Qualification > Math
Trigonometry
Ghost Of Highbury:
ko..soo the 3 identities v use are..
sin2x = 2sinx cos x
cos2x = cos2x - sin2x
sin2x + cos2x = 1
---
k now first...simplify the LHS
therefore....RHs => cos2x + 6sinx cosx + 9sin2x
now the LHS => 5 - 4cos2x + 3sin2x
----
lets break the LHS into 3 parts => 5 ; -4cos2x ; 3sin2x
now 5 = 5sin2x + 5cos2x
...-4cos2x = -4cos2x + 4sin2x
... 3sin2x = 6sinx cos x
(check the first 2 identities)
----
so LHS => 5sin2x + 5cos2x -4cos2x + 4sin2x + 6sinx cos x
now cancel them and u get the RHS
Ghost Of Highbury:
--- Quote from: nid404 on September 08, 2009, 04:23:37 am ---Express 8 sin theta-15 cos theta in the form R sin(theta-alpha), where R > 0 and 0degress < alpha < 90degrees, giving the
exact value of R and the value of alpha correct to 2 decimal places.
Can somebody help me with this....plz guide me with the steps..astar plz help
--- End quote ---
sry..cudnt solve this...
not done these types of sums in add. math
will still try though...
nid404:
--- Quote from: eddie_adi619 on September 08, 2009, 07:04:28 am ---ko..soo the 3 identities v use are..
sin2x = 2sinx cos x
cos2x = cos2x - sin2x
sin2x + cos2x = 1
---
k now first...simplify the LHS
therefore....RHs => cos2x + 6sinx cosx + 9sin2x
now the LHS => 5 - 4cos2x + 3sin2x
----
lets break the LHS into 3 parts => 5 ; -4cos2x ; 3sin2x
now 5 = 5sin2x + 5cos2x
...-4cos2x = -4cos2x + 4sin2x
... 3sin2x = 6sinx cos x
(check the first 2 identities)
----
so LHS => 5sin2x + 5cos2x -4cos2x + 4sin2x + 6sinx cos x
now cancel them and u get the RHS
--- End quote ---
hey thanx...i didn't think about splitting it and then using the identities
this will help me solve sums ahead
no probs if u can't solve it...we'll wait for astar :P
astarmathsandphysics:
Yes i will. I am also making some resource pages for my own website.
Ghost Of Highbury:
--- Quote from: nid404 on September 08, 2009, 07:11:19 am ---
--- Quote from: eddie_adi619 on September 08, 2009, 07:04:28 am ---ko..soo the 3 identities v use are..
sin2x = 2sinx cos x
cos2x = cos2x - sin2x
sin2x + cos2x = 1
---
k now first...simplify the LHS
therefore....RHs => cos2x + 6sinx cosx + 9sin2x
now the LHS => 5 - 4cos2x + 3sin2x
----
lets break the LHS into 3 parts => 5 ; -4cos2x ; 3sin2x
now 5 = 5sin2x + 5cos2x
...-4cos2x = -4cos2x + 4sin2x
... 3sin2x = 6sinx cos x
(check the first 2 identities)
----
so LHS => 5sin2x + 5cos2x -4cos2x + 4sin2x + 6sinx cos x
now cancel them and u get the RHS
--- End quote ---
hey thanx...i didn't think about splitting it and then using the identities
this will help me solve sums ahead
no probs if u can't solve it...we'll wait for astar :P
--- End quote ---
anytime..
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