Lindsey11x16
cos^x(sec^2x1)
what pythagoream identiy to use?



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hartnn
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first multiply them out.
like a(bc) = abac
so whats cos^x(sec^2x1) = ...  ... ?

jishan
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yeahhhhh right hartnn

Lindsey11x16
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cos^2x(sec^2x)  cos^2x1

hartnn
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yup, and you know what cos x * sec x = .. ?

hartnn
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and 1 won't come.

hartnn
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\(\large \cos^2x(\sec^2x1)=\cos^2x\sec^2x\cos^2x = ...?\)

hartnn
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can u simplify that ^ ?
use the fact that cos x = 1/ sec x.

Lindsey11x16
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should i use factorization in this

hartnn
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nopes.
since, cos x = 1/ sec x
what is cos x * sec x = ... ?

Lindsey11x16
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should i turn secant into cos

hartnn
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yes., do that...what u get ?

Lindsey11x16
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dw:1355024922967:dw

hartnn
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1 is correct.

hartnn
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\(\large \cos^2x\sec^2x\cos^2x = 1 \cos^2 x=...?\)

Lindsey11x16
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equals sinx^2

Lindsey11x16
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?

hartnn
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use th identity
\(\huge \color{red}{\sin^2x+\cos^2x=1}\)
yes, sin^2 x is correct :)

Lindsey11x16
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thanks!
this is hard cuz cuz i never know which identity to use :(

hartnn
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do u wanna know alternative way to solve this ?
using other identity...

Lindsey11x16
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yep

hartnn
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using \(\huge \color{red}{\sec^2x=1+\tan^2x}\)
what can you write (sec^2x1)=...?

Lindsey11x16
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sec^2x1 = tan

hartnn
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its \(\tan^2x\)

hartnn
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and you can write tan^2 x as sin^2 x/ cos^2x

hartnn
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so, put (sec^2 x1) as sin^2 x/ cos^2x
cos^2 x*( sin^2 x/ cos^2x) will again give you sin^2 x

Lindsey11x16
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dw:1355025560186:dw

hartnn
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yup.

Lindsey11x16
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thank you ! <3

hartnn
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welcome ^_^