Trig Geometry Problem: Solving sin(x)=cot(x) in Terms of cos(x)

[Sam223344]

Homework Statement

A triangle with x as the angle, 'a' as the adjacent, 8 as the opposite side and 'h' as the hypotenuse. An intermediate question was to show that sin(x)=cot(x). Now i have to write 'sin(x)=cot(x)' in terms of cos(x) only and solve the quadratic, stating correct value for cos(x)

Homework Equations

2. "sin(x)=cot(x)" in terms of cos(x). a*h=8^2

The Attempt at a Solution

3. I have used some trig identities and have got cos^2(x)+cos(x)-1=0. Is this correct? To solve i solved x^2+x-1 using quadratic formula and got 2 answers, one negative. So cos(x) is the positive value..?

 

[LCKurtz]

How can you hope to "show" that sin(x) = cot(x) when that normally isn't true?

If x was an angle such that sin(x) = cot(x), then cos(x) would satisfy your equation.
Why couldn't cos(x) be negative? And what does all this have to do with your given triangle? What are you actually trying to do?

 

[Sam223344]

For this triangle a*h=8^2. This can be shown by using a^2+b^2=c^2. To show that sin(x)=cot(x) 'in this case' : By using sin(x)=opp/hyp = 8/h and a*h=*^2. Rearrange for h=8^2/a

so sin(x)=8/(8^2/a)=a/8 and cot(x)=adj/opp=a/8

so sin(x)=cot(x) for this case.

Yeah the angle is acute(sorry forgot to mention), so that means I am correct?

 

[Bohrok]

Given what you wrote in the first post, h is the hypotenuse and a and 8 are the legs, so a² + 8² = h², not a*h = 8². Your substitution was wrong when rewriting sinx.

sinx = a/h, cotx = a/8

 

[Sam223344]

It might be helpful if i explain why a*h=8^2 for this triangle?

h^2=8^2+a^2
8=sqrt(h^2-a^2)
8=sqrt(h^2-(h^2-8^2))
8=sqrt(8^2)
so 8=8
so a*h=8^2
I'm still needing conformation that I solved the correct quadratic