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View Full Version : Trig Inverse Funt.


ar5lan
Apr 1, 2010, 08:54 PM
Solve the equation for an exact solution.

Sin(^-1) 4x + Cos (^-1)x = (pi/6)

note (^-1) indicates it's an inverse. Thank you

galactus
Apr 3, 2010, 12:07 PM
Look at the graph and try using Newton's method by staying in the domain of x as an initial guess.

x_{n+1}=x-\frac{sin^{-1}(4x)+cos^{-1}(x)-\frac{\pi}{6}}{\frac{4}{\sqrt{1-16x^{2}}}-\frac{1}{\sqrt{1-x^{2}}}}

Try an initial guess of, say, -.24

It should then converge on the exact value.

Looking at the graph, we can see we can not go less than x=-.25

By taking the derivative, we can see the slope approaches infinity as x closes in on -.25 and .25.

Therefore, the value is a little more than -.25. Around -.24