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3 years ago

Prove that (Cosx-Cos³x)/Sinx=SinxCosx

Mathematics

1 answer:

77julia77 [94]3 years ago

4 0

Answer:

(Cosx-Cos³x)/Sinx=SinxCosx

Step-by-step explanation:

sin3x−cos3x

=sinxsin2x−cosxcos2x

=sinx(1−cos2x)−cosx(1−sin2x)

=sinx−sinxcos2x−cosx+cosxsin2x

=sinx−cosx+(sinx−cosx)sinxcosx

=(sinx−cosx)(1+sinxcosx

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GUYS PLEASE HELP ME!!

Musya8 [376]

<h3>The surface area of cube is 162.24 square inches</h3>

<em><u>Solution:</u></em>

<em><u>The surface area of cube is given as:</u></em>

$Surface\ Area = 6s^2$

Where,

"s" is the length of each side of cube

From given,

s = 5.2 inches

Therefore,

$Surface\ Area = 6 \times 5.2^2\\\\Surface\ Area = 6 \times 27.04\\\\Surface\ Area = 162.24$

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Prove that (Cosx-Cos³x)/Sinx=SinxCosx​

Prove that (Cosx-Cos³x)/Sinx=SinxCosx