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]

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...

[nid404]

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

[astarmathsandphysics]

Yes i will. I am also making some resource pages for my own website.

[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

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

anytime..

[Ghost Of Highbury]

maybe astar is busy..

slvri can u try..?

[slvri]

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)

[nid404]

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 )

[slvri]

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

[astarmathsandphysics]

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

[nid404]

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