If sin(theta)=15/17 and cos(beta)=(-5/13 (both theta and beta are in quadrant II) find tan(theta+beta)
Ok i understand until you say that tan(theta) +tan(beta) needs to be divided by 1-tan(theta)tan(beta) where does that come from?
I assumed you knew the formulas, since you have know them to do these kind of problems
look under "other identities"
http://library.thinkquest.org/20991/alg2/trigi.html
I do but why did u use that one. Im trying to understand because i have a test tomorrow and i have to do well on it.
well, it asked for tan(Ø+ß) didn't it?
The letters themselves of course don't matter
e.g.
tan(A+B) =(tanA + tanB)/(1- tanAtanB)
To find the value of tan(theta + beta), we can use the trigonometric identity for the tangent of a sum:
tan(theta + beta) = (tan(theta) + tan(beta)) / (1 - tan(theta) * tan(beta))
First, let's find the values of tan(theta) and tan(beta) using the given information:
Given: sin(theta) = 15/17, cos(beta) = -5/13
In quadrant II, sin(theta) is positive and cos(theta) is negative. We can use the Pythagorean identity to find cos(theta):
cos(theta) = sqrt(1 - sin^2(theta)) = sqrt(1 - (15/17)^2) = sqrt(1 - 225/289) = sqrt(64/289) = -8/17
Now we can determine tan(theta) using the values of sin(theta) and cos(theta):
tan(theta) = sin(theta) / cos(theta) = (15/17) / (-8/17) = -15/8
Similarly, cos(beta) is negative in quadrant II, so we can find sin(beta) using the Pythagorean identity:
sin(beta) = sqrt(1 - cos^2(beta)) = sqrt(1 - (-5/13)^2) = sqrt(1 - 25/169) = sqrt(144/169) = 12/13
Using the values of sin(beta) and cos(beta), we can determine tan(beta):
tan(beta) = sin(beta) / cos(beta) = (12/13) / (-5/13) = -12/5
Now we have the values of tan(theta) and tan(beta), so we can substitute them into the formula for tan(theta + beta):
tan(theta + beta) = (tan(theta) + tan(beta)) / (1 - tan(theta) * tan(beta))
= (-15/8 + (-12/5)) / (1 - (-15/8) * (-12/5))
Now let's simplify the expression in the numerator:
-15/8 + (-12/5) = (-15/8)*5/5 + (-12/5)*8/8 = -75/40 + (-96/40) = -171/40
And simplify the expression in the denominator:
1 - (-15/8) * (-12/5) = 1 - 15/8 * 12/5 = 1 - 180/40 = 40/40 - 180/40 = -140/40
Now we can divide the numerator by the denominator:
tan(theta + beta) = (-171/40) / (-140/40) = 171/140
Therefore, tan(theta + beta) = 171/140.
You should recognize the 8,15,17 and 5,12,13 right-angled triangles
so in II if
sinØ= 15/17, then cosØ = -8/17 and tanØ = - 15/8
cosß = -5/13 , then sinß = 12/13 and tanß = -12/5
tan(Ø+ß) = (tanØ + tanß)/(1 - tanØtanß)
= ....
you finish the arithmetic.