Some Trig Questions (1 Viewer)

nrlwinner · Sep 12, 2009

If you guys could help me solve these, I'd greatly appreciate it.

1. Simplify $2cos^23x-1$ (answer is cos6x)

2. Prove $\frac{sin2\Theta +1}{cos2\Theta }=\frac{cos\Theta -sin\Theta }{cos\Theta +sin\Theta }$

3. Prove $\frac{2cos\frac{\Theta }{2}-1-cos\Theta }{2cos\frac{\Theta }{2}+1+cos\Theta }=\frac{1-cos\frac{\Theta }{2}}{1+cos\frac{\Theta }{2}}$

4. Prove $\frac{1-tan\Theta tan2\Theta }{1+tan\Theta tan2\Theta }=4cos^2\Theta -3$

5. If $tanA=\frac{p}{q}$ express in terms of p and q

a) $qsinAcosA+psin^2A$ (answer is p)

b) $psin2A+qcos2A$ (answer is q)

6. Find $tanx$ in terms of $tan\Theta$ if $tan\Theta =\frac{cos(\Theta +x)}{cos(\Theta -x)}$

(answer is $\frac{1-tan\Theta }{tan\Theta (1+tan\Theta )}$ )

7. If $tan\alpha =ktan\beta$ , show that

$(k-1)sin(\alpha +\beta )=(k+1)sin(\alpha -\beta )$

kaz1 · Sep 12, 2009

for 1 use the identity cos2A=2cos²A - 1
for 2 in the RHS multiply top and bottom by cos@-sin@
for 3 in the LHS change the cos@ into cos(@/2)

khorne · Sep 12, 2009

for number2:

2sin@ cos@ + 1/cos^2 @ sin^2 @

2sin@cos@ + cos^2 @ + sin^2 @/cos^2 @ - sin^2@

(sin@ + cos@)(sin@ + cos@)/(cos@ -sin@)(sin@ + cos@)

cancel:
cos + sin / cos - sin

Which is the answer, because this is question 42 in fitzpatrick.

3)
From the t triangle, work out that

cos@/2 = 1/sqrt[1+t^2], or sqrt[1+t^2]/1+t^2
cos@ = 1-t^2/1+t^2

Thus, that becomes

2sqrt[1+t^2] - 1 -t^2 - 1 +t^2/2sqrt[1+t^2] + 1+t^2 + 1 -t^2
simplify and cancel

you get sqrt[1+t^2] - 1/ sqrt[1+t^2] + 1

Now work on other side

1+t^2 - sqrt[1+t^2]/1+t^2 + sqrt[1+t^2]

cancel

gives sqrt[1+t^2] -1/sqrtp1+t^2] + 1

= LHS

Number 4:

This one is pretty tricky:

transform 1 into 1-tan^2/1-tan^2 and tan2@ into 2tan^2/1-tan^2

do that for both numerator/denominator, then cancel all the 1-tan^2

to be left with 1-3tan^2/1+tan^2

Now multiply through by cos^2, to get cos^2-3sin^2/cos^2 + sin^2

thus, cos^2 -3(1-cos^2)

4cos^2 - 3

=LHS

-Note: I stopped writing thetas about half way in...=/

Drongoski · Sep 12, 2009

Q4:

$\100dpi LHS\ \ =\ \ \frac{1\ -\ tan\theta \ tan2\theta }{1\ +\ tan\theta \ tan2\theta }\ \ =\ \ \frac{1-tan\theta\ tan(\theta +\theta )}{1+tan\theta \ tan(\theta +\theta )}\\ \\ =\ \ \frac{1-tan\theta \ \frac{2tan\theta }{1-tan^{2}\theta }}{1+tan\theta \ \frac{2tan\theta }{1-tan^{2}\theta }}\ \ =\ \ \frac{1-tan^{2}\theta \ -\ 2tan^{2}\theta }{1-tan^{2}\theta \ +\ 2tan^{2}\theta }\\ \\ =\ \ \frac{1-3tan^{2}\theta }{1+tan^{2}\theta }\ \ =\ \ \frac{1-3tan^{2}\theta }{sec^{2}\theta }\ \ =\ \ cos^{2}\theta -3sin^{2}\theta \\ \\ =\ \ cos^{2}\theta \ -\ 3(1-cos^{2}\theta )\ \ =\ \ 4cos^{2} \theta -3$

khorne · Sep 12, 2009

My god, what a huge oversight on Question 4...Could have made it so much simpler =/

Question 5:

If tanA = p/q, then y = p, x = q, r = sqrt[p^2+q^2]

thus:

qsinAcosA + psin^2A = q[pq/p^2 + q^2] + p^3/p^2 + q^2

thus, pq^2 + p^3/p^2 + q^2 = p(q^2 + p^2)/p^2 + q^2
= p
Do the other one yourself...it's easy

Question 6:

tan@ = cos@cosx - sin@sinx/cos@cosx + sin@sinx divide by cos@cosx to get 1-tan@tanx/1+tan@tanx

multiply both sides by denominator:

tan@ + tan^2@tanx = 1-tan@tanx

move, and factorise

tan@ [1+tan@tanx + tanx] = 1

1/tan@ = 1+ tanx[tan@ + 1]

1-tan@/tan@ = tanx[tan@ +1]

1-tan@/tan@[tan@ +1] = tanx

xMaFF · Sep 12, 2009

O_________O

Looking at threads like these make me think twice about doing 3 Unit Math next year... *gulp*

You guys are really something.

