Trigonometric Equation problem

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Sabellic
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Homework Statement


If cos x = a/b and tan x = c/d; where a, b, c do not equal 0, then sin x is...

(a) bd/ac
(b) bc/ad
(c) ad/bc
(d) ad/bc


Homework Equations


tri5a.gif

trig_ident_6.gif



The Attempt at a Solution



The way I see it, if cos x = a/b, then b=hypotenuse.
if tan x = c/d, then d= hypotenuse (two hypotenuse values?) and c = opposite.

Therefore sin x = c/d. I don't know why the answers in the book use products of the vairables in their numerators and denominators.

Please help.

Much thanks in advance.
 
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Hmmm... Your c) and d) choices are the same..and the tan doesn't involve the hypotenuse.
Are you sure you copied the whole thing right?
 
Last edited:
Since
[tex]tan \alpha= \frac{sin \alpha}{cos \alpha}[/tex]
multiplying on both sides by [itex]cos(\alpha)[/itex] gives

[tex]sin(\alpha)= cos(\alpha)tan(\alpha)[/tex]
That's all you need.