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# Can't get answer (simple tangent to the curve question) watch

1. Find the Coordinates of a point on the curve y= sin x + cos x at which the tangent is parallel to the line y=x

So basically you need to find the gradient function of the curve, then equate it to the gradient of the line y=x in order to find the x which wil result in f'(x)=1

f'(x)= Cos x - Sin x
Cos x = Sin x + 1
( squaring both sides )
Cos ^2 (x) = Sin^2(x) + 2sinx + 1 => 2sin^2 (x) + 2sin (x) = 0
2sin(x) [Sin(x) + 1] = 0
2Sin(x)= 0
x=ArcSin(0)= 0,180,360
Or

x=arcSin(-1)= 270

(im guessing there should be many points where the x gives a gradient of 1 to the tangent to the curve since the graph is continuous. The textbook answer is (0,1) , is there any specific reason why they took that point? )
2. (Original post by Carlos Nim)
Find the Coordinates of a point on the curve y= sin x + cos x at which the tangent is parallel to the line y=x

So basically you need to find the gradient function of the curve, then equate it to the gradient of the line y=x in order to find the x which wil result in f'(x)=1

f'(x)= Cos x - Sin x
Cos x = Sin x + 1
( squaring both sides )
Cos ^2 (x) = Sin^2(x) + 2sinx + 1 => 2sin^2 (x) + 2sin (x) = 0
2sin(x) [Sin(x) + 1] = 0
2Sin(x)= 0
x=ArcSin(0)= 0,180,360
Or

x=arcSin(-1)= 270

(im guessing there should be many points where the x gives a gradient of 1 to the tangent to the curve since the graph is continuous. The textbook answer is (0,1) , is there any specific reason why they took that point? )
Indeed there are infinitely many tangents to such that they are parallel to , and this is more to do with the fact that is periodic, not continuous.

As for why picking that point, it doesn't really justify as you could pick any point on the curve where the gradient is 1 and still answer the question, I guess it's because the closest point to the origin which makes it some sort of a 'simple' point.

Also, some of your solutions (like ) are not valid because at those points the gradient is -1. You get these falsely because you squared your equation early on, and squaring makes things positive.
3. (Original post by Carlos Nim)
Find the Coordinates of a point on the curve y= sin x + cos x at which the tangent is parallel to the line y=x

So basically you need to find the gradient function of the curve, then equate it to the gradient of the line y=x in order to find the x which wil result in f'(x)=1

f'(x)= Cos x - Sin x
Cos x = Sin x + 1
( squaring both sides )
Cos ^2 (x) = Sin^2(x) + 2sinx + 1 => 2sin^2 (x) + 2sin (x) = 0
2sin(x) [Sin(x) + 1] = 0
2Sin(x)= 0
x=ArcSin(0)= 0,180,360
Or

x=arcSin(-1)= 270

(im guessing there should be many points where the x gives a gradient of 1 to the tangent to the curve since the graph is continuous. The textbook answer is (0,1) , is there any specific reason why they took that point? )
Unless there is something in the wording of the original question, then one answer is as good as any other.

However, be aware that squaring an equation can, and in this case does, introduce additional solutions, that do not satisfy the original equation. E.g. x=180; gradient of f(x) is -1.
4. is there any other way to solve the above equation which won't involve squaring ?
5. (Original post by Carlos Nim)
is there any other way to solve the above equation which won't involve squaring ?
You can express form.
6. (Original post by RDKGames)
You can express form.
ahh yes thx , djd't realise that

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Updated: March 10, 2018
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