Question

Find the exact value of sin(cos^-1(4/5))

Answer 1

Answer:

11 squared / 4

Step to Step explaination:

Answer 2

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Let  $\mathsf{\theta=cos^{-1}\!\left(\dfrac{4}{5}\right).}$


$\mathsf{0\le \theta\le\pi,}$  because that is the range of the inverse cosine funcition.


Also,

$\mathsf{cos\,\theta=cos\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]}\\\\\\ \mathsf{cos\,\theta=\dfrac{4}{5}}\\\\\\ \mathsf{5\,cos\,\theta=4}$


Square both sides and apply the fundamental trigonometric identity:

$\mathsf{(5\,cos\,\theta)^2=4^2}\\\\ \mathsf{5^2\,cos^2\,\theta=4^2}\\\\ \mathsf{25\,cos^2\,\theta=16\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\ \mathsf{25\cdot (1-sin^2\,\theta)=16}$

$\mathsf{25-25\,sin^2\,\theta=16}\\\\ \mathsf{25-16=25\,sin^2\,\theta}\\\\ \mathsf{9=25\,sin^2\,\theta}\\\\ \mathsf{sin^2\,\theta=\dfrac{9}{25}}$

$\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{9}{25}}}\\\\\\ \mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{3^2}{5^2}}}\\\\\\ \mathsf{sin\,\theta=\pm\,\dfrac{3}{5}}$


But $\mathsf{0\le \theta\le\pi,}$ which means $\theta$ lies either in the 1st or the 2nd quadrant. So $\mathsf{sin\,\theta}$ is a positive number:

$\mathsf{sin\,\theta=\dfrac{3}{5}}\\\\\\ \therefore~~\mathsf{sin\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]=\dfrac{3}{5}\qquad\quad\checkmark}$


I hope this helps. =)


Tags:  inverse trigonometric function cosine sine cos sin trig trigonometry