How do you find the equation of the line tangent to #y=cosx# at x=π/4?

Question

Emilia Aguilar · Accepted Answer

#y-1/\sqrt{2}=(-1)/\sqrt{2}(x-\pi/4)# Given that to find the equation of a tangent, we first need to find the slope(#m#) of the equation. The given slope of the equation can be found by differentiating the function, and then substituting the value of either #x# or #y# in the differentiated function. So here(assuming you know what the derivative of cos function is), #y=cos(x)#. From the above paragraph, you'll understand that #m=\frac{d}{dx}(f(x))# #f(x)=y=cos(x)# #:. \frac{d}{dx}(y)=-sin(x)# The given slope for the tangent should be found at #x=\pi/4# So, #\frac{d}{dx}(y)|_(x=\pi/4)=-sin(\pi/4)=-1/\sqrt{2}# Now, the given slope equation is #y-y_o=m(x-x_o)# See that we need to find #y_o# and #x_o# From the given equation main we can find that out that at #x_o=\pi/4# #y_o=1/\sqrt{2}# So, finally, the answer is as provided above.

Ezra Adams · Answer

undefined To find the equation of the line tangent to y=cosx at x=π/4, we need to find the derivative of the function y=cosx and evaluate it at x=π/4. The derivative of y=cosx is dy/dx = -sinx. Evaluating this at x=π/4, we get dy/dx = -sin(π/4) = -√2/2.

The slope of the tangent line is equal to the derivative at the given point, so the slope of the tangent line is -√2/2.

To find the equation of the line, we use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point on the line and m is the slope. Plugging in the values x1=π/4, y1=cos(π/4) = √2/2, and m=-√2/2, we get the equation of the tangent line as y - √2/2 = -√2/2(x - π/4).