Author Topic: Trig Q help please? (Read 1729 times) Tweet Share

ggxxx · « **on:** June 09, 2011, 05:09:22 pm »

I've been attempting this question for the last half hour but don't know how to do it.

One solution of a equation acos(bx)=-sin(bx) is (5π/6). Possible values of a and b are?
Any suggestions would be appreciated

luffy · « **Reply #1 on:** June 09, 2011, 05:24:42 pm »

Quote from: ggxxx on June 09, 2011, 05:09:22 pm

I've been attempting this question for the last half hour but don't know how to do it.

One solution of a equation acos(bx)=-sin(bx) is (5π/6). Possible values of a and b are?
Any suggestions would be appreciated

Since you are simply looking for possible values, you just need to get an equation that gives a, in terms of b (or vice versa).

Therefore:

$a\cos{(bx)}=-\sin{(bx)}$

At $x = \frac{5\pi}{6}$ ,

$a\cos{(\frac{5b\pi}{6})}=-\sin{(\frac{5b\pi}{6})}$

Divide both sides by $\cos{(\frac{5b\pi}{6})}$ :

$\therefore a = -\tan{(\frac{5b\pi}{6})}$

You simply want one set of values, so:

At b = 1,

$a = -\tan{(\frac{5\pi}{6})}$

$= \tan{(\frac{\pi}{6})}$

$= \frac{1}{\sqrt{3}}$

Hence, a possible value of a and b is:

$a = \frac{1}{\sqrt{3}} \textup{ , }b = 1$

Hope I helped.

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Author Topic: Trig Q help please? (Read 1729 times) Tweet Share

ggxxx

Trig Q help please?

luffy

Re: Trig Q help please?

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