anonymous
  • anonymous
How do you verify Trigonometric Identities?
Trigonometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Hero
  • Hero
Do you have a specific question you are working on? There's more than one approach.
dumbcow
  • dumbcow
by changing one side of equation using other trig identities until it equals the other side of equation
anonymous
  • anonymous
Is there an easier way to find them?

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anonymous
  • anonymous
I tried to do tanx sin x +cos x =sec x
Hero
  • Hero
I would probably work with right side rather than the left side.
dumbcow
  • dumbcow
tan = sin/cos \[\frac{\sin^{2} x}{\cos x}+\cos x = \sec x\] \[\frac{\sin^{2} x}{\cos x}+\frac{\cos^{2} x}{\cos x} = \sec x\] \[\frac{1}{\cos x} = \sec x\]
anonymous
  • anonymous
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
anonymous
  • anonymous
oh so that's an easier one
Hero
  • Hero
\[\frac{1}{\cos x} = \frac{\sin^2x + \cos^2x}{\cos x} = \frac{(\sin x)(\sin x)}{\cos x} + \frac{\cos^2x}{\cos x} = \tan x \sin x + \cos x\]
dumbcow
  • dumbcow
@Hero , i call that way the working backwards approach, which i dont do very well
anonymous
  • anonymous
hmmm that explains why the book said there are no specific rules/ method and it all depends on my understanding
Hero
  • Hero
Oh there are all kinds of approaches to these @JohnCli
anonymous
  • anonymous
Still how do you find the easiest or simplest forms?
anonymous
  • anonymous
\[\cos x/1-sinx -co x/1+\sin x=2\tan x\]
anonymous
  • anonymous
from:http://www.teaching.martahidegkuti.com/shared/lnotes/trigidentities1/trigidentities1.pdf
Hero
  • Hero
\[\frac{\cos x}{1 - \sin x} - \frac{\cos x}{1 + \sin x} = 2\tan x\]
Hero
  • Hero
For this one, try to remember what you have to do in order to combine fractions.
anonymous
  • anonymous
I still don't get it that easy anyways thanks @Hero and @dumbcow

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