Trigonometric Equations needed (1 Viewer)

[shaon0]

Hi, Could someone post hard Trigonometric Equations so that i can test my knowledge.

 

[lyounamu]

sin 30 =
sin 60 =
sin 45 =
cos 60 =
cos 30 =
cos 45 =
tan 45 =
tan 30 =
tan 60 =

Let's start with the light exercises!

 

[vds700]

Hard questions??? Thats easy. Try the miscellaneous ones, from 22 onwareds, u should be able to do them.

Have fun!

 

[lyounamu]

Hard questions??? Thats easy. Try the miscellaneous ones, from 22 onwareds, u should be able to do them.

Have fun!

Andrew, that's too generous mate!

 

[shaon0]

Hard questions??? Thats easy. Try the miscellaneous ones, from 22 onwareds, u should be able to do them.

Have fun!

Hey i am having trouble on 39. and 40. Rest is easy as..

 

[shaon0]

Hey i am having trouble on 39. and 40. Rest is easy as..

solved Question 40. Can someone help me on question 39?

 

[gurmies]

guys, sorry to bother, but just a general question. When faced with sin5x and stuff of that calibre, do you sit there using the compound angle formuale? Is there a better way?

 

[shaon0]

theres an extension 2 method you can learn if youre interested

which method is that?

 

[kaz1]

tan 90
sin² x = -0.5

 

[tommykins]

回复: Re: Trigonometric Equations needed

which method is that?

Complex numbers method.

z = r cis @
z^n = r^n cis n@

Take note that cis = (cos@ + isin@)

we take r = 1

so z^5 = (cos@ + isin@)^5 = cis 5@

expand using pascal's triangle

snce we're finding sin, we use the imaginary bits, if its cos we use real bits.

i don't think they're meant to give you anything more than sin4x/cos4x in prelim.

 

[conics2008]

Re: 回复: Re: Trigonometric Equations needed

Complex numbers method.

z = r cis @
z^n = r^n cis n@

Take note that cis = (cos@ + isin@)

we take r = 1

so z^5 = (cos@ + isin@)^5 = cis 5@

expand using pascal's triangle

snce we're finding sin, we use the imaginary bits, if its cos we use real bits.

i don't think they're meant to give you anything more than sin4x/cos4x in prelim.

why use pascals triangle when you can use comb method.

 

[lolokay]

Re: 回复: Re: Trigonometric Equations needed

why use pascals triangle when you can use comb method.

same thing isn't it, since pascal's triangle is just a way of expressing combinations? for lower values it would be easier to just have rows of the triangle memorised

 

[shaon0]

Re: 回复: Re: Trigonometric Equations needed

Can someone help me on question 39?

 

[shaon0]

you need to know complex numbers, heres a run down:

a complex number has the form a + ib, where a and b are reals and i is the imaginary unit (square root of -1, or also defined as i^2 = -1)

if you have 2 complex numbers z₁, and z₂ where

z₁ = a + ib
z₂ = c + id

then they are equal if and only if a = c (a and c is the "real part"), and b = d (b and d is the "imaginary part"). this means if we are given 2 complex numbers we know are equal say z₁, and z₂, we can say the real parts are equal (i.e. a = c) and we can say the imaginary parts are equal (i.e. b = d).

eg. if z₁ = z₂, where z₁ = x + 3i, and z₂ = 4 + iy, find x and y.

answer:
by equating reals, x = 4
by equating imaginaries, y = 3

theres also a formula you will learn if you do extension 2 known as demoivres formula:
(cos x + i sin x)^n = cos(nx) + i sin(nx)

we can use this formula to expand the LHS using binomial theorem, and equate reals and imaginary parts to find expressions for cos(nx) and sin(nx).
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here are some examples:

1)
(cos x + i sin x)^2 = cos(2x) + i sin(2x) by demoivres formula

LHS = (cos x + i sin x)^2

= (cos x)^2 + 2 (cos x) (i sin x) + (i sin x)^2 (also remember i^2 = -1)

