Question

I've been having trouble proving trigonometric identities. Could someone help?

When I'm simplifying trigonometric identities, I often find myself down rabbit holes of what seems to be endless algebra when I'm sure there must be an easier way, or getting stuck unsure of what to do next.

The problem I'm stuck on right now is `2/(sqrt(3)cosx+sinx)=sec(pi/6-x)`, and I'm feeling really lost.

If someone could walk me through how to approach this or a problem like it, I would be much obliged - my course didn't explain it very well.

Answer 1

First, develop the denominator:
`f(x) = sqrt3cos x + sin x = sqrt3(cos x + (1/sqrt3)(sin x))`
Note that: `1/sqrt3 = tan (pi/6) = sin (pi/6)/(cos (pi/6))`
Therefor:
`f(x) = sqrt3/(cos (pi/6))(cos x.cos (pi/6) + sin (pi/6).sin x)`
Note that:
`cos (pi/6) = sqrt3/2`
`cos x.cos (pi/6) + sin (pi/6).sin x = cos (x - pi/6)`
There for:
`f(x) = (sqrt3)(2/sqrt3)cos (x - pi/6) = 2cos (x - pi/6)`
Finally
`2/f(x) = 2/(2cos (x - pi/6)) = 1/(cos (x - pi/6)) = 1/(cos (pi/6 - x))`
`2/f(x) = sec (pi/6 - x)`