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