Trig - Help with tan(arccos(x/3))

Thread starter TKDKicker89
Start date Jul 7, 2005
Tags

Trig

AI Thread Summary

To solve tan(arccos(x/3)), first define theta as arccos(x/3). Construct a right triangle where the adjacent side is x and the hypotenuse is 3. Use the Pythagorean theorem to determine the opposite side, allowing the calculation of sin(theta). The tangent can then be expressed as sin(theta)/cos(theta). This approach effectively simplifies the problem using trigonometric identities.

Jul 7, 2005

TKDKicker89

Can anyone help asap?

tan(arccos x/3)

THanks

Physics news on Phys.org

Jul 7, 2005

Jameson

Insights Author

Gold Member

MHB

Draw a right triangle. If arccos x/3 = theta, draw that triangle.

Jul 7, 2005

BobG

Science Advisor

if \theta=cos^{-1} \frac{x}{3}

then \frac{x}{3} is the cosine of \theta

You can use the pythagorean theorem to find the sine of \theta

\cos^2 \theta + \sin^2 \theta = 1

The tangent of \theta is \frac{sin \theta}{cos \theta}

Thread 'I've come across another fluid pressure problem I don't understand'

Neglect gravity other than that of water; neglect friction. F1=ρgsh=1000*10*1*(1+4)=50000 N; F1=ρgsh=1000*10*0.1*4=4000 N; F3=50000-4000=46000 N?

View full post »

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

View full post »

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

View full post »