Tricky Maths Q - No idea what to do.... (1 Viewer)

lookoutastroboy · Mar 24, 2011

Hey guys,

Any know how to solve question sin squared (5a) - sin squared (3a)

The answer is Sin (8a) x Sin (2a).

Cheers
l.a.

ohexploitable · Mar 24, 2011

Are you sure you posted the question correctly? What am I meant to solve? There's no equation.

ohexploitable · Mar 24, 2011

If you meant to simplify...

$\\\sin^2 5x-\sin^2 3x\\=(\sin5x-\sin3x)(\sin5x+\sin3x)\\=\left[2\sin\left(\frac{5x-3x}{2}\right)\cos\left(\frac{5x+3x}{2}\right)\right]\left[2\sin\left(\frac{5x+3x}{2}\right)\cos\left(\frac{5x-3x}{2}\right)\right]\\=(2\sin x\cos4x)(2\sin4x\cos x)\\=(2\sin x\cos x)(2\sin 4x\cos 4x)\\=\sin2x\sin8x$

lookoutastroboy · Mar 24, 2011

ohexploitable said:
If you meant to simplify...

$\\\sin^2 5x-\sin^2 3x\\=(\sin5x-\sin3x)(\sin5x+\sin3x)\\=\left[2\sin\left(\frac{5x-3x}{2}\right)\cos\left(\frac{5x+3x}{2}\right)\right]\left[2\sin\left(\frac{5x+3x}{2}\right)\cos\left(\frac{5x-3x}{2}\right)\right]\\=(2\sin x\cos4x)(2\sin4x\cos x)\\=(2\sin x\cos x)(2\sin 4x\cos 4x)\\=\sin2x\sin8x$

its an identity haha - how did u get 2nd/3rd steps? i think my teacher did a different way... but at least u got it

Omnipotence · Mar 24, 2011

Differences of two squares. Then sums/differences as product.

lookoutastroboy · Mar 24, 2011

any 3 unit method of doing it?

hscishard · Mar 24, 2011

Use oh exploitables straight use of the formula or...

cos2x = 1-2sin^2(x)
rearrange to make sin^2x the subject

(1-cos10a)/2 - (1-cos6a)/2
=(cos6a -cos10a)/2
= cos(8a-2a) - cos(8a + 2a) all over 2
And you will end up with sin2sin8a

blackops23 · Mar 25, 2011

ohexploitable said:
If you meant to simplify...

$\\\sin^2 5x-\sin^2 3x\\=(\sin5x-\sin3x)(\sin5x+\sin3x)\\=\left[2\sin\left(\frac{5x-3x}{2}\right)\cos\left(\frac{5x+3x}{2}\right)\right]\left[2\sin\left(\frac{5x+3x}{2}\right)\cos\left(\frac{5x-3x}{2}\right)\right]\\=(2\sin x\cos4x)(2\sin4x\cos x)\\=(2\sin x\cos x)(2\sin 4x\cos 4x)\\=\sin2x\sin8x$

how did you get from the 2nd step to 3rd step?

ohexploitable · Mar 25, 2011

blackops23 said:
how did you get from the 2nd step to 3rd step?

Sums to products.

It's not part of the syllabus but it's useful to know for 4U.

blackops23 · Mar 25, 2011

ohexploitable said:
Sums to products.

It's not part of the syllabus but it's useful to know for 4U.

do you learn these in tutor? also what is this sum to product rule? thanks

Drongoski · Mar 25, 2011

blackops23 said:
do you learn these in tutor? also what is this sum to product rule? thanks

These as ohex... explained are not in 4U but should have been in since they almost follow immediately from the compound angle formulae for sin(A+B), cos(A+B) etc. and are useful for some of the trickier derivations in conics etc.

blackops23 · Mar 25, 2011

what is the rule?

Drongoski · Mar 25, 2011

blackops23 said:
what is the rule?

$sin(A+B) = sinAcosB + cosAsinB . . . . . . . (1)$

$sin(A-B) = sinAcosB - cosAsinB . . . . . . . (2)$

$cos(A+B) = cosAcosB - sinAsinB . . . . . . . (3)$

$cos(A-B) = cosAcosB + sinAsinB . . . . . . . (4)$

$(1) + (2) \implies 2sinAcosB = sin(A+B) + sin(A-B) . . . . . (5)$

$(3) + (4) \implies 2cosAcosB = cos(A+B) + cos(A-B) . . . . . (6)$

$(3) - (4) \implies -2sinAsinB = cos(A+B) - cos(A-B) . . . . . (7)$

(5), (6) and (7) are equivalent to and can be rewritten respectively as:

$sin \alpha + sin \beta = 2 sin \frac{\alpha + \beta}{2}cos \frac {\alpha - \beta}{2} . . . . . (8)$

$cos \alpha + cos \beta = 2 cos \frac {\alpha + \beta}{2} cos \frac{\alpha - \beta}{2} . . . . . (9)$

$cos \alpha - cos \beta = -2 sin \frac{\alpha + \beta}{2} sin \frac {\alpha - \beta}{2} . . . . . (10)$

x_cp3 · Mar 25, 2011

blackops23 said:
what is the rule?

I'm very surprised that ur asking this question. This is yr 11 3u further trig that u should've learnt last year..

ohexploitable · Mar 25, 2011

x_cp3 said:
I'm very surprised that ur asking this question. This is yr 11 3u further trig that u should've learnt last year..

I'm surprised that you've learnt it seeing as it isn't part of the syllabus...

x_cp3 · Mar 25, 2011

oh really? hmm i don't read the syllabus for maths unfortunately xD

blackops23 · Mar 25, 2011

x_cp3 said:
oh really? hmm i don't read the syllabus for maths unfortunately xD

no one does, just that textbooks don't have these formulas

Drongoski · Mar 25, 2011

blackops23 said:
no one does, just that textbooks don't have these formulas

Hope you have copied down the formulae I derived for you.

blackops23 · Mar 26, 2011

Drongoski said:
$sin(A+B) = sinAcosB + cosAsinB . . . . . . . (1)$

$sin(A-B) = sinAcosB - cosAsinB . . . . . . . (2)$

$cos(A+B) = cosAcosB - sinAsinB . . . . . . . (3)$

$cos(A-B) = cosAcosB + sinAsinB . . . . . . . (4)$

$(1) + (2) \implies 2sinAcosB = sin(A+B) + sin(A-B) . . . . . (5)$

$(3) + (4) \implies 2cosAcosB = cos(A+B) + cos(A-B) . . . . . (6)$

$(3) - (4) \implies -2sinAsinB = cos(A+B) - cos(A-B) . . . . . (7)$

(5), (6) and (7) are equivalent to and can be rewritten respectively as:

$sin \alpha + sin \beta = 2 sin \frac{\alpha + \beta}{2}cos \frac {\alpha - \beta}{2} . . . . . (8)$

$cos \alpha + cos \beta = 2 cos \frac {\alpha + \beta}{2} cos \frac{\alpha - \beta}{2} . . . . . (9)$

$cos \alpha - cos \beta = -2 sin \frac{\alpha + \beta}{2} sin \frac {\alpha - \beta}{2} . . . . . (10)$

very very noice, these are gonna help in conics...

Tricky Maths Q - No idea what to do.... (1 Viewer)

lookoutastroboy

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