Tuesday

September 16, 2014

September 16, 2014

Posted by **Zee** on Monday, June 24, 2013 at 11:07pm.

sin^(1/2)x*cosx - sin^(5/2)*cosx = cos^3x sq root sin x

I honestly have no clue how to approach the sin^(5/2)*cosx part of the equation

- Precal -
**Steve**, Tuesday, June 25, 2013 at 10:00amsince 5/2 = 2 + 1/2, you have

u^5/2 = u^2 * u^1/2, and so,

√sinx cosx - sin^2x √sinx cosx

√sinx cosx (1-sin^2 x)

√sinx cosx cos^2x

√sinx cos^3x

**Answer this Question**

**Related Questions**

Precalculus/Trig - I can't seem to prove these trig identities and would really ...

Trigonometry. - ( tanx/1-cotx )+ (cotx/1-tanx)= (1+secxcscx) Good one! ...

Trigonometry - Prove the following trigonometric identities. please give a ...

Calculus - Find the dervatives: 1. f(x)=(3x+1)e^x^2 2. y = e^(sin x) ln(x) 3. f(...

trigonometry (please double check this) - Solve the following trig equations. ...

Confused! Pre-Cal - Verify that each equation is an identity.. tan A= sec a/csca...

trig - express this in sinx (1/ cscx + cotx )+ (1/cscx- cotx) i got 2sinx is ...

Trig....... - I need to prove that the following is true. Thanks (2tanx /1-tan^x...

pre-calc - Solve: cos(2x-180) - sin(x-90)=0 my work: cos2xcos180 + sin2xsin180= ...

trig - I need to prove equality. a) (sina + cosa)^2 -1 / ctga - sinacosa = 2tg^...