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http://filestore.aqa.org.uk/subjects...W-QP-JAN09.PDF
Q 2B, ii)
I've got sqrt(10)sin(x-1.25) = -sqrt(10)
-sqrt(10) / sqrt(10) = -1
sin(x-1.25) = -1
sin^-1(-1) = -1/2(pi)
-1/2pi = x - 1.25
x = -1/2pi + 1.25
The actual answer is 5.96 confused as to how they got there
Any help appreciated
Q 2B, ii)
I've got sqrt(10)sin(x-1.25) = -sqrt(10)
-sqrt(10) / sqrt(10) = -1
sin(x-1.25) = -1
sin^-1(-1) = -1/2(pi)
-1/2pi = x - 1.25
x = -1/2pi + 1.25
The actual answer is 5.96 confused as to how they got there
Any help appreciated
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#2
(Original post by Student20)
http://filestore.aqa.org.uk/subjects...W-QP-JAN09.PDF
Q 2B, ii)
I've got sqrt(10)sin(x-1.25) = -sqrt(10)
-sqrt(10) / sqrt(10) = -1
sin(x-1.25) = -1
sin^-1(-1) = -1/2(pi)
-1/2pi = x - 1.25
x = -1/2pi + 1.25
The actual answer is 5.96 confused as to how they got there
Any help appreciated
http://filestore.aqa.org.uk/subjects...W-QP-JAN09.PDF
Q 2B, ii)
I've got sqrt(10)sin(x-1.25) = -sqrt(10)
-sqrt(10) / sqrt(10) = -1
sin(x-1.25) = -1
sin^-1(-1) = -1/2(pi)
-1/2pi = x - 1.25
x = -1/2pi + 1.25
The actual answer is 5.96 confused as to how they got there
Any help appreciated
0
reply
Report
#5
(Original post by Student20)
Anyone?
Anyone?
You have
.So your "key-angle" or "basic-angle" is
.Then because your sine is "negative", then your solutions are in the 'third' or 'fourth quadrant.
So
or 
.Solve both of those equations for x and you should get your answer.
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