Derivatives of Trigonometric Functions (Equation of a Tangent Line)

Question

Find the equation of the line tangent to f(x)=2sin(x)cos(x) at x=5π/18. Round your slope and y-intercept to 2 decimal places.

(I've got the derivative figured out: f'(x)=-2sin²(x)+2cos²(x), but I need help evaluating the equation once x=5π/18 is substituted in, thanks).

Answer 1

When solving for the equation of a line tangent to a curve, the purpose of finding the derivative of the curve is to find the rate of change of the curve at the specified point, which is equal to the slope of the tangent line. While you correctly found the derivative of the curve, the equation for the derivative can be further simplified using the Cosine double angle trig identity: cos(2x)=cos(x)-sin²(x). Using this identity, the first derivative can be simplified:

f'(x)=-2sin²(x)+2cos²(x)=2(cos²(x)-sin²(x))=2cos(2x)

f'(x)=2cos(2x)

This simplified expression for the derivative will make substituting in x=5π/18 much easier

f'(5π/18)=2cos(5π/9)=-0.35

This value will serve as the slope of our tangent line. Now that we have the slope of the line, all we need to create the equation of the line is a point that the line intersects. Tangent lines intersect a curve at exactly one point, so that is the point we will use for the equation. In this case, the tangent line will intersect the curve at the point (5π/18, f(5π/18)).

While we were given the x value of the point of intersection, we need to find the y value using the equation of the curve.

f(5π/18)=2sin(5π/18)cos(5π/18)=0.99

Now that we have the slope of the tangent line (f'(5π/18)=-0.35) and the point on the curve that it intersects (5π/18,0.99), we can create our tangent line equation. The easiest way to create the equation is using point slope form:

(y-y₁)=m(x-x₁) where (x₁,y₁) is a point on the line and m is the slope of the line. Substituting the values we solved for earlier, we get:

y-0.99=-0.35(x-5π/18). However, judging by how your question specifies rounding the y-intercept to two decimal places, you may need to rearrange the equation so that it is in slope intercept form. To do this, just solve for y. (note, you have to calculate a decimal form of 5π/18 for this, about 0.87).

y=-0.35x+1.29

Answer 2

Find slope:

Given: f(x) = 2sin⁡(x)cos⁡(x)

We can use the double-angle identity for sine: sin⁡(2x) = 2sin⁡(x)cos⁡(x)

So: f(x) = sin⁡(2x)

The derivative f'(x) = 2cos(2x)

f'(5π/18) = 2cos(5π/9) = -0.35 (using a calculator)

The slope (m) of the tangent line is -0.35.

Find (x₁, y₁):

f(5π/18) = 0.98 (using a calculator)

(x₁, y₁) = (5π/18, 0.98)

Use point-slope form of a line:

y - y₁ = m(x - x₁)

So, y - 0.98 = -0.35(x - 5π/18) is the equation of the tangent line to f(x) at x = 5π/18.

If you want to convert point-slope form into slope-intercept form (y = mx + b), you get:

y = -0.35x + 1.28 where 1.28 is the y-intercept.

Hope this was helpful.

Answer 3

Your solution to the derivative is correct. To evaluate, you can apply the double angle formula for cosine, but since 5π/18 is not a special angle and you are told to round to 2 decimal places, I suggest plugging directly into a calculator.

I get an answer of approximately -0.35.

Then, to find your tangent line, plug into the formula y = mx + b. You are not given the y-coordinate, so plug x = 5π/18 into f(x). I get approximately 0.98.

Finally, solve for b.

0.98 = -0.35(5π/18) + b

b = 1.29

tangent line: y = -0.35x + 1.29