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Could someone please help me on the following questions? Would be much appreciated!
1) Find the smallest integer value of c for which g(x)=(x-2)(x^2-2x+c) has only one real root.
2) Integrate (sin4x)/sin2x dx
3) Solve tan^2x=1/3 for pi/2<x<pi
4) Given that sina=4/5 and sinb=2/root5 find the exact values of
(a) sin(a+b)
(b) tan(a+b)
Thanks very much!
, in reply to message 1.
Posted by Maths Teacher BBC (U15178780) on Thursday, 17th May 2012
1.
b²- 4ac < 0
(-2)² - 4 x 1 x c < 0
4 – 4c < 0
-4c < -4
-c < -1
c > 1
2.
Sin4x / sin2x
2sin2xcos2x / sin2x
2cos2x
Now you can integrate
2 x ½ sin 2x + c
= sin2x + c
3.
tan²x = 1/3
tan x = √(1/3)
tan x = ±1/√3
x = shift tan 1/√3
x = 30ᵒ
When tan 1/√3 – lies in A and T - 30ᵒ and 210ᵒ - both not in the correct range
When tan -1/√3 – lies in S and C - 150ᵒ and 330ᵒ
150ᵒ in the correct range so change to radians - 5∏/6
4.
sin a = 4/5
cos a = 3/5 – 3, 4, 5 triangle
sin b = 2/√5
cos b = 1/√5 by using Pythagoras
sin (a + b) = sina cosb + cosa sinb
= 4/5 x 1/√5 + 3/5 x 2/√5
= 4/5√5 + 6/5√5
= 10/5√5
= 2/√5
tan (a + b) = sin (a + b) / cos (a + b)
sin (a + b) = 2√5
cos(a + b) = cosa cos b – sina sinb
= 3/5 x 1/√5 – 4/5 x 2/√5
= 3/5√5 – 8/5√5
= -5/5√5
= -1/√5
tan (a + b) = 2/√5 / -1/√5
= 2 x √5 / -1 x √5
= 2√5 / -1√5
= -2
Okay thank you very much!
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