Tuesday

July 22, 2014

July 22, 2014

Posted by **aziiancaligirl** on Sunday, July 15, 2007 at 11:51pm.

Let's use cos 2A = 1 - 2sin

and we can match

cos x = 1 - 2sin

so we will need cos x

we know sin x = -.4 and x is in the fourth quadrant, so the cosine will be positive.

Drawing a right angled triangle with side 2 and hypotenuse 5 ( .4 = 4/10 = 2/5), and using Pythagoras it is easy to see that cos x = √21/5

then:

√21/5 = 1 - 2sin

.

.

.

.

I got sin (x/2) = √(5-√21)/√10 or appr.2043

**Related Questions**

calc - Where do I start to prove this identity: sinx/cosx= 1-cos2x/sin2x please ...

Maths/Right Triangle - There must be a relationship (formula) between the ...

Derivatives - What is the fourth derivative of f(x)= e^-x^2. I got f''(x)=-2e^-x...

Calculus problem - Evaulate: integral 3x (sinx/cos^4x) dx I think it's sec3 x , ...

Trig - Given: cos u = 3/5; 0 < u < pi/2 cos v = 5/13; 3pi/2 < v < ...

Calculus - Okay so I have a question on my assignment that says: You are given ...

trig - Solve cos x-1 = sin^2 x Find all solutions on the interval [0,2pi) a. x=...

find dy/ds - y = s*square root of(1-s^2) + cos inverse(s) Just give me some ...

Precal - I do not understand how to do this problem ((sin^3 A + cos^3 A)/(sin A...

TRIG! - Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6...