Differentiation of trigonometric functions and differentiation rules — Question 109

Original QCE Vault practice · 3 marks

Q109 · Practice questionTechnology-freeSimple familiar3 marks

QUESTION 109 (3 marks)

Determine the equation of the tangent to

y=\cos(2x)

x=\dfrac{\pi}{6}

WORKED SOLUTION

3 marks

ANSWER

y-\dfrac12=-\sqrt3\left(x-\dfrac{\pi}{6}\right)

Worked solution

x=\pi/6

y=\cos(\pi/3)=1/2

. Since

y'=-2\sin(2x)

, the gradient is

-2\sin(\pi/3)=-\sqrt3

Finds the point

\left(\pi/6,1/2\right)

[1 mark]

Finds the gradient

-\sqrt3

[1 mark]

Writes the tangent equation.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.

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