Trig Questions.. (1 Viewer)

[GaDaMIt]

sin 3x = 3sin x - 4 sin^3 x

cos 3x = 4cos^3 x - 3 cos x

Proove tan 3x = (3tanx - tan^3x) / (1 - 3tan^2 x)







Prove
(2 sin^3 x + 2 cos^3 x) / (sin x + cos x) = 2 - sin2x

AND prove
cos^2x cos^2y - sin^2x sin^2y = 1/2(cos 2x + cos 2y)

 

[_ShiFTy_]

1) tan 3x = sin3x/cos3x
=(3sin x - 4 sin^3 x) / (4cos^3 x - 3 cos x)....divide both sides by cos^3 x
=(3tanx.sec^2x - 4tan^3x) / (4 - 3sec^2x)
=[3tanx(1+tan^2x) - 4tan^3x] / [4 - 3(1+tan^2x)]
=(3tanx - tan^3x) / (1 - 3tan^2x)



2) (2 sin^3 x + 2 cos^3x) / (sinx + cosx)
= 2(sin^3x + cos^3x) / (sinx + cosx)
=2(sinx + cosx)(sin^2x - sinxcosx + cos^2x) / (sinx + cosx)
=2(1 - 1/2.sinx2x)
=2 - sin2x



3) cos^2x cos^2y - sin^2x sin^2y
=cos^2x cos^2y - (1 - cos^2x)(1 - cos^2y)
=cos^2x cos^2y - [1 - cos^2y - cos^2x + cos^2x cos^2y]
= -1 + cos^2y + cos^2x
= -1 + 1/2(cos2y + 1) + 1/2(cos2x +1)
= 1/2( cos2y + cos2x)

 

[GaDaMIt]

thanks shifty

another trig question.. could anyone help out?

use the t = tan 1/2 theta results to find the exact value of
(2 tan 10') / (1 - tan^2 10')


i figure its an easy question im just at a complete loss as to how to do it

 

[GaDaMIt]

use the t = tan 1/2 theta results to find

lol slide you missed that figured it out anyway.. it was just the first question of the sort that ive done so i had no idea how to approach it

 

[GaDaMIt]

Few more questions...

a)Given that t = tan 112.5, show that 2t / 1 - t^2 = 1

is it enough to say tan 225 = 2t / 1 - t^2

tan 225 = tan 45 = 1?

therefore 2t / 1 - t^2 = 1?



b) i) Hence show that tan 112.5 = - rt2 - 1
Using quadratics i got -1+-rt 2... 2nd quadrant as 112.5 .. therefore it is negative.. therefore -1 - rt 2 ... all right yeh??

b ii) What does the other root of the equation represent? -- Stumped? --

 

[Riviet]

Your thinking in a) and b) i) seems fine. In ii), the other root is positive, and I think it is given by tan22.5 or tan202.5.

 

[GaDaMIt]

Yet another question....

If x = tan y + sec y, use the t-formulae to show that (x^2 - 1)/(x^2 + 1) = sin y?

 

[senida]

hey i want to do that trig question but what does ^ represent

 

[GaDaMIt]

hey i want to do that trig question but what does ^ represent

to the power of..

 

[Riviet]

Yet another question....

If x = tan y + sec y, use the t-formulae to show that (x^2 - 1)/(x^2 + 1) = sin y?

let t=tan(y/2),
tany=2t/(1-t²) and secy=(1+t²)/(1-t²)
.'. x= (2t+1+t²)/(1-t²)
=[(t+1)(t+1)]/[(1+t)(1-t)]
=(t+1)/(1-t)

Substitute this into (x² - 1)/(x² + 1), simplify and you should get 2t/(1+t²) which equals siny.

 

[GaDaMIt]

Thanks.. last bunch of questions


Solve for 0 =< x =< 360. Give solutions correct to the nearest minute
2 secx - 2tan x = 5
2 cosecx + 5 cot x = 3


EDIT: Asked a stupid question
EDITEDIT: Questions still here arent the stupid ones ... just making that clear ... still need help with them

 

[Riviet]

For these two, change secx into 1/cosx, tanx into sinx/cosx, cosecx into 1/sinx and cotx into cosx/sinx. Multiply by the common denominator on the LHS to get rid of fractions. Then you get an equation in the form asinx + bcosx = c, and you can use a couple of methods to solve this equation, whichever you prefer, e.g t-results or the subsidiary angle method [asinx+bcosx = Rsin(x+@) or Rcos(x+@)]. If you use the t-results, don't forget to test angles like pi/2, 3pi/2 etc.

 

[drynxz]

little help on this question:

(1-tan15) / (1+tan² 15)

and also

given sin a =4/7 and tan b =1/3 find exact value of sin(2a-b)

 

[Riviet]

(1-tan15) / (1+tan² 15) = 1/sec¹⁵15 - tan15/sec²15, by splitting the fraction and 1+tan²@=sec²@
= cos²15 - sin15.cos15, by changing tan and sec into sin and cos and cancelling cos@
= 1/2.(1+cos30) - 1/2.sin30 since cos²@=1/2.(1+cos2@) and aasin@cos@=1/2.sin2@
= 1/2 + (rt3)/4 - 1/4
= (1+rt3)/4

For second question, draw two right triangles, label all the info with angle a for one and label all the info for angle b in the other. Expand sin(2a-b), find expressions for what you need using the triangles and substitute them in. Don't forget you further expand sin2a.

Hope that helps.