Question

How do you verify Trigonometric Identities?

Answer 1

Do you have a specific question you are working on? There's more than one approach.

Answer 2

by changing one side of equation using other trig identities until it equals the other side of equation

Answer 3

Is there an easier way to find them?

Answer 4

I tried to do tanx sin x +cos x =sec x

Answer 5

I would probably work with right side rather than the left side.

Answer 6

tan = sin/cos

\[\frac{\sin^{2} x}{\cos x}+\cos x = \sec x\]
\[\frac{\sin^{2} x}{\cos x}+\frac{\cos^{2} x}{\cos x} = \sec x\]
\[\frac{1}{\cos x} = \sec x\]

Answer 7

tan x sin x + cos x =sin x/cos x * sin x + cos x =sin2 x/cos x+ cos x =sin2 x/cos x+cos2x cos x

Answer 8

oh so that's an easier one

Answer 9

\[\frac{1}{\cos x} = \frac{\sin^2x + \cos^2x}{\cos x} = \frac{(\sin x)(\sin x)}{\cos x} + \frac{\cos^2x}{\cos x} = \tan x \sin x + \cos x\]

Answer 10

@Hero , i call that way the working backwards approach, which i dont do very well

Answer 11

hmmm that explains why the book said there are no specific rules/ method and it all depends on my understanding

Answer 12

Oh there are all kinds of approaches to these @JohnCli

Answer 13

Still how do you find the easiest or simplest forms?

Answer 14

\[\cos x/1-sinx -co x/1+\sin x=2\tan x\]

Answer 15

from: http://www.teaching.martahidegkuti.com/shared/lnotes/trigidentities1/trigidentities1.pdf

Answer 16

\[\frac{\cos x}{1 - \sin x} - \frac{\cos x}{1 + \sin x} = 2\tan x\]

Answer 17

For this one, try to remember what you have to do in order to combine fractions.

Answer 18

I still don't get it that easy anyways thanks @Hero and @dumbcow