The equation of the tangent line is  To find , the slope, calculate the derivative and plug in the desired point.

The next step is to choose a coordinate on the original  function. We can choose any  value and calculate its  value.

Let's choose .

The  value at this point is .

Plugging in those values we can solve for .

Solving for  we get =.

First we need to find the slope of  at . To do this we need the derivative of . To take the derivative we need to use the power rule for the first term and recognize that the derivative of sine is cosine.

At,

Now we need to know

.

Now we have a slope,  and a point 

so we can use the point-slope formula to find the equation of the line.

Plugging in and rearranging we find

.

First, evaluate  when .

Thus, we need a line that contains the point 

Next, find the derivative of  and evaluate it at .

To find the derivative we will use the power rule,

.

This indicates that we need a line with a slope of 8.

In point-slope form, , a line with the point  and a slope of 8 will be:

The tangent line to  at  must have the same slope as .

Applying the chain rule we get

.

Therefore the slope of the line is, 

.

In addition, the tangent line touches the graph of  at . Since , the point  lies on the line.

Plugging in the slope and point we get .

In order to find the equation of the tangent line at , we first find the slope.

To do this we need to find .

Since we have found , now we simply plug in 1.

Now we need to plug in 1, into , to find a point that the tangent line touches.

Now we can use point-slope form to figure out what the equation of the tangent line is at .

Remember that point-slope for is

where  and  is the point where the tangent line touches , and  is the slope of the tangent line.

In our case, , , and .

Thus our tangent line equation at  is

.

In order to find the equation of the tangent line at , we first find the slope.

To do this we need to find  using the power rule .

Since we have found , now we simply plug in 1.

Now we need to plug in 1, into , to find a point that the tangent line touches.

Now we can use point-slope form to figure out what the equation of the tangent line is at .

Remember that point-slope for is

where  and  is the point where the tangent line touches , and  is the slope of the tangent line.

In our case, , , and .

Thus our tangent line equation at  is

.

The equation of the tangent line has the form .

The slope  can be determined by evaluating the derivative of the function at .

Plugging this into the point slope equation, we get

 can be determined by evaluating the original function at .

Plugging this into the previous equation and simplifying gives us

To find the slope of the line tangent you must take the derivative of the function. The derivative of cosine is negative sine and the derivative of sine is cosine.

This makes the derivative of the function

.

Plug in the given x to get the slope.

To find the slope of the line tangent to the function at a point you must first find the derivative.

The power rule states that the derivative of  is .

The derivative of  is .

The derivative of the function is

.

Plugging in 1 for x gives 

.

" is decreasing at ." is incorrect.  The function is increasing at because .

" is increasing over the interval ." is possibly true, but there is not enough information to conclude that it must be true.

" has a point of inflection at ." is possibly true.  Although we know that , a requirement for an inflection point, we do not know that changes signs at .

" must have at least one relative maximum." is possibly true, but there is not enough information to conclude that it must be true.

"The line is tangent to ." must be true.  Because , the function travels through the point .  Because , the slope of the line tangent to the curve at is 5.  Use point-slope form to determine the equation of the tangent line.