silly trig question (1 Viewer)

[marianna84]

Tan(x) = 6/7 where 180 <= x <= 360

Find Sin(x) & Cos(x)

so... would it be sin(x) = 6 and cos(x) = 7

I know I saw this rule somewhere...

Thanks

 

[ssglain]

sin(x) & cos(x) can't be 6 & 7, respectively, because neither function is capable of producing values outside of the range -1<= f(x) <=1.

To solve the question, you need to construct a right-angled triangle with lengths opp=6 and adj=7 and related angle A. From this you can work out the length of the hypotenus and then work out the Pythagorean ratios of sin(A)=opp/hyp and cos(A)=adj/hyp. Note that given 180 <= x <= 360 and tan(x) positive implies that x is in the 3rd quadrant [to assist you in choosing the correct signs for sin(x) and cos(x)].

 

[kony]

i see you are familiar with the T/T method.

aka triangle trick.

gah i got stuck on a trig question in todays 4u test.

cos3x = cos4x. solve. damn it.

 

[morganforrest]

I think the rule you're thinking of is tan(x) = sin(x)/cos(x) but that doesn't apply in this case. Do what ssglain said and draw a triangle with opp 6 and adj 7 and then work it out.

 

[watatank]

cos3x = cos4x. solve. damn it.

Wrong thread.

 

[ssglain]

gah i got stuck on a trig question in todays 4u test.

cos3x = cos4x. solve. damn it.

Rewrite everything in powers of cosx? De Moivre comes to mind for some strange reason. Which topic was this under?

If that doesn't work, then for the most obvious solution:
Clearly, 3x=4x
Hence 4x-3x=0 => x=0

Edit. And don't go bossing me around. Damn you, man.