= [(cos x)^2 - (sin x)^2] + i (2 cos x. sin x)

our real part is (cos x)^2 - (sin x)^2, and our imaginary part is 2 cos x. sin x

by equating our real part from the RHS: cos(2x) = (cos x)^2 - (sin x)^2
by equating our imaginary part from the RHS: sin(2x) = 2 cos x. sin x
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2)
(cos x + i sin x)^3 = cos(3x) + i sin(3x) by demoivres formula

LHS = (cos x + i sin x)^3

= (cos x)^3 + 3 (cos x)^2 (i sin x) + 3 (cos x) (i sin x)^2 + (i sin x)^3

= (cos x)^3 - 3 (cos x) (sin x)^2 + 3 (cos x)^2 (i sin x) - i (sin x)^3
(just used the idea that i^2 = -1, i^3 = - i)

= [(cos x)^3 - 3 (cos x) (sin x)^2] + i [3 (cos x)^2 (sin x) - (sin x)^3]

our real part is (cos x)^3 - 3 (cos x) (sin x)^2, and our imaginary part is
3 (cos x)^2 (sin x) - (sin x)^3

by equating our real part from the RHS:
cos(3x) = (cos x)^3 - 3 (cos x) (sin x)^2

by equating our imaginary part from the RHS:
sin(3x) = 3 (cos x)^2 (sin x) - (sin x)^3
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heres some sites for further reading:

for complex number basics:
http://www.purplemath.com/modules/complex.htm
http://www.tech.plym.ac.uk/maths/resources/PDFLaTeX/complex.pdf

introduction to polar form of complex numbers:
http://www.intmath.com/Complex-numbers/4_Polar-form.php

hey thanks got it..Can i use that in prelim HSC/for exams or tests?

 

[3unitz]

Re: 回复: Re: Trigonometric Equations needed

Can someone help me on question 39?

for RHS use the sums to products formula for sines

sin A + sin B = 2 sin[(A + B) / 2] cos[(A - B) / 2]

for LHS factorise the 2 out and use double angle formula

 

[shaon0]

Hey i'm so sick of these freaking questions! For instance, without using 4 unit maths, show me how to best prove:

sin3x/sinx + cos3x/cosx = 4cos2x

I've tried a couple of ways expanding and stuff but it takes so damn long! If faced with this on a test what the hell to do!? I have no problems with any trig, but just sometimes chosing the correct cos2x troubles me

thats very easy..

 

[shaon0]

Re: 回复: Re: Trigonometric Equations needed

for RHS use the sums to products formula for sines

sin A + sin B = 2 sin[(A + B) / 2] cos[(A - B) / 2]

for LHS factorise the 2 out and use double angle formula

wat did u get in 4unit math...because i am out of this...this question is too long or i don't have the necessary techniques/abilities to solve it.

 

[3unitz]

Re: 回复: Re: Trigonometric Equations needed

wat did u get in 4unit math...because i am out of this...this question is too long or i don't have the necessary techniques/abilities to solve it.

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 @

Hey i'm so sick of these freaking questions! For instance, without using 4 unit maths, show me how to best prove:

sin3x/sinx + cos3x/cosx = 4cos2x

I've tried a couple of ways expanding and stuff but it takes so damn long! If faced with this on a test what the hell to do!? I have no problems with any trig, but just sometimes chosing the correct cos2x troubles me

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

 

[tommykins]

回复: Re: 回复: Re: Trigonometric Equations needed

why use pascals triangle when you can use comb method.

Haven't done that topic yet.

 

[Iruka]

I think you are thinking too deeply about the problem.

sin3x/sinx + cos3x/cosx = (sin3xcosx + cos3xsinx)/ sinxcosx
=2sin4x/sin2x
=4sin2xcos2x/sin2x
=4cos2x