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6.8.2 Trigonometric Functions Long Questions (Question 3 & 4)


Question 3:
(a) Sketch the graph of y= 3 2 cos2x for 0x 3 2 π.
(b) Hence, using the same axes, sketch a suitable straight line to find the number of solutions to the equation 4 3π xcos2x= 3 2  for 0x 3 2 π
State the number of solutions.

Solution:

(a)(b)



4 3π xcos2x= 3 2 cos2x= 4 3π x 3 2 3 2 cos2x= 3 2 ( 4 3π x 3 2 ) y= 2 π x 9 4 To sketch the graph of y= 2 π x 9 4 x=0, y= 9 4 x= 3π 2 , y= 3 4 Number of solutions  =Number of intersection points = 3



Question 4:
(a) Prove that ( cosec xsecx secx cosec x ) 2 =1sin2x  
[3 marks]
(b)(i) Sketch the graph of y = 1 – sin2x for 0 ≤ x ≤ 2π.

(b)(ii) Hence, using the same axes, sketch a suitable straight line to find the number of solutions to the equation 2 ( cosec xsecx secx cosec x ) 2 = x π  for 0 ≤ x ≤ 2π.
State the number of soultions. 
[7 marks]

Solution:
(a)
LHS = ( cosec xsecx secx cosec x ) 2 = ( cosec x secx cosec x secx secx cosec x ) 2 = ( 1 secx 1 cosec x ) 2 = ( cosxsinx ) 2 = cos 2 x2cosxsinx+ sin 2 x =1sin2x (RHS)



(b)(i)




(b)(ii)
2 ( cosec xsecx secx cosec x ) 2 = x π 2( 1sin2x )= x π From 4(a) 2y= x π From 4(b)(i) y=2 x π (suitable straight line)

x
0
π
y
2
1
0

From the graph, there is 3 number of solutions.


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