The expression (2x^2)^3 is equivalent to _____. 6x^2 6x^5 8x^6
2 answers:
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Let
because that is the range of the inverse cosine funcition.
Also,
![\mathsf{cos\,\theta=cos\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]}\\\\\\ \mathsf{cos\,\theta=\dfrac{4}{5}}\\\\\\ \mathsf{5\,cos\,\theta=4}](https://tex.z-dn.net/?f=%5Cmathsf%7Bcos%5C%2C%5Ctheta%3Dcos%5C%21%5Cleft%5Bcos%5E%7B-1%7D%5C%21%5Cleft%28%5Cdfrac%7B4%7D%7B5%7D%5Cright%29%5Cright%5D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7Bcos%5C%2C%5Ctheta%3D%5Cdfrac%7B4%7D%7B5%7D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B5%5C%2Ccos%5C%2C%5Ctheta%3D4%7D)
Square both sides and apply the fundamental trigonometric identity:
But which means 
lies either in the 1st or the 2nd quadrant. So 
is a positive number:
![\mathsf{sin\,\theta=\dfrac{3}{5}}\\\\\\ \therefore~~\mathsf{sin\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]=\dfrac{3}{5}\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7Bsin%5C%2C%5Ctheta%3D%5Cdfrac%7B3%7D%7B5%7D%7D%5C%5C%5C%5C%5C%5C%0A%5Ctherefore~~%5Cmathsf%7Bsin%5C%21%5Cleft%5Bcos%5E%7B-1%7D%5C%21%5Cleft%28%5Cdfrac%7B4%7D%7B5%7D%5Cright%29%5Cright%5D%3D%5Cdfrac%7B3%7D%7B5%7D%5Cqquad%5Cquad%5Ccheckmark%7D)
I hope this helps. =)
Tags: <em>inverse trigonometric function cosine sine cos sin trig trigonometry</em>
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Let
Also,
Square both sides and apply the fundamental trigonometric identity:
But
I hope this helps. =)
Tags: <em>inverse trigonometric function cosine sine cos sin trig trigonometry</em>