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Ghost Of Highbury:
maybe astar is busy..

slvri can u try..?

slvri:

--- 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 ---
ok.......8 sin theta-15 cos theta has the form a sin theta + b cos theta where a=8 and b=-15.this can be expressed in the form in the form R sin(theta-alpha) where R = sqrt(a2+b2) and R cos alpha = a (or R sin alpha = b)
R = sqrt(82+(-15)2)=17
now R cos alpha = 8
17 cos alpha = 8
alpha = cos-1(8/17)=61.93 degrees
so 8sin theta-15cos theta = 17 sin (theta - 61.93)

nid404:

--- Quote from: slvri on September 08, 2009, 10:20:25 am ---
--- 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 ---
ok.......8 sin theta-15 cos theta has the form a sin theta + b cos theta where a=8 and b=-15.this can be expressed in the form in the form R sin(theta-alpha) where R = sqrt(a2+b2) and R cos alpha = a (or R sin alpha = b)
R = sqrt(82+(-15)2)=17
now R cos alpha = 8
17 cos alpha = 8
alpha = cos-1(8/17)=61.93 degrees
so 8sin theta-15cos theta = 17 sin (theta - 61.93)

--- End quote ---

Thanx a lot :))

slvri:

--- Quote from: nid404 on September 08, 2009, 10:46:59 am ---
--- Quote from: slvri on September 08, 2009, 10:20:25 am ---
--- 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 ---
ok.......8 sin theta-15 cos theta has the form a sin theta + b cos theta where a=8 and b=-15.this can be expressed in the form in the form R sin(theta-alpha) where R = sqrt(a2+b2) and R cos alpha = a (or R sin alpha = b)
R = sqrt(82+(-15)2)=17
now R cos alpha = 8
17 cos alpha = 8
alpha = cos-1(8/17)=61.93 degrees
so 8sin theta-15cos theta = 17 sin (theta - 61.93)

--- End quote ---

Thanx a lot :))


--- End quote ---
no problem ;)

astarmathsandphysics:
Sorry I missed this. I am trying to do the templates for some maths notes I want to write for my own site.

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