7. the tangent to the function $y = cos^{2}x-sin^{2}x$ at $x=\frac{pi}{4}$ and the x - and y - axes form a…

Explanation:

Step1: Simplify the function

Using the double - angle formula $\cos^{2}x-\sin^{2}x=\cos2x$, so $y = \cos2x$.

Step2: Find the derivative

The derivative of $y=\cos2x$ using the chain - rule. If $y = \cos u$ and $u = 2x$, then $y^\prime=-\sin u\times2=-2\sin2x$.

Step3: Find the slope of the tangent at $x = \frac{\pi}{4}$

Substitute $x=\frac{\pi}{4}$ into $y^\prime$. $y^\prime|_{x = \frac{\pi}{4}}=-2\sin(2\times\frac{\pi}{4})=-2\sin\frac{\pi}{2}=-2$.

Step4: Find the point on the function at $x=\frac{\pi}{4}$

Substitute $x = \frac{\pi}{4}$ into $y=\cos2x$. $y|_{x=\frac{\pi}{4}}=\cos(2\times\frac{\pi}{4})=\cos\frac{\pi}{2}=0$.

Step5: Find the equation of the tangent line

Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(\frac{\pi}{4},0)$ and $m=-2$. The equation is $y-0=-2(x - \frac{\pi}{4})$, which simplifies to $y=-2x+\frac{\pi}{2}$.

Step6: Find the x - intercept and y - intercept

For the x - intercept, set $y = 0$. Then $0=-2x+\frac{\pi}{2}$, so $x=\frac{\pi}{4}$. For the y - intercept, set $x = 0$. Then $y=\frac{\pi}{2}$.

Step7: Calculate the area of the triangle

The area of a triangle with base $b$ and height $h$ is $A=\frac{1}{2}bh$. Here, $b=\frac{\pi}{4}$ and $h=\frac{\pi}{2}$. So $A=\frac{1}{2}\times\frac{\pi}{4}\times\frac{\pi}{2}=\frac{\pi^{2}}{16}$.

Answer:

$\frac{\pi^{2}}{16}$