# trig

cos(x+B)=cosxcosB-sinxsinB

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2. 2

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1. ### TRIG!

Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x +
2. ### pre-cal

Simplify the given expression........? (2sin2x)(cos6x) sin 2x and cos 6x can be expressed as a series of terms that involve sin x or cos x only, but the end result is not a simplification. sin 2x = 2 sinx cosx cos 6x = 32 cos^6 x
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Determine exact value of cos(cos^-1(19 pi)). is this the cos (a+b)= cos a cos b- sina sin b? or is it something different. When plugging it in the calculator, do we enter it with cos and then the (cos^-1(19 pi)).
4. ### math

Find the exact value of cos 300 degrees. thanks guys cos 300 = 1/2 = 0.500 how do you know? I am supposed to show my work. You ought to know the rule on 30-60-90 triangles. If the hyp is 2, the shorter side is 1, and the longer
5. ### Precalculus

Solve Cos^2(x)+cos(x)=cos(2x). Give exact answers within the interval [0,2π) Ive got the equation down to -cos^2(x)+cos(x)+1=0 or and it can be simplified too sin^2(x)+cos(x)=0 If you could tell me where to go from either of
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Find the roots of z^6 + 1 and hence resolve z^6 + 1into read quadratic factors; deduce that cos3x = 4[cos(x) -cos(pi/6)][(cos(x) -cos(pi/2)][(cos(x) -cos(5pi/6)]
7. ### maths

Find the roots of z^6 + 1 and hence resolve z^6 + 1into read quadratic factors; deduce that cos3x = 4[cos(x) -cos(pi/6)][(cos(x) -cos(pi/2)][(cos(x) -cos(5pi/6)]
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Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x)
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Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x)
10. ### Math - Solving Trig Equations

What am I doing wrong? Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) = cos(2x) sin(x) cos(x) = 2cos^2(x) - 1 cos(x) (+/-)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1 cos^2(x)(1 -

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