Drongoski · Sep 12, 2009

Q5

Sorry . . didn't notice Khorne has done this)

$\100dpi using\ right-angled\ \ \Delta: opp = p\ \ \ adj=q \ \ \ hyp= \sqrt{p^{2}+q^{2}}\ \ =\ \ \sqrt{*} \\ \\ \therefore \ \ sinA=\frac{p}{\sqrt{*}}\ \ cosA=\frac{q}{\sqrt{*}}\\ \\ a)\ \ qsinAcosA+psin^{2}A\ \ =\ \ q\cdot \frac{p}{\sqrt{*}}\cdot \frac{q}{\sqrt{*}}\ +\ p\cdot \frac{p^{2}}{(\sqrt{*})^{2}}\\ \\ =\ \ \frac{pq^{2}}{p^{2}+q^{2}}\ +\ \frac{p^{3}}{p^{2}+q^{2}}\ =\ \frac{p(p^{2}+q^{2})}{p^{2}+q^{2}}\ \ =\ \ p\\ \\ \\ b)\ \ psin2A\ +\ qcos2A\ \ =\ \ p(2sinAcosA)\ +\ q(1-2sin^{2}A)\\ \\ =\ \ 2p\cdot \frac{p}{\sqrt{*}}\cdot \frac{q}{\sqrt{*}}\ +\ q(1-\frac{2p^{2}}{*})\ \ =\ \ \frac{2p^{2}q+qp^{2}+q^{3}-2p^{2}q}{p^{2}+q^{2}}\ =\ q$

khorne · Sep 12, 2009

xMaFF said:
O_________O

Looking at threads like these make me think twice about doing 3 Unit Math next year... *gulp*

You guys are really something.

Not as hard as it looks.

Drongoski · Sep 12, 2009

khorne said:
Not as hard as it looks.

Indeed. Proving trig identities of this type is really easy.

Trebla · Sep 12, 2009

For Q7:

$\tan \alpha = k\tan \beta\\\\\Rightarrow \dfrac{\sin \alpha}{\cos \alpha}= k\dfrac{\sin \beta}{\cos \beta}\\\\\sin \alpha \cos \beta = k \sin \beta \cos \alpha\\\\LHS = (k-1)\sin (\alpha+\beta)\\\\= (k - 1)(\sin \alpha \cos \beta + \sin \beta \cos \alpha)\\\\= k\sin \alpha \cos \beta + k\sin \beta \cos \alpha - \sin \alpha \cos \beta - \sin \beta \cos \alpha\\\\=k\sin \alpha \cos \beta + k\sin \beta \cos \alpha - k \sin \beta \cos \alpha - \sin \beta \cos \alpha\\\\= k\sin \alpha \cos \beta - k\sin \beta \cos \alpha + k\sin \beta \cos \alpha - \sin \beta \cos \alpha\\\\= k(\sin \alpha \cos \beta - \sin \beta \cos \alpha) + k\sin \beta \cos \alpha - \sin \beta \cos \alpha\\\\= k\sin (\alpha-\beta)+ \sin \alpha \cos \beta - \sin \beta \cos \alpha\\\\= k\sin (\alpha-\beta) + \sin (\alpha-\beta)\\\\= (k+1)\sin (\alpha-\beta)\\\\= RHS$

nrlwinner · Sep 12, 2009

thanks guys. can someone explain question 3 for me cauz i still dont get it.

Drongoski · Sep 12, 2009

nrlwinner said:
thanks guys. can someone explain question 3 for me cauz i still dont get it.

OK. I'll do it now.

Drongoski · Sep 12, 2009

To understand you need remember the double-angle identities below:
[1st one not here required]

$\100dpi 1 - cos\theta\ \equiv 1-cos(2\times \frac{\theta }{2})\ \equiv \ 2sin^{2}\frac{\theta }{2}\\ \\ Similarly:\ \ 1+cos\theta \ \equiv\ 2cos^{2}\frac{\theta }{2}\\ \\ \therefore \ \ LHS\ =\ \frac{2cos\frac{\theta }{2}-2cos^{2}\ \frac{\theta }{2}}{2cos\frac{\theta }{2}+2cos^{2}\ \frac{\theta }{2}}\ \ =\ \ \frac{2cos\frac{\theta }{2}(1-cos\frac{\theta }{2})}{2cos\frac{\theta }{2}(1+cos\frac{\theta }{2})}\ =\ \frac{1-cos\frac{\theta }{2}}{1+cos\frac{\theta }{2}}\ =\ RHS$

ps: khorne's solution uses the half-angle t-formulae (a version of the double angle formulae) which you
may not have learnt yet.That may be why you were unable to follow his solution.

Some Trig Questions (1 Viewer)

nrlwinner

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