solve inverse-cos(x) + inverse-cos(sq. rt. of 15x) = (pi/2)

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.

solve inverse-cos(x) + inverse-cos(sq. rt. of 15x) = (pi/2)

solve inverse-cos(x) + inverse-cos(square root of 15x) = (pi/2). No one was able to solve this problem. Please help!
car.man

Posts: 7
Joined: Tue Mar 24, 2009 6:18 pm

car.man wrote:solve inverse-cos(x) + inverse-cos(square root of 15x) = (pi/2).

By definition of inverse cosine, the above means the following:

. . . . .$\alpha\, +\, \beta\, =\, \frac{\pi}{2}$

...for:

. . . . .$\cos^{-1}(x)\, =\, \alpha\, \mbox{ and }\, \cos^{-1}(\sqrt{15}x)\, =\, \beta$

This is a sum of two angles, and involves cosines. What if we now take the cosine of both sides of the above?

. . . . .$\cos(\alpha\, +\, \beta)\, =\, 0$

Using an angle-sum identity, we get:

. . . . .$\cos(\alpha)\cos(\beta)\, -\, \sin(\alpha)\sin(\beta)\, =\, 0$

We can simplify the factors in the first product by using the definition of "inverse cosine":

. . . . .$\cos\left(\cos^{-1}(x)\right)\cos\left(\cos^{-1}(\sqrt{15}x)\right)\, =\, (x)\left(\sqrt{15}x\right)\, =\, \sqrt{15}x^2$

To find the values of the sines, draw right triangles for each of the angles $\alpha$ and $\beta$. Using the Pythagorean Theorem, find the length of the "opposite" side for each triangle. Then read off the values of the sines. You should end up with an equation, after squaring, that looks like:

. . . . .$15x^4\, =\, 1\, -\, 16x^2\, +\, 15x^4$

Then:

. . . . .$0\, =\, 1\, -\, 16x^2\, =\, (1\, -\, 4x)(1\, +\, 4x)$

...and so forth.

stapel_eliz

Posts: 1782
Joined: Mon Dec 08, 2008 4:22 pm

Re: solve inverse-cos(x) + inverse-cos(sq. rt. of 15x) = (pi/2)

We got x=+-1/4 and x=+1/4 worked. Thank you!
car.man

Posts: 7
Joined: Tue Mar 24, 2009 6:18 pm