SOLUTION: Heres 3 questions I was curious about
Given sin A = 1/3 A in Quad II, CosB = -3/5 B in Quad III,
find Sin(A+B) and tan2B
___________ ___________ ___________ ___________ __
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-> SOLUTION: Heres 3 questions I was curious about
Given sin A = 1/3 A in Quad II, CosB = -3/5 B in Quad III,
find Sin(A+B) and tan2B
___________ ___________ ___________ ___________ __
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Question 251050: Heres 3 questions I was curious about
Given sin A = 1/3 A in Quad II, CosB = -3/5 B in Quad III,
find Sin(A+B) and tan2B
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Solve for x : 0 < or = x < or = 360
3cos^2x = cosx AND 3cos^2x - 5sinx = 1
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prove this identity:
sin2x/sinx - cos2x/cosx = secx Answer by drk(1908) (Show Source):
You can put this solution on YOUR website! Lets take them one at a time:
Given sin A = 1/3 A in Quad II, CosB = -3/5 B in Quad III,
find Sin(A+B) and tan2B
Sin A = 1/3, Cos A = -2*sqrt(2)/3
Sin B = -4/5 Cos B = -3/5
Sin(A+B) = SinACosB + CosASinB = (1/3)(-3/5) + (2*sqrt(2)/3)(-4/5) = -3/15 - 8*sqrt(2)/15.
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Solve for x : 0 < or = x < or = 360
3cos^2x = cosx AND 3cos^2x - 5sinx = 1
3cos^2(x) - cos(x) = 0
cos(x) [3cos(x) - 1] = 0
cos(x) = 0 - -> X = pi/2 or 3pi/2
3cos(x) - 1 = 0
cos(x) = 1/3 - -> ~ 70.5 degrees
3cos^2(x) - 5Sin(x) = 1
we use an identity: sin^2x + cos^2x = 1.
3[1 - sin^2(x)] - 5sin(x) = 1
3Sin^2(x) + 5sin(x) -2 = 0
(3Sin(x) - 1)(Sin(x) + 2) = 0
3sin(x) - 1 = 0
sin(x) = 1/3 - -> ~19.5 degrees
sin(x) - 2 = 0 - - > no solution.
"AND" needs over lap, there is none here. However if you used substitution and set cos(x) = 5sin(x) + 1. The only values that work are 0 degrees and 360 degrees.
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prove this identity:
First we want to turn the more complicated side into the less complicated side. I will start with the left side.
An identity we need is: sin2x = 2sinxcosx
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2cos^2x - cos2x / cosx
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Another identity we need is: cos2x = cos^2x - sin^2x = 2cos^2x - 1
2cos^2x - (2cos^2x - 1 / cosx
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sec(x)
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