Trig Identity, Need Help?

Question

This problem is giving me problems....

The easier side is the left side, as you can see it's a double angle identity. Can you help me solve it? Work one side.

sin2xsecx = 2tanxcosx

Answer 1

sin(2x)sec(x) = 2tan(x)cos(x)

The more complex side is the left side (in my opinion); I disagree with you. The complex side is always the one you work with, to get to the simple side.

LHS = sin(2x)sec(x)

Using the double angle identity sin(2x) = 2sin(x)cos(x),

LHS = 2sin(x)cos(x)sec(x)

Change everything to sines and cosines.

LHS = 2sin(x)cos(x)[1/cos(x)]

Note the cancellation between cos(x) and 1/cos(x).

LHS = 2sin(x)

Now, let's begin with the right hand side.

RHS = 2tan(x)cos(x)

Convert everything to sines and cosines,

RHS = 2[sin(x)/cos(x)]cos(x)

Note the cancellation

RHS = 2sin(x) = LHS

Answer 2

well um tan x = sin x / cos x, so the right side is just 2 sin x.

the left side, um sec x is 1/cos x, and sin 2x is 2 sin x cos x
so the the left side becomes 2 sin x and LS = RS, so its proved.

If you really want, on the LHS, once u get sin 2x = 2 sinx cosx, then make the sinx secx = sinx/cosx = tanx and u get 2 tanxcosx.

Answer 3

The best approach to solving any trig identity question is to change everything to sine and cosine.

Guido

Answer 4

sin2x secx = 2 sinx cosx sec x = 2 sinx = 2 (sinx/cosx)cosx
= 2 tanx cos x

Answer 5

sin2xsecx = 2tanxcosx
2sinxcosxsecx = 2tanxcosx
2sinxcosx/cosx = (2sinx/cosx)cosx
2sinx = 2sinx