In need of trig help..love ya!

**Sinsane** · December 2nd 2008, 12:57 AM

I couldnt figure out some of my trig problems...so Im having a bit of trouble wasting time on these problems. Any help would be great. Thanks

Evaluate the expressions:
cos(arcsin 1/4) couldnt figure out what arcsin 1/4 was...maybe 22.5 degrees?

csc(arccsc -2)

arccos(cos pi/2)

cos(arcsin 3/5 +arccos sqrt3/2)

sec^2x-2=tan^2x

sin^2+sin x=0

Use a calculator to find the value. Give answers as real numbers.

sin(arccos .8324)

tan(arcsin 2)

Write the following as an algebraic expression in u, u>0

cos(arctan u)

sin 2theta =-1/2

Solve the equation for the interval [0,2pi)

cos^2x + 2cos x+1-0

Solve the equation for solutions in the interval [0, 360). Round to the nearest degree.

sin 2theta =cos theta

sin 2theta= -1/2

sqrt3 sec2theta =2

Solve the equation for x

y=8 cos 3x

Solve the equation for exact solutions.

arcsin(y-pi/6) = pi/3

arctanx = arcsin 7/25

Solve the equation.

arcsin 2x +2 arccos x=pi

**Prove It** · December 2nd 2008, 01:12 AM

$\sec^2x - 2 = \tan^2x$ .

Use the identity $1+\tan^2x = \sec^2x$

$1+\tan^2x - 2 = \tan^2x$

$\tan^2x - 1 = \tan^2x$

$-1 = 0$ .

This equation is nonsense, so there does not exist an x that satisfies this equation.

$\arccos(\cos{\frac{\pi}{2}}) = \frac{\pi}{2}$ .

$\sin^2x + \sin x = 0$

$\sin x(\sin x + 1) = 0$

$\sin x = 0$ or $\sin x + 1 = 0$ .

If $\sin x = 0, x = \{0, \pi\} + 2n\pi, n \in \mathbf{Z}$ .

If $\sin x + 1 = 0, \sin x = -1, x = \frac{3\pi}{2} + 2n\pi, n \in \mathbf{Z}$ .

So the solution to this equation is $\{0, \pi, \frac{3\pi}{2}\} + 2n\pi$ .

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