{\pi}{4}$, $y = \sqrt{2}\sin\left(\dfrac{\pi}{2} + \dfrac{\pi}{4}\right) = \sqrt{2}\cos\dfrac{\pi}{4} = \sqrt{2}\cdot\dfrac{1}{\sqrt{2}} = 1$. So the point is $\left(\dfrac{\pi}{4}, 1\right)$.
Step 2: Differentiate the curve.
$\dfrac{dy}{dx} = \sqrt{2}\cos\left(2x + \dfrac{\pi}{4}\right)\cdot 2 = 2\sqrt{2}\cos\left(2x + \dfrac{\pi}{4}\right)$.
Step 3: Compute the slope at the point.
At $x = \dfrac{\pi}{4}$: $\cos\left(\dfrac{\pi}{2} + \dfrac{\pi}{4}\right) = -\sin\dfrac{\pi}{4} = -\dfrac{1}{\sqrt{2}}$, so slope $m = 2\sqrt{2}\left(-\dfrac{1}{\sqrt{2}}\right) = -2$.
Step 4: Apply point-slope form.
$y - 1 = -2\left(x - \dfrac{\pi}{4}\right)$.
Step 5: Expand the right side.
$y - 1 = -2x + \dfrac{\pi}{2}$.
Step 6: Bring everything to one side.
$2x + y - \dfrac{\pi}{2} - 1 = 0$.
\[ \boxed{2x + y - \dfrac{\pi}{2} - 1 = 0} \]