# a level maths yr 2 simplyifying trig identities

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#1
given that sin (x+y) = 4 cos (x-y), write an expression for tan x in terms of tan y.

my working:
sinx cosy + cosx siny = 4 (cosx cosy + sinx siny)
sinx cosy + cosx siny = 4 cosx cosy + 4 sinx siny
sinx cosy - 4 sinx siny = 4 cosx cosy - cosx siny
sinx (cosy - 4 siny) = cosx (4 cosy - siny)
tanx (cosy - 4 siny) = 4 cosy - siny
tanx = (4 cosy - siny) / (cosy - 4 siny)

I'm stuck here and can't see where I may have gone wrong.

tanx = (4 - tany) / (1 - 4 tany)

I'll be grateful for any help.
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1 month ago
#2
about half way down divide through by cos(x)cos(y)

If youd worked back from the answer a couple of steps, it should have been reasonably clear you'd need to do it?
Last edited by mqb2766; 1 month ago
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#3
(Original post by mqb2766)
about half way down divide through by cos(x)cos(y)

If youd worked back from the answer a couple of steps, it should have been reasonably clear you'd need to do it?
thanks a lot.
0
1 month ago
#4
(Original post by NaBrO)
given that sin (x+y) = 4 cos (x-y), write an expression for tan x in terms of tan y.

my working:
sinx cosy + cosx siny = 4 (cosx cosy + sinx siny)
sinx cosy + cosx siny = 4 cosx cosy + 4 sinx siny
sinx cosy - 4 sinx siny = 4 cosx cosy - cosx siny
sinx (cosy - 4 siny) = cosx (4 cosy - siny)
tanx (cosy - 4 siny) = 4 cosy - siny
tanx = (4 cosy - siny) / (cosy - 4 siny)

I'm stuck here and can't see where I may have gone wrong.

tanx = (4 - tany) / (1 - 4 tany)

I'll be grateful for any help.
Just to say, you could also just divide top and bottom of the RHS by cos y at the point you got stuck.
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#5
(Original post by DFranklin)
Just to say, you could also just divide top and bottom of the RHS by cos y at the point you got stuck.
thank you!
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