Author Topic: Trigonometry  (Read 4925 times)

Offline Ghost Of Highbury

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Re: Trigonometry
« Reply #15 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
divine intervention!

Offline Ghost Of Highbury

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Re: Trigonometry
« Reply #16 on: September 08, 2009, 07:07:46 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


sry..cudnt solve this...
not done these types of sums in add. math

will still try though...
divine intervention!

nid404

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Re: Trigonometry
« Reply #17 on: September 08, 2009, 07:11:19 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

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

Offline astarmathsandphysics

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Re: Trigonometry
« Reply #18 on: September 08, 2009, 08:36:36 am »
Yes i will. I am also making some resource pages for my own website.

Offline Ghost Of Highbury

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Re: Trigonometry
« Reply #19 on: September 08, 2009, 09:36:56 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

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
anytime..
divine intervention!

Offline Ghost Of Highbury

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Re: Trigonometry
« Reply #20 on: September 08, 2009, 10:14:20 am »
maybe astar is busy..

slvri can u try..?
divine intervention!

Offline slvri

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Re: Trigonometry
« Reply #21 on: September 08, 2009, 10:20:25 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


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)
i hate A level...........

nid404

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Re: Trigonometry
« Reply #22 on: September 08, 2009, 10:46:59 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


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)

Thanx a lot :))

Offline slvri

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Re: Trigonometry
« Reply #23 on: September 08, 2009, 11:09:02 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


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)

Thanx a lot :))

no problem ;)
i hate A level...........

Offline astarmathsandphysics

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Re: Trigonometry
« Reply #24 on: September 08, 2009, 11:22:12 am »
Sorry I missed this. I am trying to do the templates for some maths notes I want to write for my own site.

nid404

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Re: Trigonometry
« Reply #25 on: September 08, 2009, 11:23:54 am »
Sorry I missed this. I am trying to do the templates for some maths notes I want to write for my own site.

It's ok sir...you're so worked up..u need a break