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AleksandrR [38]
3 years ago
9

Prove the identity: cos(3t)=cos^3(t)-3sin^2(t)cos(t)

Mathematics
1 answer:
Pepsi [2]3 years ago
3 0
Solutions 

To solve the given equation lets use double-angle theorem. Double-angle formulas allow the expression of trigonometric functions of angles equal to 2α<span> in terms of </span>α<span>, which can simplify the functions and make it easier to perform more complex calculations, such as integration, on them.

</span>cos(3t)=cos^3(t)-3sin^2(t)cos(t)

cos(A+B) = cosAcosB - sinAsinB 

<span>Now cos(3t) = cos(2t+t) </span>

⇒<span>cos(2t)cos(t) - sin(2t)sin(t) </span>

⇒<span>(2cos^2(t)-1)cos(t) - 2sin(t)cos(t)sin(t) </span>

⇒<span> 2cos^(3)t - cos(t) - 2sin^2(t)cos(t) </span>

⇒<span> 2cos^3(t) - cos(t) - 2(1-cos^2(t))cos(t) </span>

⇒<span>2cos^3(t) - cos(t) - 2cos(t) + 2cos^3(t) </span>

⇒4cos^3(t) - 3cos(t)
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Answer:

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and this value represents 0.9987

we subtract:

0.9987 - 0.1587 = 0.84

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