Trigonometric Equations needed (1 Viewer)

[shaon0]

Re: 回复: Re: Trigonometric Equations needed

LHS = sin(a - @) + sin(b - @)
= 2 sin[(a - @ + b - @) / 2] cos[(a - @ - b + @) / 2] (using sums to product formula)
= 2 sin[(a + b)/2 - @]cos[(a - b)/2]

RHS = sin(a + b - 2@)
= sin{2[(a+b)/2 - @]}
= 2 sin[(a + b)/2 - @] cos[(a+b)/2 - @] (using double angle formula)

cos[(a - b)/2] = cos[(a+b)/2 - @]

from here you can solve for @



sin3x = sin(2x + x) = sin2x cosx + cos2x sinx
cos3x = cos(2x + x) = cos2x cosx - sin2x sinx

LHS = (sin2x cosx + cos2x sinx)/sin x + (cos2x cosx - sin2x sinx)/ cosx
= 2sinx cosx cos x/ sin x + cos2x + cos2x - 2sinx cosx sinx/cosx (expanding sin2x)
= 2 cos^2x + 2cos2x - 2sin^2x
= 2(cos^2x - sin^2x) + 2cos2x
= 2cos2x + 2cos2x
= 4cos2x
= RHS

hey thanks...
btw wat did u get in 4unit?

 

[shaon0]

Re: 回复: Re: Trigonometric Equations needed

what is this? FBI agency?

just curious

 

[conics2008]

Re: 回复: Re: 回复: Re: Trigonometric Equations needed

Haven't done that topic yet.

its alot easier to expand using combantion..

what topics have you guys done at your school.

we finished 3unit and we have only 2 topics to go for 4units.. mechanics and harder 3unit.. might do harder 3unit over the holidays thats coming up.

 

[Sezenator]

omg can anyone helpo me with 2u maths formulasss
i fell behind durin the beginning of the year and im finding it hard to understand nytin
my prelim exam is next wednesday =[

 

[lyounamu]

omg can anyone helpo me with 2u maths formulasss
i fell behind durin the beginning of the year and im finding it hard to understand nytin
my prelim exam is next wednesday =[

Some 2 Unit Formulas:

Sin^2(x) + cos^2(x) = 1 ...(1)

From (1), dividing everything by cos^2(x), you get:
tan^2(x) + 1 = sec^2(x)

From (1), dividing everything by sin^2(x), you get:
cot^2(x) + 1 = cosec^2(x)

From (1), moving cos^2(x) you get:
Sin^2(x) = 1 - cos^2(x)

From (1), moving sin^2(x) you get:
Cos^2(x) = 1-sin^2(x)

 

[Tsylana]

Theres also a pretty tricky method solving by general solutions for stuff like sin 9A = cos4A or similar stuff xD...

but i can't be bothered going into that xD

 

[bored of sc]

Here's some questions:

1) Solve 4cosx + 4sin²x = 5, -180 < x < 180^o

2) Solve cosx = sin2x, 0 < x < 360^o

3) 3cos2x = 1 - sinx, 0 < x < 360^o

4) tan2x = 3tanx, 0 < x < 360^o

5) By expressing cos3x as cos(2x + x), show that cos3x = 4cos³x - 3cosx. Hence solve cos3x + 2cosx = 0, where 0 < x < 180^o

I'll post up the worked solutions soon.

 

[youngminii]

Here's some questions:

1) Solve 4cosx + 4sin²x = 5, -180 < x < 180^o

2) Solve cosx = sin2x, 0 < x < 360^o

3) 3cos2x = 1 - sinx, 0 < x < 360^o

4) tan2x = 3tanx, 0 < x < 360^o

5) By expressing cos3x as cos(2x + x), show that cos3x = 4cos³x - 3cosx. Hence solve cos3x + 2cosx = 0, where 0 < x < 180^o

I'll post up the worked solutions soon.

Phew, took me a while to get 5
Kept screwing up cos3x ==;;
Tell me if I'm wrong but:
(I'm not writing down the method 'cause I can't be bothered ==;
1. x=-60deg, 60deg
2. x=30deg, 90deg, 150deg, 270deg
3. x=42deg, 138deg, 210deg, 330deg
4. x=0deg, 30deg, 150deg, 180deg, 210deg, 330deg
5. x=60deg, 90deg, 120deg

 

[conics2008]

Find all @ such that

cos@ + cos2@ + cos3@ + cos4@ = 0