### Higher Help Please!

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Messages: 1 - 3 of 3
• #### Message 1.

Posted by U15219824 (U15219824) on Tuesday, 15th May 2012

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!

• #### Message 2

, 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

• #### Message 3

, in reply to message 2.

Posted by U15219824 (U15219824) on Thursday, 17th May 2012

Okay thank you very much!

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